problem_id int64 1 978 | question stringlengths 86 2.11k | source stringlengths 19 76 | solution stringlengths 94 14.7k | asymptote_code stringlengths 44 17.8k | solution_image_url stringclasses 16
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957 | Equilateral $\triangle ABC$ with side length $14$ is rotated about its center by angle $\theta$, where $0 < \theta < 60^{\circ}$, to form $\triangle DEF$. See the figure. The area of hexagon $ADBECF$ is $91\sqrt{3}$. What is $\tan\theta$?
$\textbf{(A)}~\frac{3}{4}\qquad\textbf{(B)}~\frac{5\sqrt{3}}{11}\qquad\textbf{(C)}~\frac{4}{5}\qquad\textbf{(D)}~\frac{11}{13}\qquad\textbf{(E)}~\frac{7\sqrt{3}}{13}$ | 2024 AMC 12B Problem 19 | Let the circumcenter of the circle inscribing this polygon be $O$. The area of the equilateral triangle is $\frac{\sqrt{3}}{4}*196=49\sqrt{3}$. The area of one of the three smaller triangles, say $\triangle{DBE}$ is $14\sqrt{3}$. Let $BH$ be the altitude of $\triangle{DBE}$, so if we extend $BH$ to point $M$ where $MO\perp{BM}$, we get right triangle $\triangle{OMB}$. Note that the height $BH=2\sqrt{3}$, computed given the area and side length $14$, so $MB=MH+HB=\frac{7\sqrt{3}}{3}+2\sqrt{3}=\frac{13\sqrt{3}}{3}$. $OB=\frac{14\sqrt{3}}{3}$ so Pythag gives $OM=\sqrt{OB^2-MB^2}=3$. This means that $HE=7-OM=4$, so Pythag gives $BE=2\sqrt{7}$. Let $\frac{\theta}{2}=\alpha$ and the midpoint of $BE$ be $P$ so that $BP=PE=\sqrt{7}$, so that Pythag on $\triangle{OPE}$ gives $OP=\sqrt{OE^2-PE^2}=\sqrt{\frac{175}{3}}$. Then $\tan{\alpha}=\frac{\sqrt{7}}{\sqrt{\frac{175}{3}}}=\frac{\sqrt{3}}{5}$. Then $\tan{2\alpha}=\tan{\theta}=\frac{\frac{2\sqrt{3}}{5}}{1-\frac{3}{25}}=\boxed{\frac{5\sqrt{3}}{11}}$.
-Magnetoninja | // Block 1
// Modified Asymptote diagram with correct right angle mark and alpha label.
import olympiad;
defaultpen(fontsize(13));
size(200);
// Circumradius R = 14*sqrt(3)/3
real R = 14*sqrt(3)/3;
// Define original points with scaling
pair O = (0,0);
pair A = R * dir(225);
pair B = R * dir(-15);
pair C = R * dir(105);
pair D = rotate(38.21, O) * A;
pair E = rotate(38.21, O) * B;
pair F = rotate(38.21, O) * C;
// Point H: Foot of altitude from B to DE
pair DE_vec = E - D;
pair perp_DE = (DE_vec.y, -DE_vec.x); // Rotate -90 degrees
perp_DE = perp_DE / length(perp_DE); // Unit vector
pair H = B - (2*sqrt(3)) * perp_DE; // Corrected direction
// Point M: Extend BH to M, BM = 13*sqrt(3)/3
pair BH_vec = H - B;
pair BM_unit = BH_vec / length(BH_vec); // Unit vector along BH
pair M = B + (13*sqrt(3)/3) * BM_unit; // BM = 13*sqrt(3)/3
// Point P: Midpoint of BE
pair P = (B + E) / 2;
// Draw original triangles and polygon
draw(A--B--C--A, gray+0.4);
draw(D--E--F--D, gray+0.4);
draw(A--D--B--E--C--F--A, black+0.9);
// Draw new dashed segments
draw(B--H, black+0.7+dashed);
draw(H--M, black+0.7+dashed);
draw(O--M, black+0.7+dashed);
draw(O--P, black+0.7+dashed);
draw(P--E, black+0.7+dashed);
draw(O--B, black+0.7+dashed);
// Draw right angle marker at OMB
draw(rightanglemark(B, M, O, 15)); // Right angle at M, size 15
// Draw dots and labels
dot(O);
dot("$O$", O, dir(180));
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$D$", D, dir(D));
dot("$E$", E, dir(E));
dot("$F$", F, dir(F));
dot("$H$", H, dir(270));
dot("$M$", M, dir(90));
dot("$P$", P, dir(0));
// Label angle POB as alpha, to the right of O
draw(anglemark(B, O, P, 50)); // Angle mark at O
label("$\alpha$", O + (1.5, -.23), dir(0)); // Position to the right of O
// Block 2
// Modified Asymptote diagram with correct right angle mark and alpha label. import olympiad; defaultpen(fontsize(13)); size(200); // Circumradius R = 14*sqrt(3)/3 real R = 14*sqrt(3)/3; // Define original points with scaling pair O = (0,0); pair A = R * dir(225); pair B = R * dir(-15); pair C = R * dir(105); pair D = rotate(38.21, O) * A; pair E = rotate(38.21, O) * B; pair F = rotate(38.21, O) * C; // Point H: Foot of altitude from B to DE pair DE_vec = E - D; pair perp_DE = (DE_vec.y, -DE_vec.x); // Rotate -90 degrees perp_DE = perp_DE / length(perp_DE); // Unit vector pair H = B - (2*sqrt(3)) * perp_DE; // Corrected direction // Point M: Extend BH to M, BM = 13*sqrt(3)/3 pair BH_vec = H - B; pair BM_unit = BH_vec / length(BH_vec); // Unit vector along BH pair M = B + (13*sqrt(3)/3) * BM_unit; // BM = 13*sqrt(3)/3 // Point P: Midpoint of BE pair P = (B + E) / 2; // Draw original triangles and polygon draw(A--B--C--A, gray+0.4); draw(D--E--F--D, gray+0.4); draw(A--D--B--E--C--F--A, black+0.9); // Draw new dashed segments draw(B--H, black+0.7+dashed); draw(H--M, black+0.7+dashed); draw(O--M, black+0.7+dashed); draw(O--P, black+0.7+dashed); draw(P--E, black+0.7+dashed); draw(O--B, black+0.7+dashed); // Draw right angle marker at OMB draw(rightanglemark(B, M, O, 15)); // Right angle at M, size 15 // Draw dots and labels dot(O); dot("$O$", O, dir(180)); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$E$", E, dir(E)); dot("$F$", F, dir(F)); dot("$H$", H, dir(270)); dot("$M$", M, dir(90)); dot("$P$", P, dir(0)); // Label angle POB as alpha, to the right of O draw(anglemark(B, O, P, 50)); // Angle mark at O label("$\alpha$", O + (1.5, -.23), dir(0)); // Position to the right of O | [] |
957 | Equilateral $\triangle ABC$ with side length $14$ is rotated about its center by angle $\theta$, where $0 < \theta < 60^{\circ}$, to form $\triangle DEF$. See the figure. The area of hexagon $ADBECF$ is $91\sqrt{3}$. What is $\tan\theta$?
$\textbf{(A)}~\frac{3}{4}\qquad\textbf{(B)}~\frac{5\sqrt{3}}{11}\qquad\textbf{(C)}~\frac{4}{5}\qquad\textbf{(D)}~\frac{11}{13}\qquad\textbf{(E)}~\frac{7\sqrt{3}}{13}$ | 2024 AMC 12B Problem 19 | \[[\triangle ABD] = \frac{[ADBECF]-[\triangle ABC]}{3} = \frac{91\sqrt{3}-49\sqrt{3}}{3} = 14\sqrt{3}\]
Let $M$ be the intersection of $AB$ and $DE$. Since $DMB$ is isosceles and $\angle AMD = \theta$, we have $\angle ABD = \theta/2$. Also, all of the hexagon's internal angles are equal, so $\angle ADB = 120^\circ$.
Using the side-angle-side area formula:
\[14\sqrt{3} = \frac{1}{2} \cdot 14 \cdot BD \cdot \sin\left(\frac{\theta}{2}\right) \Rightarrow BD = \frac{2\sqrt{3}}{\sin(\theta/2)}.\]
Apply Law of Sines on $\triangle ABD$ with $\angle DAB = 60^\circ - \theta/2$:
\[\frac{14}{\sin 120^\circ} = \frac{BD}{\sin(60^\circ - \theta/2)}\]
\[\frac{28}{\sqrt{3}} = \frac{2\sqrt{3}}{\sin(\theta/2)\sin(60^\circ - \theta/2)}.\]
\[\sin(\theta/2)\sin(60^\circ - \theta/2) = \frac{3}{14}.\]
Using trig identities:
\[\sqrt{\frac{1-\cos(\theta)}{2}} \cdot (\sin(60^\circ)\cos(\theta/2) - \sin(\theta/2)\cos(60^\circ)) = \frac{3}{14}\]
\[\sqrt{\frac{1-\cos(\theta)}{2}} \cdot \left(\frac{\sqrt{3}}{2}\cos(\theta/2) - \frac{1}{2}\sin(\theta/2)\right) = \frac{3}{14}\]
\[\sqrt{\frac{1-\cos(\theta)}{2}} \cdot \left(\frac{\sqrt{3}}{2} \sqrt{\frac{1+\cos(\theta)}{2}} - \frac{1}{2} \sqrt{\frac{1-\cos(\theta)}{2}} \right) = \frac{3}{14}\]
\[\left(\frac{\sqrt{3}}{2}\sqrt{\frac{1-\cos^2(\theta)}{4}} - \frac{1}{2}\left(\frac{1-\cos(\theta)}{2}\right)\right) = \frac{3}{14}\]
\[\frac{\sqrt{3}\sin\theta}{4} - \frac{(1-\cos\theta)}{4} = \frac{3}{14}\]
\[\sqrt{3}\sin\theta + \cos\theta = \frac{13}{7}\]
\[\cos\theta = \frac{13}{7} - \sqrt{3}\sin\theta.\]
Substitute into $\sin^2\theta + \cos^2\theta = 1$:
\[\left(\frac{13}{7} - \sqrt{3}\sin\theta\right)^2 + \sin^2\theta = 1.\]
\[4\sin^2\theta - \frac{26\sqrt{3}}{7}\sin\theta + \frac{120}{49} = 0.\]
Solving:
\[\sin\theta = \frac{5\sqrt{3}}{14} \quad (\text{valid root}), \quad \cos\theta = \frac{11}{14} \Rightarrow \tan\theta = \frac{5\sqrt{3}}{11}.\]
\[\boxed{\text{(B) }\frac{5\sqrt{3}}{11}}\]
~sparkycat (ttm) | // Block 1
defaultpen(fontsize(13));
size(220);
// Base points and rotation
pair O = (0,0);
pair A = dir(225);
pair B = dir(-15);
pair D = rotate(38.21, O)*A;
pair E = rotate(38.21, O)*B;
// Intersection point of AB and DE
pair M = extension(A, B, D, E);
// Triangle and segments
defaultpen(fontsize(13));
size(220);
// Base triangle from rotated figure
pair O = (0,0);
pair A = dir(225);
pair B = dir(-15);
pair D = rotate(38.21, O)*A;
pair E = rotate(38.21, O)*B;
// M is intersection of AB and DE
pair M = extension(A, B, D, E);
// Draw triangle and segment
draw(A--B--D--cycle, black+0.9);
draw(D--M, gray + dashed);
// Draw angle theta at AMD
real r = 0.1;
path thetaArc = arc(M, r, degrees(D - M), degrees(A - M));
draw(thetaArc, gray);
label("$\theta$", M + (r+0.1)*dir((degrees(D - M) + degrees(A - M))/2), gray);
// Draw congruence ticks on MB and MD
pair u1 = unit(B - M), u2 = unit(D - M);
pair tick1a = midpoint(M--B) + rotate(90)*u1 * 0.04;
pair tick1b = midpoint(M--B) + rotate(-90)*u1 * 0.04;
pair tick2a = midpoint(M--D) + rotate(90)*u2 * 0.04;
pair tick2b = midpoint(M--D) + rotate(-90)*u2 * 0.04;
draw(tick1a--tick1b, gray);
draw(tick2a--tick2b, gray);
// Labels
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$D$", D, dir(D));
dot("$M$", M, N);
// Block 2
defaultpen(fontsize(13)); size(220); // Base points and rotation pair O = (0,0); pair A = dir(225); pair B = dir(-15); pair D = rotate(38.21, O)*A; pair E = rotate(38.21, O)*B; // Intersection point of AB and DE pair M = extension(A, B, D, E); // Triangle and segments defaultpen(fontsize(13)); size(220); // Base triangle from rotated figure pair O = (0,0); pair A = dir(225); pair B = dir(-15); pair D = rotate(38.21, O)*A; pair E = rotate(38.21, O)*B; // M is intersection of AB and DE pair M = extension(A, B, D, E); // Draw triangle and segment draw(A--B--D--cycle, black+0.9); draw(D--M, gray + dashed); // Draw angle theta at AMD real r = 0.1; path thetaArc = arc(M, r, degrees(D - M), degrees(A - M)); draw(thetaArc, gray); label("$\theta$", M + (r+0.1)*dir((degrees(D - M) + degrees(A - M))/2), gray); // Draw congruence ticks on MB and MD pair u1 = unit(B - M), u2 = unit(D - M); pair tick1a = midpoint(M--B) + rotate(90)*u1 * 0.04; pair tick1b = midpoint(M--B) + rotate(-90)*u1 * 0.04; pair tick2a = midpoint(M--D) + rotate(90)*u2 * 0.04; pair tick2b = midpoint(M--D) + rotate(-90)*u2 * 0.04; draw(tick1a--tick1b, gray); draw(tick2a--tick2b, gray); // Labels dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$D$", D, dir(D)); dot("$M$", M, N); | [] |
958 | The measures of the smallest angles of three different right triangles sum to $90^\circ$. All three triangles have side lengths that are primitive Pythagorean triples. Two of them are $3-4-5$ and $5-12-13$. What is the perimeter of the third triangle?
$\textbf{(A) } 40 \qquad\textbf{(B) } 126 \qquad\textbf{(C) } 154 \qquad\textbf{(D) } 176 \qquad\textbf{(E) } 208$ | 2024 AMC 12B Problem 21 | We can start by scaling the $3-4-5$ right triangle up by a factor of $3$ and "gluing" it to the $5-12-13$ triangle's longer leg. Let $\alpha$, $\beta$, and $\theta$ be the smallest angles in the $3-4-5$, $5-12-13$ and unknown triangle respectively. We can construct the following diagram of the two known triangles.
We can also construct a diagram for the third, unknown triangle like so. We know that $\alpha+\beta+\theta=90$, and therefore that $\theta=90-\alpha-\beta$. We also know that the other acute angle in this third triangle will have a measure of $\alpha+\beta$ by the triangle angle sum theorem.
We can use the law of cosines on the triangle in the first diagram to get the equation $14^2=15^2+13^2-2(13)(15)\cos{(\alpha+\beta)}$. Isolating $\cos{(\alpha+\beta)}$, we get $-198=-390\cos{(\alpha+\beta)}$, which further simplifies to $\cos{(\alpha+\beta)}=\frac{198}{390}$. Since the third triangle has to be a primitive Pythagorean triple, we must take this trig ratio into its most simple form, namely $\cos{(\alpha+\beta)}=\frac{33}{65}$. Using this information in our second diagram, we know that the smaller, adjacent leg must have length $33$, and the hypotenuse must have length $65$. We can then use the Pythagorean theorem to find the other, unknown leg, which has length $56$. Adding these three lengths together, we find that the perimeter of this right triangle is $33+56+65=\boxed{\textbf{(C)\ 154}}$.
~Phinetium
Solution 6.1 (Faster Ending)
Instead of computing $\sqrt{65^2-33^2}$ to find the second leg in the unknown triangle by hand, we can use process of elimination. $\textbf{(B)}\ 126$ and $\textbf{(C)}\ 154$ are the only answers within the realm of possibility, because $\textbf{(A)}\ 40$ would entitle a triangle with a negative side length, and $\textbf{(D)}\ 176$ and $\textbf{(E)}\ 208$ would require legs greater than the length of the hypotenuse. The answer choice $\textbf{(B)}\ 126$ would force the second leg to have a length of $28$, which is smaller than the smallest leg in the triangle. (We know that the leg with length $33$ is the smallest leg in the triangle by the side-angle relationship theorem, because it is opposite the smallest angle in the triangle.) Therefore, the only valid answer choice remaining is $\boxed{\textbf{(C)\ 154}}$.
~Phinetium (again) | // Block 1
pair A = (0,0);
pair B = (5,0);
pair C = (14,0);
pair D = (5,12);
draw(A--C--D--cycle);
draw(B--D);
draw(rightanglemark(A,B,D,20));
label("5", A--B, S);
label("9", B--C, S);
label("15", C--D, NE);
label("13", D--A, NW);
label("12", B--D, E);
label("$\beta$", D, 6*dir(257));
label("$\alpha$", D, 4*dir(287));
// Block 2
pair A = (0,0);
pair B = (56,0);
pair C = (56,33);
draw(A--B--C--cycle);
draw(rightanglemark(A,B,C,50));
label("$90-\alpha-\beta$", A, 8*dir(12));
label("$\alpha+\beta$", C, 6*dir(242));
// Block 3
pair A = (0,0); pair B = (5,0); pair C = (14,0); pair D = (5,12); draw(A--C--D--cycle); draw(B--D); draw(rightanglemark(A,B,D,20)); label("5", A--B, S); label("9", B--C, S); label("15", C--D, NE); label("13", D--A, NW); label("12", B--D, E); label("$\beta$", D, 6*dir(257)); label("$\alpha$", D, 4*dir(287));
// Block 4
pair A = (0,0); pair B = (56,0); pair C = (56,33); draw(A--B--C--cycle); draw(rightanglemark(A,B,C,50)); label("$90-\alpha-\beta$", A, 8*dir(12)); label("$\alpha+\beta$", C, 6*dir(242)); | [] |
959 | A semicircle has diameter $\overline{AB}$ and chord $\overline{CD}$ of length $16$ parallel to $\overline{AB}$. A smaller semicircle
with diameter on $\overline{AB}$ and tangent to $\overline{CD}$ is cut from the larger semicircle, as shown below.
What is the area of the resulting figure, shown shaded?
$\textbf{(A) } 16\pi \qquad\textbf{(B) } 24\pi \qquad\textbf{(C) } 32\pi \qquad\textbf{(D) } 48\pi \qquad\textbf{(E) } 64\pi$ | 2025 AMC 10A Problem 10 | The problem doesn't define where the chord is within the circle. So, let's place the chord such that when lines are drawn from its ends to the center of the larger semicircle, an equilateral triangle with side length $16$ is formed.
In this definition, the radius of the larger semicircle is $16,$ giving it an area of $\frac{\pi\cdot16^2}{2} = 128\pi.$
The height of the equilateral triangle is $8\sqrt{3},$ which is equal to the radius of the smaller semicircle. This gives the smaller semicircle an area of $\frac{\pi\cdot(8\sqrt{3})^2}{2} = 96\pi.$
The total shaded area is the difference of these semicircles, or $128\pi - 96\pi = \boxed{\text{(C) } 32\pi}.$
~chisps | // Block 1
import graph; unitsize(14mm);
defaultpen(linewidth(.8pt)+fontsize(10pt));
dotfactor=4;
fill(Arc((0,0),3,0,180)--cycle,palered);
fill(Arc((0,0),2.598,0,180)--cycle,white);
draw(Arc((0,0),3,0,180),black);
draw(Arc((0,0),2.598,0,180),black);
draw((-2.598,0) -- (2.598,0), white);
draw((-1.5,2.598) -- (1.5,2.598),black);
draw((0,0) -- (-1.5, 2.598), black);
draw((0,0) -- (1.5, 2.598), black);
draw((0,0) -- (0, 2.598), black);
draw((-3,0) -- (-2.598, 0), black);
draw((3,0) -- (2.598, 0), black);
dot((0,0));
dot((0,2.598));
dot((-1.5,2.598));
dot((1.5,2.598));
dot((3,0));
dot((-3,0));
label("$16$",midpoint((0, 2.598)--(0, 2.598)),N);
label("$8\sqrt{3}$",midpoint((0, 1.4)--(0, 1.4)),E);
label("$16$",midpoint((-0.9, 1.3)--(-0.9, 1.3)),W);
label("$16$",midpoint((0.9, 1.3)--(0.9, 1.3)),E);
label("$A$",midpoint((-3, 0)--(-3, 0)),W);
label("$B$",midpoint((3, 0)--(3, 0)),E);
label("$C$",midpoint((-1.5,2.7)--(-1.5,2.7)),N);
label("$D$",midpoint((1.5,2.7)--(1.5,2.7)),N);
// Block 2
import graph; unitsize(14mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; fill(Arc((0,0),3,0,180)--cycle,palered); fill(Arc((0,0),2.598,0,180)--cycle,white); draw(Arc((0,0),3,0,180),black); draw(Arc((0,0),2.598,0,180),black); draw((-2.598,0) -- (2.598,0), white); draw((-1.5,2.598) -- (1.5,2.598),black); draw((0,0) -- (-1.5, 2.598), black); draw((0,0) -- (1.5, 2.598), black); draw((0,0) -- (0, 2.598), black); draw((-3,0) -- (-2.598, 0), black); draw((3,0) -- (2.598, 0), black); dot((0,0)); dot((0,2.598)); dot((-1.5,2.598)); dot((1.5,2.598)); dot((3,0)); dot((-3,0)); label("$16$",midpoint((0, 2.598)--(0, 2.598)),N); label("$8\sqrt{3}$",midpoint((0, 1.4)--(0, 1.4)),E); label("$16$",midpoint((-0.9, 1.3)--(-0.9, 1.3)),W); label("$16$",midpoint((0.9, 1.3)--(0.9, 1.3)),E); label("$A$",midpoint((-3, 0)--(-3, 0)),W); label("$B$",midpoint((3, 0)--(3, 0)),E); label("$C$",midpoint((-1.5,2.7)--(-1.5,2.7)),N); label("$D$",midpoint((1.5,2.7)--(1.5,2.7)),N); | [] |
959 | A semicircle has diameter $\overline{AB}$ and chord $\overline{CD}$ of length $16$ parallel to $\overline{AB}$. A smaller semicircle
with diameter on $\overline{AB}$ and tangent to $\overline{CD}$ is cut from the larger semicircle, as shown below.
What is the area of the resulting figure, shown shaded?
$\textbf{(A) } 16\pi \qquad\textbf{(B) } 24\pi \qquad\textbf{(C) } 32\pi \qquad\textbf{(D) } 48\pi \qquad\textbf{(E) } 64\pi$ | 2025 AMC 10A Problem 10 | We can move this small semicircle to the middle of the big semicircle. Let $r$ be the radius of the small semicircle. By Pythagorean Theorem, the line draw from the midpoint of the diameter of the big semicircle to the big semicircle at the height of the small semicircle (see the diagram) has length $\sqrt{r^2 + 64}$. This is a radius of the big semicircle. So the shaded area would be
\[\frac{\pi \cdot (\sqrt{r^2+64})^2 - \pi r^2}{2} = \frac{\pi(r^2 + 64 - r^2)}{2} = \frac{64\pi}{2} = \boxed{\text{(C) } 32\pi}.\]
Diagram:
~JerryZYang | // Block 1
import graph;
unitsize(14mm);
defaultpen(linewidth(.8pt)+fontsize(10pt));
dotfactor=4;
real R = 3;
pair A = (-R,0), B = (R,0);
fill(Arc((0,0),R,0,180)--cycle,palered);
draw(Arc((0,0),R,0,180),black);
real h = 0.75;
pair C = (-sqrt(R^2 - h^2), h);
pair D = ( sqrt(R^2 - h^2), h);
draw(C--D,black);
label("$C$",C,W);
label("$D$",D,E);
label("$16$", midpoint(C--D), N);
real r = h;
pair O = (0,0);
pair Q = (0,r);
fill(Arc(O,r,0,180)--cycle,white);
draw(Arc(O,r,0,180),black);
draw(A--B,black);
draw(O--Q);
draw(O--D);
draw(Q--D, dashed);
label("$A$",A,S);
label("$B$",B,S);
label("$r$", midpoint(O--Q), W);
label("$\sqrt{r^{2}+64}$", midpoint(O--D), S);
label("$8$", midpoint(Q--D), N);
// Block 2
import graph; unitsize(14mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; real R = 3; pair A = (-R,0), B = (R,0); fill(Arc((0,0),R,0,180)--cycle,palered); draw(Arc((0,0),R,0,180),black); real h = 0.75; pair C = (-sqrt(R^2 - h^2), h); pair D = ( sqrt(R^2 - h^2), h); draw(C--D,black); label("$C$",C,W); label("$D$",D,E); label("$16$", midpoint(C--D), N); real r = h; pair O = (0,0); pair Q = (0,r); fill(Arc(O,r,0,180)--cycle,white); draw(Arc(O,r,0,180),black); draw(A--B,black); draw(O--Q); draw(O--D); draw(Q--D, dashed); label("$A$",A,S); label("$B$",B,S); label("$r$", midpoint(O--Q), W); label("$\sqrt{r^{2}+64}$", midpoint(O--D), S); label("$8$", midpoint(Q--D), N); | [] |
960 | In the figure below, $ABEF$ is a rectangle, $\overline{AD}\perp\overline{DE}$, $AF=7$, $AB=1$, and $AD=5$.
What is the area of $\triangle ABC$?
$\textbf{(A) } \frac{3}{8} \qquad\textbf{(B) } \frac{4}{9} \qquad\textbf{(C) } \frac{1}{8}\sqrt{13} \qquad\textbf{(D) } \frac{7}{15} \qquad\textbf{(E) } \frac{1}{8}\sqrt{15}$ | 2025 AMC 10A Problem 15 | As shown in the figure, establish a coordinate system with point A as the origin.
Suppose $C = (a, 1)$, then the equation of $AC$ is $y = \dfrac{1}{a}x$. Since $DE \perp AC$, the slope of $DE$ is $-a$, the equation of $DE$ is $y - 1 = -a(x - 7) \to ax + y - 7a - 1 = 0$.
Notice that the length of $AD$ is the distance between point $A$ and line $DE$, due to the distance formula, we get:
\[\frac{|a \cdot 0 + 0 - 7a - 1|}{\sqrt{a^2 + 1}} = 5\]
\[(7a + 1)^2 = 25(a^2 + 1)\]
\[(4a - 3)(3a + 4) = 0\]
\[a = \frac{3}{4}\]
The area of $\triangle BCA$ is therefore $\frac{a \cdot 1}{2}=\frac{3}{8}=\boxed{A}$.
~reda_mandymath | // Block 1
unitsize(1cm);
pair _A, _B, _C, _D, _E, _F;
_A = (0, 0);
_B = (0, 1);
_E = (7, 1);
_F = (7, 0);
_C = (3/4, 1);
_D = extension(_A, _C, _E, _E + rotate(90)*(_A - _C));
fill(_A -- _B -- _C -- cycle, gray);
draw(_A -- _B -- _E -- _D -- _A -- _F -- _E);
draw((-1, 0) -- (8, 0), Arrows);
draw((0, -1) -- (0, 5), Arrows);
label("$A$", _A, SW);
label("$B$", _B, W);
label("$C$", _C, N);
label("$D$", _D, N);
label("$E$", _E, NE);
label("$F$", _F, SE);
label("$x$", (8, 0), S);
label("$y$", (0, 5), E);
label("$1$", _B, NE);
label("$7$", _F, SW);
label("$(a, 1)$", _C, SE);
label(scale(0.8) * rotate(aTan(4/3)) * "$y =\frac{1}{a}x$", (_A + _D)/2, NW);
label(scale(0.8) * rotate(-aTan(3/4)) * "$y - 1 = -a(x - 7)$", (_D + _E)/2, NE);
// Block 2
unitsize(1cm); pair _A, _B, _C, _D, _E, _F; _A = (0, 0); _B = (0, 1); _E = (7, 1); _F = (7, 0); _C = (3/4, 1); _D = extension(_A, _C, _E, _E + rotate(90)*(_A - _C)); fill(_A -- _B -- _C -- cycle, gray); draw(_A -- _B -- _E -- _D -- _A -- _F -- _E); draw((-1, 0) -- (8, 0), Arrows); draw((0, -1) -- (0, 5), Arrows); label("$A$", _A, SW); label("$B$", _B, W); label("$C$", _C, N); label("$D$", _D, N); label("$E$", _E, NE); label("$F$", _F, SE); label("$x$", (8, 0), S); label("$y$", (0, 5), E); label("$1$", _B, NE); label("$7$", _F, SW); label("$(a, 1)$", _C, SE); label(scale(0.8) * rotate(aTan(4/3)) * "$y =\frac{1}{a}x$", (_A + _D)/2, NW); label(scale(0.8) * rotate(-aTan(3/4)) * "$y - 1 = -a(x - 7)$", (_D + _E)/2, NE); | [] |
960 | In the figure below, $ABEF$ is a rectangle, $\overline{AD}\perp\overline{DE}$, $AF=7$, $AB=1$, and $AD=5$.
What is the area of $\triangle ABC$?
$\textbf{(A) } \frac{3}{8} \qquad\textbf{(B) } \frac{4}{9} \qquad\textbf{(C) } \frac{1}{8}\sqrt{13} \qquad\textbf{(D) } \frac{7}{15} \qquad\textbf{(E) } \frac{1}{8}\sqrt{15}$ | 2025 AMC 10A Problem 15 | Start by assigning the variable $x$ to the length of side $\overline{AC}$. Since $\overline{AD}=5$, $\overline {CD}=5-x$. Additionally, the diagonal $\overline{AE}=5\sqrt{2}$ by the Pythagorean Theorem. Notice that with this diagonal, we've created a right triangle $\triangle ADE$.
Now, you could manually use the Pythagorean Theorem, but you can just spot the $45^\circ-45^\circ-90^\circ$ triangle and see $\overline{ED}=5$. After this, we can use the Pythagorean Theorem to find the length of $\overline{CE}$. \[5^2+(5-x)^2 = (\overline{CE})^2\]
\[25+25+x^2-10x = (\overline{CE})^2\]
\[x^2-10x+50 = (\overline{CE})^2\] \[\sqrt{x^2-10x+50}=\overline{CE}\]
From this, we can see that $\overline{CB} = 7-\overline{CE} \implies \overline{CB} = 7 - \sqrt{x^2-10x+50}$
Now, we know that $\left[CDE\right] + \left[ABE\right] -\left[ABC\right] = \left[ ADE\right]= 12.5$. Therefore,
\[\frac{5\cdot(5-x)}{2} + \frac{7}{2} - \frac{7 - \sqrt{x^2-10x+50}}{2}=12.5\]
\[5\cdot(5-x) + 7 - (7 - \sqrt{x^2-10x+50})=25\]
\[25-5x + \sqrt{x^2-10x+50}=25\]
\[\sqrt{x^2-10x+50}=5x\]
\[x^2-10x+50=25x^2\]
\[0=24x^2+10x-50\]
\[0=12x^2+5x-25\]
\[0=(4x-5)(3x+5)\]
Now, we know that $x$ must be positive so our only valid solution is
\[x=\frac{5}{4}\]
Finally, we can substitute this value of $x$ into our expression for length $\overline {CB}$. The other leg would just be $1$ making for a total area of $\frac{1}{2} \cdot\overline{CB}$. Hence,
\[\left[ ABC\right]= \frac{7 - \sqrt{x^2-10x+50}}{2}\]
\[\left[ ABC\right]= \frac{7 - \sqrt{\frac{5}{4}^2-10\cdot \frac{5}{4}+50}}{2}\]
\[\left[ ABC\right]= \frac{7 - \sqrt{\frac{25}{16}-\frac{50}{4}+\frac{200}{4}}}{2}\]
\[\left[ ABC\right]= \frac{7 - \sqrt{\frac{25}{16}+\frac{150}{4}}}{2}\]
\[\left[ ABC\right]= \frac{7 - \sqrt{\frac{25}{16}+\frac{600}{16}}}{2}\]
\[\left[ ABC\right]= \frac{7 - \sqrt{\frac{625}{16}}}{2}\]
\[\left[ ABC\right]= \frac{7 - \frac{25}{4}}{2}\]
\[\left[ ABC\right]= \frac{\frac{28}{4} - \frac{25}{4}}{2}\]
\[\left[ ABC\right]= \frac{\frac{3}{4}}{2}\]
\[\left[ ABC\right]=\frac{3}{8} \implies \boxed{\textbf{(A)} \frac{3}{8}}\]
~Shockfront99 | // Block 1
unitsize(1cm);
pen small=fontsize(9pt);
pair A, B, C, D, E, F;
draw(brace((5,5),(5,0)),1bp+black);
label(" $ $ $5$",brace((5,5),(5,0)), E);
A = (5, 5);
B = (5.6, 4.2);
C = (5, 3.75);
D = (5, 0);
E = (0, 0);
F = (-0.6, 0.8);
fill(A--B--C--cycle, palered);
dot(A);
dot(B);
dot(C);
dot(D);
dot(E);
dot(F);
label("$A$", A, N);
label("$B$", B, (1,0));
label("$C$", C, SE);
label("$D$", D, (1,0));
label("$E$", E, S);
label("$F$", F, W);
label("$5\sqrt{2}$", A--E, W, small);
draw(A--D--E);
draw(A--B--E--F--A);
draw(A--E);
draw(rightanglemark(C, D, E));
// Block 2
unitsize(1cm);
pen small=fontsize(9pt);
pair A, B, C, D, E, F;
draw(brace((5,0),(5,5)),black+1bp);
label("$5$ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ ",brace((5,5),(5,0)), W);
A = (5, 5);
B = (5.6, 4.2);
C = (5, 3.75);
D = (5, 0);
E = (0, 0);
F = (-0.6, 0.8);
fill(A--B--C--cycle, palered);
dot(A);
dot(B);
dot(C);
dot(D);
dot(E);
dot(F);
label("$A$", A, N);
label("$B$", B, (1,0));
label("$C$", C, SE);
label("$D$", D, (1,0));
label("$E$", E, S);
label("$F$", F, W);
label("$5\sqrt{2}$", A--E, W, small);
label(rotate(36.5) * "$\quad \sqrt{x^2-10x+50}$", E--C, S);
label(rotate(306.5) * "$\leftarrow7-\sqrt{x^2-10x+50}$", B--C, SE);
draw(A--D--E);
draw(A--B--E--F--A);
draw(A--E);
draw(C--E);
draw(rightanglemark(C, D, E));
// Block 3
unitsize(1cm); pen small=fontsize(9pt); pair A, B, C, D, E, F; draw(brace((5,5),(5,0)),1bp+black); label(" $ $ $5$",brace((5,5),(5,0)), E); A = (5, 5); B = (5.6, 4.2); C = (5, 3.75); D = (5, 0); E = (0, 0); F = (-0.6, 0.8); fill(A--B--C--cycle, palered); dot(A); dot(B); dot(C); dot(D); dot(E); dot(F); label("$A$", A, N); label("$B$", B, (1,0)); label("$C$", C, SE); label("$D$", D, (1,0)); label("$E$", E, S); label("$F$", F, W); label("$5\sqrt{2}$", A--E, W, small); draw(A--D--E); draw(A--B--E--F--A); draw(A--E); draw(rightanglemark(C, D, E));
// Block 4
unitsize(1cm); pen small=fontsize(9pt); pair A, B, C, D, E, F; draw(brace((5,0),(5,5)),black+1bp); label("$5$ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ ",brace((5,5),(5,0)), W); A = (5, 5); B = (5.6, 4.2); C = (5, 3.75); D = (5, 0); E = (0, 0); F = (-0.6, 0.8); fill(A--B--C--cycle, palered); dot(A); dot(B); dot(C); dot(D); dot(E); dot(F); label("$A$", A, N); label("$B$", B, (1,0)); label("$C$", C, SE); label("$D$", D, (1,0)); label("$E$", E, S); label("$F$", F, W); label("$5\sqrt{2}$", A--E, W, small); label(rotate(36.5) * "$\quad \sqrt{x^2-10x+50}$", E--C, S); label(rotate(306.5) * "$\leftarrow7-\sqrt{x^2-10x+50}$", B--C, SE); draw(A--D--E); draw(A--B--E--F--A); draw(A--E); draw(C--E); draw(rightanglemark(C, D, E)); | [] |
961 | A silo (right circular cylinder) with diameter 20 meters stands in a field. MacDonald is located 20 meters west and 15 meters south of the center of the silo. McGregor is located 20 meters east and $g > 0$ meters south of the center of the silo. The line of sight between MacDonald and McGregor is tangent to the silo. The value of g can be written as $\frac{a\sqrt{b}-c}{d}$, where $a,b,c,$ and $d$ are positive integers, $b$ is not divisible by the square of any prime, and $d$ is relatively prime to the greatest common divisor of $a$ and $c$. What is $a+b+c+d$?
$\textbf{(A) } 119 \qquad\textbf{(B) } 120 \qquad\textbf{(C) } 121 \qquad\textbf{(D) } 122 \qquad\textbf{(E) } 123$ | 2025 AMC 10A Problem 20 | Let the silo center be $O$, let the point MacDonald is situated at be $A$, and let the point $20$ meters west of the silo center be $B$. $ABO$ is then a right triangle with side lengths $15, 20,$ and $25$.
Let the point $20$ meters east of the silo center be $C$, and let the point McGregor is at be $D$ with $CD=g>0$. Also let $AD$ be tangent to circle $O$ at $E$.
Extend $BC$ and $AD$ to meet at point $F$. This creates $3$ similar triangles, $\triangle ABF\sim \triangle DCF \sim \triangle OEF$. Let the distance between point $C$ and $F$ be $x$. The similarity ratio between triangles $ABF$ and $DCF$ is then $\frac{longer\;leg}{shorter\;leg} = \frac{40+x}{15} = \frac{x}{g}$
This is currently unsolvable so we bring in triangle $OEF$. The hypotenuse of triangle $OEF$ is $OF=20+x$ and its shorter leg is the radius of the silo $=10$. We can then establish a second similarity relationship between triangles $OEF$ and $ABF$ with $\frac{shorter\; leg}{hypotenuse}=\frac{10}{20+x}=\frac{15}{AF}$
Now we find the hypotenuse of $ABF$ in terms of $x$ using the Pythagorean theorem. $AF^2=15^2+(40+x)^2$. Which simplifies to $AF^2=225+1600+80x+x^2=1825+80x+x^2$ So $AF=\sqrt{x^2+80x+1825}$
Plugging back in we get $\frac{10}{20+x}=\frac{15}{\sqrt{x^2+80x+1825}}$. Now we can begin to break this down by multiplying both sides by both denominators. $10(\sqrt{x^2+80x+1825})=15(20+x)$ Dividing both sides by $5$ then squaring yields, $4x^2+320x+7300=9x^2+360x+3600$ This furthermore simplifies to $5x^2+40x-3700=0$ At which point we can divide off a $5$ and then apply the quadratic formula on $x^2+8x-740=0$ which we take the positive root of.
\[x = \frac{-8+\sqrt{64+2960}}{2} = \frac{-8+\sqrt{3024}}{2} =\frac{-8+\sqrt{144 \cdot 21}}{2}.\]
Simplifying yields that $x=6\sqrt{21}-4$
Then to solve for $g$ we simply plug $6\sqrt{21}-4$ back into the first similarity ratio to get $\frac{36+6\sqrt{21}}{15}=\frac{6\sqrt{21}-4}{g}$
Multiply both sides by $15g$ and dividing by $36+6\sqrt{21}$ will let us solve for $g=\frac{15(6\sqrt{21}-4)}{36+6\sqrt{21}}$ and after rationalizing the denominator we get $\frac{20\sqrt{21}-75}{3}$. $20+21+75+3=\boxed{\textbf{(A)}~119}$
$\sim$ Nioronean
~ happyfish0922 (minor edit regarding what was previously point T changed to point E)
~ Avs2010 (diagram)
~JerryZYang (minor edit in latex and style) | // Block 1
import olympiad;
size(340);
// numeric setup for reliable AoPS rendering
real R = 10;
pair O = (0,0);
pair B = (-20,0);
pair A = (-20,-15);
pair C = (20,0);
real g = 5.55; // approximate numeric version of (20β21β75)/3
pair D = (20,-g);
pair E = (2.30,-9.74);
pair F = intersectionpoint(A--(A+4*(D-A)), B--(B+4*(C-B)));
filldraw(circle(O,R), white, heavyblue);
draw(A--F, heavygreen);
draw(B--F);
draw(A--B--O--cycle);
draw(O--E);
draw((C+2*(C-D))--(C + 3*(D-C)), dashed);
dot(A); label("$A$", A, SW);
dot(B); label("$B$", B, NW);
dot(C); label("$C$", C, NE);
dot(D); label("$D$", D, SE);
dot(E); label("$E$", E, NE);
dot(O); label("$O$", O, N);
dot(F); label("$F$", F, S);
draw(rightanglemark(A,B,O,50));
label("$20$", midpoint(B--O), dir(90));
label("$15$", midpoint(A--B), W);
label("$25$", midpoint(A--O), SE);
label("$g$", (C+D)/2, W);
label(β?β, midpoint(A--F), SE);
currentpicture.fit();
// Block 2
import olympiad; size(340); // numeric setup for reliable AoPS rendering real R = 10; pair O = (0,0); pair B = (-20,0); pair A = (-20,-15); pair C = (20,0); real g = 5.55; // approximate numeric version of (20β21β75)/3 pair D = (20,-g); pair E = (2.30,-9.74); pair F = intersectionpoint(A--(A+4*(D-A)), B--(B+4*(C-B))); filldraw(circle(O,R), white, heavyblue); draw(A--F, heavygreen); draw(B--F); draw(A--B--O--cycle); draw(O--E); draw((C+2*(C-D))--(C + 3*(D-C)), dashed); dot(A); label("$A$", A, SW); dot(B); label("$B$", B, NW); dot(C); label("$C$", C, NE); dot(D); label("$D$", D, SE); dot(E); label("$E$", E, NE); dot(O); label("$O$", O, N); dot(F); label("$F$", F, S); draw(rightanglemark(A,B,O,50)); label("$20$", midpoint(B--O), dir(90)); label("$15$", midpoint(A--B), W); label("$25$", midpoint(A--O), SE); label("$g$", (C+D)/2, W); label(β?β, midpoint(A--F), SE); currentpicture.fit(); | [] |
961 | A silo (right circular cylinder) with diameter 20 meters stands in a field. MacDonald is located 20 meters west and 15 meters south of the center of the silo. McGregor is located 20 meters east and $g > 0$ meters south of the center of the silo. The line of sight between MacDonald and McGregor is tangent to the silo. The value of g can be written as $\frac{a\sqrt{b}-c}{d}$, where $a,b,c,$ and $d$ are positive integers, $b$ is not divisible by the square of any prime, and $d$ is relatively prime to the greatest common divisor of $a$ and $c$. What is $a+b+c+d$?
$\textbf{(A) } 119 \qquad\textbf{(B) } 120 \qquad\textbf{(C) } 121 \qquad\textbf{(D) } 122 \qquad\textbf{(E) } 123$ | 2025 AMC 10A Problem 20 | We use the point names given in Solution 1.
By Pythagorean Theorem (twice), we have \[AE=\sqrt{AO^2-OE^2}=\sqrt{AB^2+BO^2-10^2}=\sqrt{525}=5\sqrt{21}.\] Similarly, we find that \[DE=\sqrt{DO^2-OE^2}=\sqrt{CD^2+CO^2-10^2}=\sqrt{300+g^2}.\] Note that the altitude from $D$ to $AB$ (call the foot $H$) has length $40$. Also, $AD=AE+DE=5\sqrt{21}+\sqrt{300+g^2}$ and $AH=15-g$. By Pythagoras on $\triangle ADH$, we have \[\left(5\sqrt{21}+\sqrt{300+g^2}\right)^2=40^2+(15-g)^2.\] Expanding, \[525+10\sqrt{21}\cdot\sqrt{300+g^2}+300+g^2=1600+225-30g+g^2\] and simplifying gives \[10\sqrt{6300+21g^2}=1000-30g.\] Dividing by $10$ and squaring, \[6300+21g^2=10000-600g+9g^2\] and putting it in standard form, \[0=12g^2+600g-3700.\] Dividing by $4$ and applying the quadratic formula, \[g=\frac{-150\pm\sqrt{22500-4(3)(-925)}}{6}=\frac{-75\pm20\sqrt{21}}3.\] From the answer form, we take the positive root, so \[g=\frac{20\sqrt{21}-75}3.\] Finally, we have $20+21+75+3=\boxed{\text{(A) }119}.$
~Waddles2010
(Minor calculation error fixed by hellokikki)
~AH2025
minor typo fix | // Block 1
import olympiad;
size(340);
// numeric setup for reliable AoPS rendering
real R = 10;
pair O = (0,0);
pair B = (-20,0);
pair A = (-20,-15);
pair C = (20,0);
real g = 5.55; // approximate numeric version of (20β21β75)/3
pair D = (20,-g);
pair E = (2.30,-9.74);
pair F = intersectionpoint(A--(A+4*(D-A)), B--(B+4*(C-B)));
filldraw(circle(O,R), white, heavyblue);
draw(A--F, heavygreen);
draw(B--F);
draw(A--B--O--cycle);
draw(O--E);
draw((C+2*(C-D))--(C + 3*(D-C)), dashed);
dot(A); label("$A$", A, SW);
dot(B); label("$B$", B, NW);
dot(C); label("$C$", C, NE);
dot(D); label("$D$", D, SE);
dot(E); label("$E$", E, NE);
dot(O); label("$O$", O, N);
dot(F); label("$F$", F, S);
draw(rightanglemark(A,B,O,50));
label("$20$", midpoint(B--O), dir(90));
label("$15$", midpoint(A--B), W);
label("$25$", midpoint(A--O), SE);
label("$g$", (C+D)/2, W);
currentpicture.fit();
// Block 2
import olympiad; size(340); // numeric setup for reliable AoPS rendering real R = 10; pair O = (0,0); pair B = (-20,0); pair A = (-20,-15); pair C = (20,0); real g = 5.55; // approximate numeric version of (20β21β75)/3 pair D = (20,-g); pair E = (2.30,-9.74); pair F = intersectionpoint(A--(A+4*(D-A)), B--(B+4*(C-B))); filldraw(circle(O,R), white, heavyblue); draw(A--F, heavygreen); draw(B--F); draw(A--B--O--cycle); draw(O--E); draw((C+2*(C-D))--(C + 3*(D-C)), dashed); dot(A); label("$A$", A, SW); dot(B); label("$B$", B, NW); dot(C); label("$C$", C, NE); dot(D); label("$D$", D, SE); dot(E); label("$E$", E, NE); dot(O); label("$O$", O, N); dot(F); label("$F$", F, S); draw(rightanglemark(A,B,O,50)); label("$20$", midpoint(B--O), dir(90)); label("$15$", midpoint(A--B), W); label("$25$", midpoint(A--O), SE); label("$g$", (C+D)/2, W); currentpicture.fit(); | [] |
961 | A silo (right circular cylinder) with diameter 20 meters stands in a field. MacDonald is located 20 meters west and 15 meters south of the center of the silo. McGregor is located 20 meters east and $g > 0$ meters south of the center of the silo. The line of sight between MacDonald and McGregor is tangent to the silo. The value of g can be written as $\frac{a\sqrt{b}-c}{d}$, where $a,b,c,$ and $d$ are positive integers, $b$ is not divisible by the square of any prime, and $d$ is relatively prime to the greatest common divisor of $a$ and $c$. What is $a+b+c+d$?
$\textbf{(A) } 119 \qquad\textbf{(B) } 120 \qquad\textbf{(C) } 121 \qquad\textbf{(D) } 122 \qquad\textbf{(E) } 123$ | 2025 AMC 10A Problem 20 | First, taking the Pythagorean Theorem on $\Delta BOD$ and $\Delta AOE$ gives $AO = 25, AE = 5\sqrt{21}$. By cosine angular addition, $\cos{(\angle BOD + \angle AOE)} = \cos{\angle BOE} = \frac{8-3\sqrt21}{25}$, so $\cos{\angle EOC} = \frac{3\sqrt21-8}{25}$.
Dropping the altitude from $E$ to $BC$ and labelling the foot $P$, we have that $OP = 10 \cdot \frac{3\sqrt21-8}{25} = \frac{6\sqrt21-16}{5}$, so the ratio of $BP$ to $BC$ is $\frac{3\sqrt{21}+42}{100}$. This is the same as the ratio of $AE$ to $AD$, so $AD = 5\sqrt{21} \cdot \frac{100}{3\sqrt{21}+42} = \frac{20}{3} \cdot (2\sqrt{21}-3)$.
Dropping the altitude from $D$ to $AB$ and labelling the foot $Q$, $AQ$ can be (tediously) computed to be $\frac{20}{3} \cdot \sqrt{(2\sqrt{21}-3)^2-6^2} = \frac{20}{3} \cdot \sqrt{57-12\sqrt{21}} = \frac{20}{3} \cdot (6-\sqrt{21}) =\frac{120-20\sqrt{21}}{3}$, and subtracting this from $AB$ gives the length of $BQ$ equivalent to $CD$, which is our desired length, so our answer is $\frac{20\sqrt{21}-75}{3} \implies \boxed{\text{(A) }119}.$
~ tiguhbabehwo
~Sakura_kitty (copy and pasted diagram, plz add point p, idk how to use latex.) | // Block 1
import olympiad;
size(340);
// numeric setup for reliable AoPS rendering
real R = 10;
pair O = (0,0);
pair B = (-20,0);
pair A = (-20,-15);
pair C = (20,0);
real g = 5.55; // approximate numeric version of (20β21β75)/3
pair D = (20,-g);
pair E = (2.30,-9.74);
pair F = intersectionpoint(A--(A+4*(D-A)), B--(B+4*(C-B)));
filldraw(circle(O,R), white, heavyblue);
draw(A--F, heavygreen);
draw(B--F);
draw(A--B--O--cycle);
draw(O--E);
draw((C+2*(C-D))--(C + 3*(D-C)), dashed);
dot(A); label("$A$", A, SW);
dot(B); label("$B$", B, NW);
dot(C); label("$C$", C, NE);
dot(D); label("$D$", D, SE);
dot(E); label("$E$", E, NE);
dot(O); label("$O$", O, N);
dot(F); label("$F$", F, S);
draw(rightanglemark(A,B,O,50));
label("$20$", midpoint(B--O), dir(90));
label("$15$", midpoint(A--B), W);
label("$25$", midpoint(A--O), SE);
label("$g$", (C+D)/2, W);
currentpicture.fit();
// Block 2
import olympiad; size(340); // numeric setup for reliable AoPS rendering real R = 10; pair O = (0,0); pair B = (-20,0); pair A = (-20,-15); pair C = (20,0); real g = 5.55; // approximate numeric version of (20β21β75)/3 pair D = (20,-g); pair E = (2.30,-9.74); pair F = intersectionpoint(A--(A+4*(D-A)), B--(B+4*(C-B))); filldraw(circle(O,R), white, heavyblue); draw(A--F, heavygreen); draw(B--F); draw(A--B--O--cycle); draw(O--E); draw((C+2*(C-D))--(C + 3*(D-C)), dashed); dot(A); label("$A$", A, SW); dot(B); label("$B$", B, NW); dot(C); label("$C$", C, NE); dot(D); label("$D$", D, SE); dot(E); label("$E$", E, NE); dot(O); label("$O$", O, N); dot(F); label("$F$", F, S); draw(rightanglemark(A,B,O,50)); label("$20$", midpoint(B--O), dir(90)); label("$15$", midpoint(A--B), W); label("$25$", midpoint(A--O), SE); label("$g$", (C+D)/2, W); currentpicture.fit(); | [] |
961 | A silo (right circular cylinder) with diameter 20 meters stands in a field. MacDonald is located 20 meters west and 15 meters south of the center of the silo. McGregor is located 20 meters east and $g > 0$ meters south of the center of the silo. The line of sight between MacDonald and McGregor is tangent to the silo. The value of g can be written as $\frac{a\sqrt{b}-c}{d}$, where $a,b,c,$ and $d$ are positive integers, $b$ is not divisible by the square of any prime, and $d$ is relatively prime to the greatest common divisor of $a$ and $c$. What is $a+b+c+d$?
$\textbf{(A) } 119 \qquad\textbf{(B) } 120 \qquad\textbf{(C) } 121 \qquad\textbf{(D) } 122 \qquad\textbf{(E) } 123$ | 2025 AMC 10A Problem 20 | $F$ is the midpoint of $AD$, $DG \perp AB$. Notice that $\triangle OEF \sim \triangle DGA$, we have
\[\frac{OE}{OF} = \frac{DG}{DA}\]
Given the length of $CD$ is $g$, $OF$ is the midline, $CD = BG$, therefore $GA = 15 - g$, $DA = \sqrt{40^2 + (15 - g)^2}$, $OF = \dfrac{15 + g}{2}$
\[\frac{OE}{OF} = \frac{DG}{GA}\]
\[\cfrac{10}{\cfrac{15 + g}{2}} = \frac{40}{\sqrt{40^2 + (15 - g)^2}}\]
\[2(15 + g) = \sqrt{40^2 + (15 - g)^2}\]
\[3g^2 + 150g - 925 = 0\]
\[g = \frac{-150 + \sqrt{22500 + 11100}}{6}\]
\[g = \frac{20\sqrt{21}-75}{3}\]
The answer is $20+21+75+3=\boxed{\textbf{(A)}~119}$
~reda_mandymath | // Block 1
import geometry;
unitsize(7);
pair _O = (0,0);
pair _B = (-20,0);
pair _A = (-20,-15);
pair _C = (20,0);
pair _D = (20, -(20sqrt(21)-75)/3);
pair _E = extension(_A, _D, _O, _O + rotate(90)*(_A - _D));
pair _F = (_A + _D)/2;
pair _G = (-20, _D.y);
draw(arc(_O, 10, 0, -180));
draw(_A -- _D -- _C -- _B -- _A ^^ _E -- _O -- (_A + _D)/2);
draw(_D -- _G, dashed);
dot(_A ^^ _B ^^ _C ^^ _D ^^ _E ^^ _F ^^ _G ^^ _O);
markrightangle(_A, _B, _O, 0.5*markangleradius());
label("$A$", _A, SW);
label("$B$", _B, NW);
label("$C$", _C, NE);
label("$D$", _D, SE);
label("$E$", _E, S);
label("$F$", _F, S);
label("$G$", _G, SE);
label("$O$", _O, N);
label("$20$", midpoint(_B -- _O), N);
label("$20$", midpoint(_C -- _O), N);
label("$15$", midpoint(_A -- _B), W);
label("$10$", (_E + 2_O)/3, SE);
label("$g$", midpoint(_C -- _D), E);
// Block 2
import geometry; unitsize(7); pair _O = (0,0); pair _B = (-20,0); pair _A = (-20,-15); pair _C = (20,0); pair _D = (20, -(20sqrt(21)-75)/3); pair _E = extension(_A, _D, _O, _O + rotate(90)*(_A - _D)); pair _F = (_A + _D)/2; pair _G = (-20, _D.y); draw(arc(_O, 10, 0, -180)); draw(_A -- _D -- _C -- _B -- _A ^^ _E -- _O -- (_A + _D)/2); draw(_D -- _G, dashed); dot(_A ^^ _B ^^ _C ^^ _D ^^ _E ^^ _F ^^ _G ^^ _O); markrightangle(_A, _B, _O, 0.5*markangleradius()); label("$A$", _A, SW); label("$B$", _B, NW); label("$C$", _C, NE); label("$D$", _D, SE); label("$E$", _E, S); label("$F$", _F, S); label("$G$", _G, SE); label("$O$", _O, N); label("$20$", midpoint(_B -- _O), N); label("$20$", midpoint(_C -- _O), N); label("$15$", midpoint(_A -- _B), W); label("$10$", (_E + 2_O)/3, SE); label("$g$", midpoint(_C -- _D), E); | [] |
962 | Triangle $\triangle ABC$ has side lengths $AB = 80$, $BC = 45$, and $AC = 75$. The bisector of $\angle B$ and the altitude to side $\overline{AB}$ intersect at point $P$. What is $BP$?
$\textbf{(A)}~18\qquad\textbf{(B)}~19\qquad\textbf{(C)}~20\qquad\textbf{(D)}~21\qquad\textbf{(E)}~22$ | 2025 AMC 10A Problem 23 | Preface: When we have a problem as such, involving a simple diagram with minimal instructions, I use a method I named "rulerbash". Rulerbash should only be used in specific cases and as a last resort, mainly in the event of a time crunch or a difficult problem.
Start with the longest side, drawing a line with a length of $8\,\text{cm}$ ($AB$). Then, using a compass, draw 2 circles centered around points $A$ and $B$, $7.5\,\text{cm}$ and $4.5\,\text{cm}$ radiuses respectfully. At the point of intersection of these 2 circles, we have point $C$, completing a perfectly scaled drawing of $\triangle ABC$. (Note the circles are not necessary with a bit of trial and error with the side lengths, they simply offer a way to get it done first try).
After this, dropping the altitude to $AB$ is simple with a ruler and careful placement, and angle bisector can be estimated quite accurately. (Note: Alternately, a compass can be used to accurately place the altitude and the bisector, which are both trivial constructions ~ZingiberiMaracandae)
After all of this, we can reuse our ruler and measure $BP = 2.1\,\text{cm}$, and using our scale of $80=8\,\text{cm}$, our final answer is $\boxed{\text{(D) }21}.$
Other note: you kind of have to draw the angle bisector really accurately for this to work (I tried this)~ Sakura_kitty
~shreyan.chethan (notes by curryswish) | // Block 1
unitsize(0.25cm); pair A=(0,0), B=(8,0); path cA=circle(A,7.5), cB=circle(B,4.5); pair C=intersectionpoint(cA,cB); draw(A--B); draw(cA); draw(cB); draw(A--C--B--cycle); dot(A); dot(B); dot(C); label("A",A,S, fontsize(8)); label("B",B,S, fontsize(8)); label("C",C,N, fontsize(8)); label("8 cm",(A+B)/2, S, fontsize(6)); label("7.5 cm",(A+C)/2 + (-0.85,0.8),fontsize(6)); label("4.5 cm",(B+C)/2 + (1.75,-0.1),fontsize(6));
// Block 2
unitsize(0.55cm); import olympiad; pair A=(0,0), B=(8,0); pair C=intersectionpoint(circle(A,7.5),circle(B,4.5)); draw(A--B--C--cycle); pair F=foot(C,A,B); draw(C--F); draw(rightanglemark(A,F,C,2)); pair U=unit(unit(A-B)+unit(C-B)); pair P=intersectionpoint(B--(B+100*U), C--F); draw(B--P,dashed); dot(A);dot(B);dot(C);dot(P); label("A",A,S,fontsize(12)); label("B",B,S,fontsize(12)); label("C",C,N,fontsize(12)); label("P",P,NE,fontsize(12)); | [] |
962 | Triangle $\triangle ABC$ has side lengths $AB = 80$, $BC = 45$, and $AC = 75$. The bisector of $\angle B$ and the altitude to side $\overline{AB}$ intersect at point $P$. What is $BP$?
$\textbf{(A)}~18\qquad\textbf{(B)}~19\qquad\textbf{(C)}~20\qquad\textbf{(D)}~21\qquad\textbf{(E)}~22$ | 2025 AMC 10A Problem 23 | Due to Angle Bisector Theorem
\[\frac{CE}{AE} = \frac{BC}{BA} = \frac{45}{80} = \frac{9}{16}\]
\[CE = AC \times \frac{9}{25} = 27\]
\[AE = AC \times \frac{16}{25} = 48\]
Notice that $\dfrac{CE}{CB} = \dfrac{CB}{CA} = \dfrac{3}{5}$, $\triangle CBE \sim \triangle CAB$, $\angle A = \angle CBE = \angle EBF$, so $\triangle ABE$ is isosceles triangle, hence
\[AF = BF\]
Notice that $AF:FD = AE:EC = 16:9$, $DB = FB - FD$, then
\[AF:DB = 16:7\]
Notice that $\triangle PBD \sim \triangle EAF$, $PB = AE \times \dfrac{BD}{AF} = 48 \times \dfrac{7}{16} = \boxed{\text{(D) }21}.$
~θ_mandymath | /* Figure drawn by reda*/ import geometry; unitsize(2.5); pair _A = (0, 0); pair _B = (80, 0); pair _C = (75*(75^2+80^2-45^2)/(2*75*80), 75*sqrt(1-((75^2+80^2-45^2)/(2*75*80))^2)); pair _D = (_C.x, 0); pair _E = (_A * 45 + _C * 80)/(45 + 80); pair _F = (_E.x, 0); pair _P = extension(_C, _D, _B, _E); draw(_B -- _A -- _C -- _B -- _E ^^ _C -- _D); draw(_E -- _F, dashed); /*dot(_A ^^ _B ^^ _C ^^ _D ^^ _E ^^ _F ^^ _P);*/ markrightangle(_B, _D, _P, 0.2*markangleradius()); markrightangle(_B, _F, _E, 0.2*markangleradius()); markangle(_P, _B, _D, radius = 0.25*markangleradius()); markangle(_P, _B, _D, radius = 0.3*markangleradius()); markangle(_C, _B, _P, radius = 0.25*markangleradius()); markangle(_C, _B, _P, radius = 0.3*markangleradius()); markangle(_B, _A, _C, radius = 0.25*markangleradius()); markangle(_B, _A, _C, radius = 0.3*markangleradius()); label("$A$", _A, S); label("$B$", _B, S); label("$C$", _C, N); label("$D$", _D, S); label("$E$", _E, NW); label("$F$", _F, S); label("$P$", _P, NE); label("$27$", (_C + _E)/2, N); label("$48$", (_A + _E)/2, N); label("$45$", (_B + _C)/2, E); label("$80$", (_A + _B)/2, 4S); | [] |
963 | A point $P$ is chosen at random inside square $ABCD$. The probability that $\overline{AP}$ is neither the shortest nor the longest side of $\triangle APB$ can be written as $\frac{a + b \pi - c \sqrt{d}}{e}$, where $a, b, c, d,$ and $e$ are positive integers, $\text{gcd}(a, b, c, e) = 1$, and $d$ is not divisible by the square of a prime. What is $a+b+c+d+e$?
$\textbf{(A) }25 \qquad \textbf{(B) }26 \qquad \textbf{(C) }27 \qquad \textbf{(D) }28 \qquad \textbf{(E) } 29 \qquad$ | 2025 AMC 10A Problem 25 | Say WLOG that $AB$ is the top side of the square, and the square is of side length 1. Let us say that the midpoint of $AB$ is $M$, while the midpoint of $CD$ is $Q$. Drawing a vertical line to split the square in half, we notice that if $P$ is to the left of the line, $AP < BP$, and if P is to the right of the line, $AP > BP$. Also, drawing a quarter circle of radius 1 from point $A$, we can split the area into points P for which $AP < AB$ and $AP > AB$. Because of our constraints, there are 2 cases:
Case 1: $AB > AP > BP$
In this case, $P$ will be to the right of the vertical line and inside of the quarter circle. Let us say that the intersection of the vertical line and quarter circle is $N$. The distance from $N$ to $AD$ is 1/2, and we can say that $\angle BAN$ is $60^\circ$. Sector $BAN$ of circle $A$ would therefore have an area of $\frac{\pi}{6}$. Because $\triangle AMN$ is a 30-60-90 triangle, the area of $AMN$ is $\frac{\sqrt{3}}{8}$. The probability of case 1 happening should then be $\frac{\pi}{6}-\frac{\sqrt{3}}{8}$.
Case 2: $AB < AP < BP$
In this case, $P$ will be to the left of the vertical line and outside of the quarter circle. Knowing that the quarter circle's area is $\frac{\pi}{4}$, we can subtract the probability of Case 1 happening to get the chance that $P$ is on the left of the vertical line and in circle $A$. Doing this would give $\frac{\pi}{12}+\frac{\sqrt{3}}{8}$. To get the probability of Case 2 happening, we can subtract this from the area of rectangle $AMQD$. This would give us $\frac{1}{2}-\frac{\pi}{12}-\frac{\sqrt{3}}{8}$.
Adding both cases, we get the total probability as $\frac{1}{2}+\frac{\pi}{12}-\frac{\sqrt{3}}{4} = \frac{6+\pi-3\sqrt{3}}{12}$. Formatting this gives us $6+1+3+3+12 = \boxed{\textbf{(A) } 25}$.
~AVS2010
~Sakura_kitty (very minor capitalization error. {Literally just one letter}) | // Block 1
size(200);
pair A = (0,1), B = (1,1), C = (1,0), D = (0,0);
pair M = (0.5,1), Q = (0.5,0);
draw((0.5,0)--(0.5,1));
draw(A--B--C--D--cycle);
draw(arc(A,1,270,360));
pair N = intersectionpoint(arc(A,1,270,360), (0.5,0)--(0.5,1));
dot(N); label("$N$", N, NE);
dot(M); label("$M$", M, N);
dot(Q); label("$Q$", Q, S);
label("$A$", A, NW);
label("$B$", B, NE);
label("$C$", C, SE);
label("$D$", D, SW);
// Block 2
size(200); pair A = (0,1), B = (1,1), C = (1,0), D = (0,0); pair M = (0.5,1), Q = (0.5,0); draw((0.5,0)--(0.5,1)); draw(A--B--C--D--cycle); draw(arc(A,1,270,360)); pair N = intersectionpoint(arc(A,1,270,360), (0.5,0)--(0.5,1)); dot(N); label("$N$", N, NE); dot(M); label("$M$", M, N); dot(Q); label("$Q$", Q, S); label("$A$", A, NW); label("$B$", B, NE); label("$C$", C, SE); label("$D$", D, SW); | [] |
963 | A point $P$ is chosen at random inside square $ABCD$. The probability that $\overline{AP}$ is neither the shortest nor the longest side of $\triangle APB$ can be written as $\frac{a + b \pi - c \sqrt{d}}{e}$, where $a, b, c, d,$ and $e$ are positive integers, $\text{gcd}(a, b, c, e) = 1$, and $d$ is not divisible by the square of a prime. What is $a+b+c+d+e$?
$\textbf{(A) }25 \qquad \textbf{(B) }26 \qquad \textbf{(C) }27 \qquad \textbf{(D) }28 \qquad \textbf{(E) } 29 \qquad$ | 2025 AMC 10A Problem 25 | Assume the sides of this square is 1, hence we only need to find the area of the desired regions. From Solution 1, it is easy to see that the regions are the bottom left region c and the top right region b, hence we must compute $b+c$. Also, define $k$ to be the circular segment. We have two equations right off the bat:
$2a+2b+2c=1 \implies a+b+c=\frac{1}{2}$ since the sum of all regions is just the area of the square and also, $2b+a=\frac{\pi}{4}$, just the area of a quarter-circle.
Next, $\triangle ABN$ has a area of $\frac{\sqrt{3}}{4}$ since it is just an equilateral triangle with length 1 (each side is a radius of a circle with radius of 1). From the diagram, $2k+[ABN]=2k+\frac{\sqrt3}{4}=2b$. Subsequently, we see that sector $ADN$ has an angle of $90-60=30$ and is the sum of $a+k$. Therefore, $a+k=\frac{\pi}{12}$.
Multiply this equation by 2: $2a+2k=\frac{\pi}{6}$ and combining it with $2b-2k=\frac{\sqrt{3}}{4}$ yields $2a+2b=\frac{\pi}{6}+\frac{\sqrt{3}}{4}$. Since we also have that $2b+a=\frac{\pi}{4}$, subtracting this from the equation yields $a=(\frac{\pi}{6}+\frac{\sqrt{3}}{4}) - (\frac{\pi}{4}) = \frac{-\pi}{12}+\frac{\sqrt{3}}{4}$. We are to find $b+c=\frac{1}{2}-a=\frac{1}{2} - (\frac{\sqrt{3}}{4} - \frac{\pi}{12}) = \frac{1}{2} - \frac{\sqrt{3}}{4} + \frac{\pi}{12} = \frac{6+\pi-3\sqrt{3}}{12}$.
At last, $a+b+c+d+e=6+1+3+3+12=\boxed{\textbf{(A) } 25}$.
~hxve | // Block 1
size(200);
import olympiad;
pair A = (0,1), B = (1,1), C = (1,0), D = (0,0);
pair M = (0.5,1), Q = (0.5,0);
draw((0.5,0)--(0.5,1));
draw(A--B--C--D--cycle);
draw(arc(A,1,270,360));
pair N = intersectionpoint(arc(A,1,270,360), (0.5,0)--(0.5,1));
dot(N); label("$N$", N, NE);
draw(arc(B,1,180,270));
dot(M); label("$M$", M, N);
dot(Q); label("$Q$", Q, S);
draw(A--N--cycle);
label("$A$", A, NW);
label("$B$", B, NE);
label("$C$", C, SE);
label("$D$", D, SW);
label("$a$", (0.15,0.2), W);
label("$a$", (0.85,0.2), N);
label("$b$", (0.65,0.6), E);
label("$c$", (0.35,0.01), M);
label("$c$", (0.65,0.03), Q);
label("$k$", (0.2,0.45), B);
// Block 2
size(200); import olympiad; pair A = (0,1), B = (1,1), C = (1,0), D = (0,0); pair M = (0.5,1), Q = (0.5,0); draw((0.5,0)--(0.5,1)); draw(A--B--C--D--cycle); draw(arc(A,1,270,360)); pair N = intersectionpoint(arc(A,1,270,360), (0.5,0)--(0.5,1)); dot(N); label("$N$", N, NE); draw(arc(B,1,180,270)); dot(M); label("$M$", M, N); dot(Q); label("$Q$", Q, S); draw(A--N--cycle); label("$A$", A, NW); label("$B$", B, NE); label("$C$", C, SE); label("$D$", D, SW); label("$a$", (0.15,0.2), W); label("$a$", (0.85,0.2), N); label("$b$", (0.65,0.6), E); label("$c$", (0.35,0.01), M); label("$c$", (0.65,0.03), Q); label("$k$", (0.2,0.45), B); | [] |
964 | Pentagon $ABCDE$ is inscribed in a circle, and $\angle BEC = \angle CED = 30^\circ$. Let line $AC$ and line $BD$ intersect at point $F$, and suppose that $AB = 9$ and $AD = 24$. What is $BF$?
$\textbf{(A) } \frac{57}{11} \qquad\textbf{(B) } \frac{59}{11} \qquad\textbf{(C) } \frac{60}{11} \qquad\textbf{(D) } \frac{61}{11} \qquad\textbf{(E) } \frac{63}{11}$ | 2025 AMC 12A Problem 8 | We will scale down the diagram by a factor of $3$ so that $AB = 3$ and $AD = 8.$ Since $\angle BEC = 30^{\circ},$ it follows that $\angle BAC = \angle BDC = 30^{\circ}$ as they all subtend the same arc. Similarly, since $\angle CED = 30^{\circ},$ it follows that $\angle CAD = \angle CBD = 30^{\circ}$ as well.
We obtain the following diagram:
Note that $\triangle ABD$ has $\angle BAD = 60^{\circ}.$ Applying Law of Cosines, we get
\begin{align*} BD^2 &= AB^2+AD^2-2AB\cdot AD \cdot\cos{60^{\circ}} \\ &= 9 + 64 - 2 \cdot 3 \cdot 8 \cdot \frac{1}{2} \\ &= 49, \end{align*}
from which $BD = 7.$
From here, we wish to find $BF.$ As $AF$ is the angle bisector of $\angle BAD,$ we apply the Angle Bisector Theorem:
\begin{align*} \frac{AB}{BF} &= \frac{AD}{DF} \\ \frac{3}{BF} &= \frac{8}{7-BF}. \end{align*}
Solving for $BF,$ we get $BF = \frac{21}{11}.$ Remember to scale the figure back up by a factor of $3,$ so our answer is $\frac{21}{11}\cdot 3 = \boxed{\textbf{(E) } \frac{63}{11}}.$
~lprado ~MRENTHUSIASM | // Block 1
/* Made by MRENTHUSIASM */
size(200);
real r = 7*sqrt(3);
pair O, A, B, C, D, E, F;
O = origin;
B = r*dir(30);
C = r*dir(-30);
D = r*dir(-90);
E = r*dir(180);
A = intersectionpoints(Circle(O,r),Circle(B,9))[0];
F = intersectionpoint(A--C,B--D);
draw(Circle(O,r)^^B--C--D--E--A^^B--E--C--F);
draw(A--D--B--cycle^^A--F,red);
dot("$B$",B,1.5*B/r,linewidth(4));
dot("$C$",C,1.5*C/r,linewidth(4));
dot("$D$",D,1.5*D/r,linewidth(4));
dot("$E$",E,1.5*E/r,linewidth(4));
dot("$A$",A,1.5*A/r,linewidth(4));
dot("$F$",F,1.5*F/r,linewidth(4));
label("$30^{\circ}$",E,6*(1,0),fontsize(8));
label("$30^{\circ}$",E,7*dir(-32),fontsize(8));
label("$30^{\circ}$",A,9*dir(-56),red+fontsize(8));
label("$30^{\circ}$",A,9*dir(-84),red+fontsize(8));
label("$3$",1.1*midpoint(A--B),red);
label("$8$",0.4*midpoint(A--D),red);
// Block 2
/* Made by MRENTHUSIASM */ size(200); real r = 7*sqrt(3); pair O, A, B, C, D, E, F; O = origin; B = r*dir(30); C = r*dir(-30); D = r*dir(-90); E = r*dir(180); A = intersectionpoints(Circle(O,r),Circle(B,9))[0]; F = intersectionpoint(A--C,B--D); draw(Circle(O,r)^^B--C--D--E--A^^B--E--C--F); draw(A--D--B--cycle^^A--F,red); dot("$B$",B,1.5*B/r,linewidth(4)); dot("$C$",C,1.5*C/r,linewidth(4)); dot("$D$",D,1.5*D/r,linewidth(4)); dot("$E$",E,1.5*E/r,linewidth(4)); dot("$A$",A,1.5*A/r,linewidth(4)); dot("$F$",F,1.5*F/r,linewidth(4)); label("$30^{\circ}$",E,6*(1,0),fontsize(8)); label("$30^{\circ}$",E,7*dir(-32),fontsize(8)); label("$30^{\circ}$",A,9*dir(-56),red+fontsize(8)); label("$30^{\circ}$",A,9*dir(-84),red+fontsize(8)); label("$3$",1.1*midpoint(A--B),red); label("$8$",0.4*midpoint(A--D),red); | [] |
965 | Let $w$ be the complex number $2+i$, where $i=\sqrt{-1}$. What real number $r$ has the property that $r$, $w$, and $w^2$ are three collinear points in the complex plane?
$\textbf{(A)}~\frac34\qquad\textbf{(B)}~1\qquad\textbf{(C)}~\frac75\qquad\textbf{(D)}~\frac32\qquad\textbf{(E)}~\frac53$ | 2025 AMC 12A Problem 9 | Recall that the slope of a line is $m=\tan\phi,$ where $\phi$ is the angle formed by the line and the positive $x$-axis.
Note that $|w|=\sqrt{2^2+1^2}=\sqrt{5}.$ In polar coordinates, let $w=\sqrt{5}\operatorname{cis}\theta.$ It follows that $\tan\theta=\frac12.$
By De Moivre's Theorem, we have $w^2=5\operatorname{cis}(2\theta),$ from which \[\tan(2\theta)=\frac{2\tan\theta}{1-\tan^2\theta}=\frac{4}{3}.\]
We obtain the following diagram:
Since $\left|w^2\right|=5,$ we have $w^2=3+4i$ by a $3$-$4$-$5$ triangle. The complex numbers $w$ and $w^2$ correspond to the points $(2,1)$ and $(3,4),$ respectively, and $r$ corresponds to the $x$-intercept of this line. In slope-intercept form, the line containing these two points is $y=3x-5.$ Therefore, the $x$-intercept is $r=\boxed{\textbf{(E)}~\frac53}.$
~MRENTHUSIASM | // Block 1
/* Made by MRENTHUSIASM */
size(200);
int xMin = -2;
int xMax = 5;
int yMin = -2;
int yMax = 5;
//Draws the horizontal gridlines
void horizontalLines()
{
for (int i = yMin+1; i < yMax; ++i)
{
draw((xMin,i)--(xMax,i), mediumgray+linewidth(0.4));
}
}
//Draws the vertical gridlines
void verticalLines()
{
for (int i = xMin+1; i < xMax; ++i)
{
draw((i,yMin)--(i,yMax), mediumgray+linewidth(0.4));
}
}
horizontalLines();
verticalLines();
draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5));
draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5));
label("Re",(xMax,0),(2,0));
label("Im",(0,yMax),(0,2));
pair W1, W2, R;
W1 = (2,1);
W2 = (3,4);
R = (5/3,0);
label("$w=2+i$", W1, 2*dir(W1));
label("$w^2$", W2, 2*dir(W2));
label("$r$", R, 2*dir(R-W2));
label("$\theta$", origin, 6*dir(14));
label("$\theta$", origin, 6*dir(40));
draw(W1--origin--W2);
draw(W2--R,dashed);
dot(W1^^W2^^R, linewidth(4));
// Block 2
/* Made by MRENTHUSIASM */ size(200); int xMin = -2; int xMax = 5; int yMin = -2; int yMax = 5; //Draws the horizontal gridlines void horizontalLines() { for (int i = yMin+1; i < yMax; ++i) { draw((xMin,i)--(xMax,i), mediumgray+linewidth(0.4)); } } //Draws the vertical gridlines void verticalLines() { for (int i = xMin+1; i < xMax; ++i) { draw((i,yMin)--(i,yMax), mediumgray+linewidth(0.4)); } } horizontalLines(); verticalLines(); draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label("Re",(xMax,0),(2,0)); label("Im",(0,yMax),(0,2)); pair W1, W2, R; W1 = (2,1); W2 = (3,4); R = (5/3,0); label("$w=2+i$", W1, 2*dir(W1)); label("$w^2$", W2, 2*dir(W2)); label("$r$", R, 2*dir(R-W2)); label("$\theta$", origin, 6*dir(14)); label("$\theta$", origin, 6*dir(40)); draw(W1--origin--W2); draw(W2--R,dashed); dot(W1^^W2^^R, linewidth(4)); | [] |
966 | The orthocenter of a triangle is the concurrent intersection of the three (possibly extended) altitudes. What is the sum of the coordinates of the orthocenter of the triangle whose vertices are $A(2,31), B(8,27),$ and $C(18,27)$?
$\textbf{(A)}~5\qquad\textbf{(B)}~17\qquad\textbf{(C)}~10+4\sqrt{17} +2\sqrt{13}\qquad\textbf{(D)}~\frac{113}{3}\qquad\textbf{(E)}~54$ | 2025 AMC 12A Problem 11 | Let the intersection of the heights and their corresponding sides be $H_1$ and $H_2$, respectively, and let the intersection of line $AH_1$ and $H_2B$ be $E$. Since $\angle EH_2A = \angle EH_1B = 90^{\circ}$, $\angle EBH_1 = 90^{\circ}- \angle BEH_1 = \angle EAH_2$, we have
\[\tan{\angle EBH_1} = \tan{\angle EAH_2} = \tan{\angle H_1AC} = \frac{CH_1}{AH_1} = 4\]
Consequently,
\[\frac{EH_1}{B_H1}=\frac{EH_1}{6}=\tan{\angle EBH_1}=4\]
so $EH_1=24$, and the coordinates of $E$ is exactly $(2, 31-4-24)=(2,3)$. Hence, the answer is $2+3=\boxed{\text{(A) }5}$.
~Bloggish | // Block 1
size(200);
pair A = (2,31);
pair B = (8, 27);
pair C = (18, 27);
pair H1 = (2, 27);
pair H2 = (8.5882352941176, 29.3529411764706);
pair D = (2,3);
label("$A$", A, NW);
label("$B$", (8.5, 27), S);
label("$C$", C, SE);
label("$H_1$", H1, SW);
label("$H_2$", H2, NE);
label("$E$", E, S);
draw(A--B--C--cycle);
draw(A--E, dashed);
draw(H1--B, dashed);
draw(H2--E, dashed);
// Block 2
size(200); pair A = (2,31); pair B = (8, 27); pair C = (18, 27); pair H1 = (2, 27); pair H2 = (8.5882352941176, 29.3529411764706); pair D = (2,3); label("$A$", A, NW); label("$B$", (8.5, 27), S); label("$C$", C, SE); label("$H_1$", H1, SW); label("$H_2$", H2, NE); label("$E$", E, S); draw(A--B--C--cycle); draw(A--E, dashed); draw(H1--B, dashed); draw(H2--E, dashed); | [] |
967 | Triangle $\triangle ABC$ has side lengths $AB = 80$, $BC = 45$, and $AC = 75$. The bisector of $\angle B$ and the altitude to side $\overline{AB}$ intersect at point $P$. What is $BP$?
$\textbf{(A)}~18\qquad\textbf{(B)}~19\qquad\textbf{(C)}~20\qquad\textbf{(D)}~21\qquad\textbf{(E)}~22$ | 2025 AMC 12A Problem 16 | Preface: When we have a problem as such, involving a simple diagram with minimal instructions, I use a method I named "rulerbash". Rulerbash should only be used in specific cases and as a last resort, mainly in the event of a time crunch or a difficult problem.
Start with the longest side, drawing a line with a length of $8\,\text{cm}$ ($AB$). Then, using a compass, draw 2 circles centered around points $A$ and $B$, $7.5\,\text{cm}$ and $4.5\,\text{cm}$ radiuses respectfully. At the point of intersection of these 2 circles, we have point $C$, completing a perfectly scaled drawing of $\triangle ABC$. (Note the circles are not necessary with a bit of trial and error with the side lengths, they simply offer a way to get it done first try).
After this, dropping the altitude to $AB$ is simple with a ruler and careful placement, and angle bisector can be estimated quite accurately. (Note: Alternately, a compass can be used to accurately place the altitude and the bisector, which are both trivial constructions ~ZingiberiMaracandae)
After all of this, we can reuse our ruler and measure $BP = 2.1\,\text{cm}$, and using our scale of $80=8\,\text{cm}$, our final answer is $\boxed{\text{(D) }21}.$
Other note: you kind of have to draw the angle bisector really accurately for this to work (I tried this)~ Sakura_kitty
~shreyan.chethan (notes by curryswish) | // Block 1
unitsize(0.25cm);
pair A=(0,0), B=(8,0);
path cA=circle(A,7.5), cB=circle(B,4.5);
pair C=intersectionpoint(cA,cB);
draw(A--B);
draw(cA);
draw(cB);
draw(A--C--B--cycle);
dot(A); dot(B); dot(C);
label("A",A,S, fontsize(8));
label("B",B,S, fontsize(8));
label("C",C,N, fontsize(8));
label("8 cm",(A+B)/2, S, fontsize(6));
label("7.5 cm",(A+C)/2 + (-0.85,0.8),fontsize(6));
label("4.5 cm",(B+C)/2 + (1.75,-0.1),fontsize(6));
// Block 2
unitsize(0.55cm);
import olympiad;
pair A=(0,0), B=(8,0);
pair C=intersectionpoint(circle(A,7.5),circle(B,4.5));
draw(A--B--C--cycle);
pair F=foot(C,A,B);
draw(C--F);
draw(rightanglemark(A,F,C,2));
pair U=unit(unit(A-B)+unit(C-B));
pair P=intersectionpoint(B--(B+100*U), C--F);
draw(B--P,dashed);
dot(A);dot(B);dot(C);dot(P);
label("A",A,S,fontsize(12));
label("B",B,S,fontsize(12));
label("C",C,N,fontsize(12));
label("P",P,NE,fontsize(12));
// Block 3
unitsize(0.25cm); pair A=(0,0), B=(8,0); path cA=circle(A,7.5), cB=circle(B,4.5); pair C=intersectionpoint(cA,cB); draw(A--B); draw(cA); draw(cB); draw(A--C--B--cycle); dot(A); dot(B); dot(C); label("A",A,S, fontsize(8)); label("B",B,S, fontsize(8)); label("C",C,N, fontsize(8)); label("8 cm",(A+B)/2, S, fontsize(6)); label("7.5 cm",(A+C)/2 + (-0.85,0.8),fontsize(6)); label("4.5 cm",(B+C)/2 + (1.75,-0.1),fontsize(6));
// Block 4
unitsize(0.55cm); import olympiad; pair A=(0,0), B=(8,0); pair C=intersectionpoint(circle(A,7.5),circle(B,4.5)); draw(A--B--C--cycle); pair F=foot(C,A,B); draw(C--F); draw(rightanglemark(A,F,C,2)); pair U=unit(unit(A-B)+unit(C-B)); pair P=intersectionpoint(B--(B+100*U), C--F); draw(B--P,dashed); dot(A);dot(B);dot(C);dot(P); label("A",A,S,fontsize(12)); label("B",B,S,fontsize(12)); label("C",C,N,fontsize(12)); label("P",P,NE,fontsize(12)); | [] |
967 | Triangle $\triangle ABC$ has side lengths $AB = 80$, $BC = 45$, and $AC = 75$. The bisector of $\angle B$ and the altitude to side $\overline{AB}$ intersect at point $P$. What is $BP$?
$\textbf{(A)}~18\qquad\textbf{(B)}~19\qquad\textbf{(C)}~20\qquad\textbf{(D)}~21\qquad\textbf{(E)}~22$ | 2025 AMC 12A Problem 16 | Due to Angle Bisector Theorem
\[\frac{CE}{AE} = \frac{BC}{BA} = \frac{45}{80} = \frac{9}{16}\]
\[CE = AC \times \frac{9}{25} = 27\]
\[AE = AC \times \frac{16}{25} = 48\]
Notice that $\dfrac{CE}{CB} = \dfrac{CB}{CA} = \dfrac{3}{5}$, $\triangle CBE \sim \triangle CAB$, $\angle A = \angle CBE = \angle EBF$, so $\triangle ABE$ is isosceles triangle, hence
\[AF = BF\]
Notice that $AF:FD = AE:EC = 16:9$, $DB = FB - FD$, then
\[AF:DB = 16:7\]
Notice that $\triangle PBD \sim \triangle EAF$, $PB = AE \times \dfrac{BD}{AF} = 48 \times \dfrac{7}{16} = \boxed{\text{(D) }21}.$
~θ_mandymath | // Block 1
/* Figure drawn by reda*/
import geometry;
unitsize(2.5);
pair _A = (0, 0);
pair _B = (80, 0);
pair _C = (75*(75^2+80^2-45^2)/(2*75*80), 75*sqrt(1-((75^2+80^2-45^2)/(2*75*80))^2));
pair _D = (_C.x, 0);
pair _E = (_A * 45 + _C * 80)/(45 + 80);
pair _F = (_E.x, 0);
pair _P = extension(_C, _D, _B, _E);
draw(_B -- _A -- _C -- _B -- _E ^^ _C -- _D);
draw(_E -- _F, dashed);
/*dot(_A ^^ _B ^^ _C ^^ _D ^^ _E ^^ _F ^^ _P);*/
markrightangle(_B, _D, _P, 0.2*markangleradius());
markrightangle(_B, _F, _E, 0.2*markangleradius());
markangle(_P, _B, _D, radius = 0.25*markangleradius());
markangle(_P, _B, _D, radius = 0.3*markangleradius());
markangle(_C, _B, _P, radius = 0.25*markangleradius());
markangle(_C, _B, _P, radius = 0.3*markangleradius());
markangle(_B, _A, _C, radius = 0.25*markangleradius());
markangle(_B, _A, _C, radius = 0.3*markangleradius());
label("$A$", _A, S);
label("$B$", _B, S);
label("$C$", _C, N);
label("$D$", _D, S);
label("$E$", _E, NW);
label("$F$", _F, S);
label("$P$", _P, NE);
label("$27$", (_C + _E)/2, N);
label("$48$", (_A + _E)/2, N);
label("$45$", (_B + _C)/2, E);
label("$80$", (_A + _B)/2, 4S);
// Block 2
/* Figure drawn by reda*/ import geometry; unitsize(2.5); pair _A = (0, 0); pair _B = (80, 0); pair _C = (75*(75^2+80^2-45^2)/(2*75*80), 75*sqrt(1-((75^2+80^2-45^2)/(2*75*80))^2)); pair _D = (_C.x, 0); pair _E = (_A * 45 + _C * 80)/(45 + 80); pair _F = (_E.x, 0); pair _P = extension(_C, _D, _B, _E); draw(_B -- _A -- _C -- _B -- _E ^^ _C -- _D); draw(_E -- _F, dashed); /*dot(_A ^^ _B ^^ _C ^^ _D ^^ _E ^^ _F ^^ _P);*/ markrightangle(_B, _D, _P, 0.2*markangleradius()); markrightangle(_B, _F, _E, 0.2*markangleradius()); markangle(_P, _B, _D, radius = 0.25*markangleradius()); markangle(_P, _B, _D, radius = 0.3*markangleradius()); markangle(_C, _B, _P, radius = 0.25*markangleradius()); markangle(_C, _B, _P, radius = 0.3*markangleradius()); markangle(_B, _A, _C, radius = 0.25*markangleradius()); markangle(_B, _A, _C, radius = 0.3*markangleradius()); label("$A$", _A, S); label("$B$", _B, S); label("$C$", _C, N); label("$D$", _D, S); label("$E$", _E, NW); label("$F$", _F, S); label("$P$", _P, NE); label("$27$", (_C + _E)/2, N); label("$48$", (_A + _E)/2, N); label("$45$", (_B + _C)/2, E); label("$80$", (_A + _B)/2, 4S); | [] |
968 | A circle of radius $r$ is surrounded by $12$ circles of radius $1,$ externally tangent to the central circle and sequentially tangent to each other, as shown. Then $r$ can be written as $\sqrt a + \sqrt b + c,$ where $a, b, c$ are integers. What is $a+b+c?$
$\textbf{(A) } 3 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 7 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 11$ | 2025 AMC 12A Problem 24 | We connect the centers of the smaller circles at $0^\circ$ and $90^\circ$ as follows:
Note that connecting the centers of the 12 smaller circles forms a regular 12-gon. A regular 12-gon has angles
\[\frac{10\cdot 180}{12} = 150^\circ,\]
so $\angle OAB = \angle ODC = 75^\circ.$ Furthermore, since $OA=OD$ and $\angle AOD = 90^\circ$, we have $\angle OAD = 45^\circ$, so $\angle DAB = 30^\circ$. Next, we drop the perpendiculars from $B$ and $C$ to $AD$. This creates a $1 \times 2$ rectangle and two $30-60-90$ triangles. Thus,
\[AD = (r+1)\sqrt{2} = 2+2\sqrt{3},\]
and manipulating yields
\[r = -1 + \sqrt{2}+\sqrt{6},\]
so the desired answer is $-1+2+6 = \boxed{\text{(C) } 7.}$
P.S. I posted the asymptote code here. Anyone that wishes to use it can steal~Β :)
~ABC09090927 | // Block 1
import graph;
size(250);
real R = -1 + sqrt(2) + sqrt(6);
real r = 1;
pair O = (0,0);
label("$O$", 0, W);
pair C0 = (R + r) * dir(0);
label("$A$", C0, E);
pair C15 = (R + r) * dir(15);
label("2", C15, NE);
pair C30 = (R + r) * dir(30);
label("$B$", C30, E);
pair C45 = (R + r) * dir(45);
label("2", C45, NE);
pair C60 = (R + r) * dir(60);
label("$C$", C60, NE);
pair C75 = (R + r) * dir(75);
label("2", C75, NE);
pair C90 = (R + r) * dir(90);
label("$D$", C90, N);
pair project(pair P, pair A, pair B) {
pair AB = B - A;
real t = dot(P - A, AB) / dot(AB, AB);
return A + t * AB;
}
pair foot30 = project(C30, C0, C90);
pair foot60 = project(C60, C0, C90);
draw(Circle(O, R), black + linewidth(1bp));
for(int i = 0; i < 12; ++i) {
real theta = i * 360 / 12;
pair center = (R + r) * dir(theta);
draw(Circle(center, r), blue + linewidth(1bp));
}
draw(O -- C0, black + linewidth(1bp));
draw(O -- C90, black + linewidth(1bp));
draw(C0 -- C30, red + linewidth(1bp));
draw(C30 -- C60, red + linewidth(1bp));
draw(C60 -- C90, red + linewidth(1bp));
draw(C0 -- C90, red + linewidth(1bp));
draw(C30 -- foot30, blue + linewidth(1bp));
draw(C60 -- foot60, blue + linewidth(1bp));
// Block 2
import graph; size(250); real R = -1 + sqrt(2) + sqrt(6); real r = 1; pair O = (0,0); label("$O$", 0, W); pair C0 = (R + r) * dir(0); label("$A$", C0, E); pair C15 = (R + r) * dir(15); label("2", C15, NE); pair C30 = (R + r) * dir(30); label("$B$", C30, E); pair C45 = (R + r) * dir(45); label("2", C45, NE); pair C60 = (R + r) * dir(60); label("$C$", C60, NE); pair C75 = (R + r) * dir(75); label("2", C75, NE); pair C90 = (R + r) * dir(90); label("$D$", C90, N); pair project(pair P, pair A, pair B) { pair AB = B - A; real t = dot(P - A, AB) / dot(AB, AB); return A + t * AB; } pair foot30 = project(C30, C0, C90); pair foot60 = project(C60, C0, C90); draw(Circle(O, R), black + linewidth(1bp)); for(int i = 0; i < 12; ++i) { real theta = i * 360 / 12; pair center = (R + r) * dir(theta); draw(Circle(center, r), blue + linewidth(1bp)); } draw(O -- C0, black + linewidth(1bp)); draw(O -- C90, black + linewidth(1bp)); draw(C0 -- C30, red + linewidth(1bp)); draw(C30 -- C60, red + linewidth(1bp)); draw(C60 -- C90, red + linewidth(1bp)); draw(C0 -- C90, red + linewidth(1bp)); draw(C30 -- foot30, blue + linewidth(1bp)); draw(C60 -- foot60, blue + linewidth(1bp)); | [] |
969 | In $\triangle ABC$, $AB = 10$, $AC = 18$, and $\angle B = 130^\circ$. Let $O$ be the center of the circle containing points $A, B, C$. What is the degree measure of $\angle CAO$?
$\textbf{(A)}\ 20^\circ \qquad \textbf{(B)}\ 30^\circ \qquad \textbf{(C)}\ 40^\circ \qquad \textbf{(D)}\ 50^\circ \qquad \textbf{(E)}\ 60^\circ$ | 2025 AMC 10B Problem 5 | Since $\angle ABC = 130^\circ$, by the Inscribed Angle Theorem, it intercepts the major arc $AC$, so that arc has measure $260^\circ$. Point $O$ is the circumcenter, and is therefore outside of triangle ABC. Therefore, the minor arc $AC$ has measure $360^\circ - 260^\circ = 100^\circ$. The central angle $\angle AOC$ subtends the minor arc $AC$, so $\angle AOC = 100^\circ$. Because $OA = OC$ are radii of the circumcircle, $\triangle AOC$ is isosceles, so $\angle CAO = \angle ACO$. Let $\angle CAO = \angle ACO = x$. Then $x + x + 100^\circ = 180^\circ$, so $2x = 80^\circ$ and $x = 40^\circ$. Thus, $\angle CAO = \boxed{40^\circ}$.
~Strickenox
~EZ123 (Minor edits)
Below is a diagram to help visualize the problemΒ : | // Block 1
size(220);
// Choose coordinates so that angle B is obtuse (~130Β°)
pair B = (0,0);
pair C = (3,0);
pair A = 3*dir(130);
// Circumcenter of triangle ABC (outside since angle at B is obtuse)
pair O = circumcenter(A,B,C);
// Draw circumcircle
draw(circle(O, abs(O - A)), gray(0.7));
// Draw triangle ABC
draw(A--B--C--cycle);
// Draw radii OA and OC
draw(O--A, dashed);
draw(O--C, dashed);
// Label points
dot("$A$", A, NE);
dot("$B$", B, S);
dot("$C$", C, SE);
dot("$O$", O, N);
// Block 2
size(220); // Choose coordinates so that angle B is obtuse (~130Β°) pair B = (0,0); pair C = (3,0); pair A = 3*dir(130); // Circumcenter of triangle ABC (outside since angle at B is obtuse) pair O = circumcenter(A,B,C); // Draw circumcircle draw(circle(O, abs(O - A)), gray(0.7)); // Draw triangle ABC draw(A--B--C--cycle); // Draw radii OA and OC draw(O--A, dashed); draw(O--C, dashed); // Label points dot("$A$", A, NE); dot("$B$", B, S); dot("$C$", C, SE); dot("$O$", O, N); | [] |
970 | The figure below shows an equilateral triangle, a rhombus with a $60^\circ$ angle, and a regular hexagon, each of them containing some mutually tangent congruent disks. Let $T, R,$ and $H,$ respectively, denote the ratio in each case of the total area of the disks to the area of the enclosing polygon. Which of the following is true?
$\textbf{(A)}\ T=H=R \qquad \textbf{(B)}\ H<R=T \qquad \textbf{(C)}\ H=R<T\qquad \textbf{(D)}\ H<R<T \qquad \textbf{(E)}\ H<T<R \qquad$ | 2025 AMC 10B Problem 12 | To solve this problem, we can dissect the different shapes into equilateral triangles. We can keep figure $T$ the same, and then divide the rhombus $R$ into two equilateral triangles (as a result of one of the angles equaling $60^\circ$), with one circle in each that is the incircle to its triangle. We can do the same for the hexagon $H$, dividing it into six equilateral triangles with one circle in each, and each circle is tangent to the triangle that encloses it.
We can see that $H = R$, since the proportion of the shaded circles to the unshaded portions in the rhombus and the hexagon is the same. Both are equal to an equilateral triangle with a circle inscribed in it; therefore they have an equal ratio. We can use this to estimate that the ratio of the circle to the triangle is the largest for the first figure, as we can obviously see this through intuition and observation.
Furthermore, there is no answer choice for $T < H = R$, and it is evidently not $H = R = T$. Hence, the answer to the question is as follows:
$\boxed{\textbf{(C)}\ H = R < T}.$
~EZ123
~imbetterthanyou (made some wording clearer)
~Strickenox (Minor edits)
~EZ123 (More minor edits, don't try to one-up me)
~Eggopancakes (Minor useful grammar edits 0_0)
~Wen_Liang (Even more minor edits :D)
~EZ123 (EVEN MORE MINOR EDITS)
~Galactic_Saber (Minor edits, thought I would join)
~NumberNinja1234 (Minor grammar edit, had to do something)
~EZ123 (EVEN MORE MINOR EDITS --- Edited the solution label -- Don't try to one-up me)
~MathyPikachu (Minor edits to the people's edits above me)
~larryliao (I just filled in the problem statement)
~Boywithnuke (I'll join too, fixed grammar and looks)
~Wen_Liang (EVEN MORE :D)
~Hamburger_ (Capitalized the βIβs in the edits above)
~NumberNinja1234 (More grammar :D)
~Boywithnuke (Capitalized the M and the C in the edits above)
~Wen_Liang (definitely did something [wink wink])
~ChickensEatGrass (removed a comma)
~moonwatcher0201 (edited grammar of edits)
~AquaNotFoundPlayz (added colon)
~wawasd (best edit of all time)
~dragonmather (added "basically")
~JeremyLikesCats (removed "basically")
~InvalidResponse (grammar conciseness)
~Okpere_Jr (Blundered the question, and edited)
~vinceS (changed "a" to "an")
~LUCKYOKXIAO (specified "equilateral" triangle twice)
~vinceS (specified "equilateral" triangle in the solution label)
~sixtetris (minor edit)
~NumberNinja1234 (changed (-) to (~) at the start of 2 people's edits.).
.
~AAaAa_Aa (added 1 space) really necessary edit I know.
~zoyshaikh (figured i'd join this and add the word "clearly").
..
~Wen_Liang (added space in front of parentheses).
~Wen_Liang again (removed a space in front of people's names).
~pisquared64 (specified "use this" to estimate).
~ToxicWaste (Changed "Therefore" to "Hence" :D) Note: Holy bandwagon, the list of peoples usernames is probably longer than the explanation at this point..
~zoyashaikh (Changed "this" to "the") ^above^ - i doubt it, but we might be able to get there soon.
~ToxicWaste (Capitalized the "c" in "changed) ^above^ - Don't worry zoyashaikh, it will go fast.
~zoyashaikh (Changed "Basic" to "Simple" in the title) ^above^ ok the list of people is 1963 characters while the solution is 984. oh hooray!.
~piquared64 (Added "can" to the explanation).
~ToxicWaste (Capitalized the "a" and "c") ^^above^^ -See I was right!.
~Soo9 (Deleted two words).
~PinkbIbtoy (Gave reasoning for splitting up the rhombus)
~tΒ :) (ChangedΒ ; to ,)
~reda_mandymath (add figure).
~zoyashaikh(changed "is" to "is as follows").
~NumberNinja1234 (Deleted a single word).
~NumberNinja1234 (FYI this is the 47th edit, all of these edits combined contain 350 words, and a total of 2,395 charactersΒ :).
~zoyashaikh (edited to add this commentΒ : D).
~PinkbIbtoy (Very minor edits to title).
~zoyashaikh (edited the title a little).
~piquared64 (replaced "by" with "through").
~zoyashaikh (replaced "also" with "furthermore").
~imbetterthanyou (replaced "ratio" with "proportion" and changed "greatest" to "largest").
~NumberNinja1234 (changed "clearly" to "obviously").
~kfclover (minor $\LaTeX$ formatting).
~kfclover (minor $\LaTeX$ formatting).
~kfclover (minor $\LaTeX$ formatting).
~ldrcs (changed "obviously" to "evidently").
~ (added βcanβ and βasβ).
~WaxyBoy (removed HumanBeingThatBreathesAir).
~Wen_Liang (removed unessential period). Note: OMG ive left for like a week and what happened.
~apple256 (cleared some wordiness).
~Toferlamopher (changed diagram).
~wr4th0fsh4dow (changed the title).
~wr4th0fsh4dow (closed the bracket on my previous edit).
~wr4th0fsh4dow (edited the previous edit because I realized I accidentally edited it without specifying what I did).
~wr4th0fsh4dow (edited the grammar of the previous edit).
~vinceS (added degree symbol to "60").
~SqueakyHamster2 (added the preamble of constitution: We the People of the United States, in Order to form a more perfect Union, establish Justice, insure domestic Tranquility, provide for the common defence, promote the general Welfare, and secure the Blessings of Liberty to ourselves and our Posterity, do ordain and establish this Constitution for the United States of America.).
~mathwizard123123 (changed tangent to incircle of).
~ lemon_ice (capitalized stuff)
~powerpanda24 (added period to previous edit).
~cheerfulfrog (added period to all previous edits).
~kfclover (added current edit count: 71)
~zzzzzzzzzzzz (edited the flow of the title) | // Block 1
/* Figure by reda */
unitsize(15);
real r = 1 + sqrt(3);
draw(r*dir(90) -- r*dir(210) -- r*dir(330) -- cycle);
filldraw(circle(dir(90), sqrt(3)/2), rgb("#CCC9E6"));
filldraw(circle(dir(210), sqrt(3)/2), rgb("#CCC9E6"));
filldraw(circle(dir(330), sqrt(3)/2), rgb("#CCC9E6"));
draw(dir(90)+dir(-30)*sqrt(3) -- dir(90)+dir(210)*sqrt(3) ^^
dir(210)+dir(90)*sqrt(3) -- dir(210)+dir(-30)*sqrt(3) ^^
dir(330)+dir(90)*sqrt(3) -- dir(330)+dir(210)*sqrt(3), red+dashed);
label("$T$", (0, -3.5));
draw(shift(8, 0)*(r*dir(90) -- r*dir(180)/sqrt(3) -- r*dir(270) -- r*dir(0)/sqrt(3) -- cycle));
filldraw(shift(8, 0)*circle(dir(90)*r/3, r/3), rgb("#CCC9E6"));
filldraw(shift(8, 0)*circle(dir(270)*r/3, r/3), rgb("#CCC9E6"));
draw(shift(8, 0)*(r*dir(180)/sqrt(3) -- r*dir(0)/sqrt(3)), red+dashed);
label("$R$", (8, -3.5));
draw(shift(16, 0)*(r*dir(0) -- r*dir(60) -- r*dir(120) -- r*dir(180) -- r*dir(240) -- r*dir(300) -- cycle));
for (int i = 30; i <= 330; i += 60) {
filldraw(shift(16, 0)*circle(r*dir(i)/sqrt(3), r/2/sqrt(3)), rgb("#CCC9E6"));
draw(shift(16, 0)*((0, 0) -- r*dir(i - 30)), red+dashed);
}
label("$H$", (16, -3.5));
// Block 2
/* Figure by reda */ unitsize(15); real r = 1 + sqrt(3); draw(r*dir(90) -- r*dir(210) -- r*dir(330) -- cycle); filldraw(circle(dir(90), sqrt(3)/2), rgb("#CCC9E6")); filldraw(circle(dir(210), sqrt(3)/2), rgb("#CCC9E6")); filldraw(circle(dir(330), sqrt(3)/2), rgb("#CCC9E6")); draw(dir(90)+dir(-30)*sqrt(3) -- dir(90)+dir(210)*sqrt(3) ^^ dir(210)+dir(90)*sqrt(3) -- dir(210)+dir(-30)*sqrt(3) ^^ dir(330)+dir(90)*sqrt(3) -- dir(330)+dir(210)*sqrt(3), red+dashed); label("$T$", (0, -3.5)); draw(shift(8, 0)*(r*dir(90) -- r*dir(180)/sqrt(3) -- r*dir(270) -- r*dir(0)/sqrt(3) -- cycle)); filldraw(shift(8, 0)*circle(dir(90)*r/3, r/3), rgb("#CCC9E6")); filldraw(shift(8, 0)*circle(dir(270)*r/3, r/3), rgb("#CCC9E6")); draw(shift(8, 0)*(r*dir(180)/sqrt(3) -- r*dir(0)/sqrt(3)), red+dashed); label("$R$", (8, -3.5)); draw(shift(16, 0)*(r*dir(0) -- r*dir(60) -- r*dir(120) -- r*dir(180) -- r*dir(240) -- r*dir(300) -- cycle)); for (int i = 30; i <= 330; i += 60) { filldraw(shift(16, 0)*circle(r*dir(i)/sqrt(3), r/2/sqrt(3)), rgb("#CCC9E6")); draw(shift(16, 0)*((0, 0) -- r*dir(i - 30)), red+dashed); } label("$H$", (16, -3.5)); | [] |
971 | The altitude to the hypotenuse of a $30^\circ{-}60^\circ{-}90^\circ$ is divided into two segments of lengths $x<y$ by the median to the shortest side of the triangle. What is the ratio $\tfrac{x}{x+y}$?
$\textbf{(A) } \dfrac{3}{7} \qquad \textbf{(B) } \dfrac{\sqrt3}{4} \qquad \textbf{(C) } \dfrac{4}{9} \qquad \textbf{(D) } \dfrac{5}{11} \qquad \textbf{(E) } \dfrac{4\sqrt3}{15}$ | 2025 AMC 10B Problem 13 | Without loss of generality, let $\triangle ABC$ have side-lengths $AB=2, BC=2\sqrt{3},$ and $AC=4.$ Let $D$ be the foot of the perpendicular from $B$ to $\overline{AC}, \ E$ be the midpoint of $\overline{AB},$ and $F$ be the intersection of $\overline{CE}$ and $\overline{BD}.$ Note that $\triangle ADB$ and $\triangle BDC$ are both $30^\circ{-}60^\circ{-}90^\circ$ triangles. From the side-length ratio, we get $AD=1$ and $DC=3.$
We obtain the following diagram:
From here, we will proceed with mass points. Throughout this solution, we will use $W_P$ to denote the weight of point $P.$
Let $W_C=1.$ Since $3AD=DC,$ it follows that $W_A=3$ and $W_D=W_C+W_A=4.$ Since $AE=EB$ and $W_A=3,$ it follows that $W_B=3.$
Now we focus on $\overline{BD}:$ Since $W_B=3$ and $W_D=4,$ we have $\frac{DF}{FB}=\frac xy=\frac34.$ Therefore, the answer is \[\frac{x}{x+y}=\frac{3}{3+4}=\boxed{\textbf{(A) } \dfrac{3}{7}}.\]
~lprado
~MRENTHUSIASM | /* Made by MRENTHUSIASM */ size(250); pair A, B, C, D, E, F; A = (0,1); B = origin; C = (sqrt(3),0); D = foot(B,A,C); E = midpoint(A--B); F = intersectionpoint(B--D,C--E); markscalefactor=0.008; draw(rightanglemark(A,B,C)^^rightanglemark(B,D,C),red+linewidth(1.25)); draw(A--B--C--cycle,linewidth(1.25)); draw(B--D^^C--E,linewidth(1.25)); label("$1$",midpoint(A--E),1.5*W,red); label("$1$",midpoint(E--B),1.5*W,red); label("$2\sqrt{3}$",midpoint(B--C),1.5*S,red); label("$x$",midpoint(D--F),1.25*dir(-45),red); label("$y$",midpoint(F--B),1.5*dir(-10),red); label("$30^{\circ}$",C,7*dir(165),red); label("$60^{\circ}$",A,4*dir(-63),red); label("$1$",midpoint(A--D),1.5*midpoint(A--D),red); label("$3$",midpoint(D--C),1.5*dir(70),red); label("$A$",A,1.5*NW); label("$B$",B,1.5*SW); label("$C$",C,1.5*SE); label("$D$",D,1.5*NE); label("$E$",E,1.5*W); label("$F$",F,2*dir(-75)); | [] |
971 | The altitude to the hypotenuse of a $30^\circ{-}60^\circ{-}90^\circ$ is divided into two segments of lengths $x<y$ by the median to the shortest side of the triangle. What is the ratio $\tfrac{x}{x+y}$?
$\textbf{(A) } \dfrac{3}{7} \qquad \textbf{(B) } \dfrac{\sqrt3}{4} \qquad \textbf{(C) } \dfrac{4}{9} \qquad \textbf{(D) } \dfrac{5}{11} \qquad \textbf{(E) } \dfrac{4\sqrt3}{15}$ | 2025 AMC 10B Problem 13 | This solution refers to the diagram in Solution 1, with the definitions slightly modified.
Let $AB=2$. Label the foot of the median $M$ and the foot of the altitude $H$. In $\triangle ABH$, we construct the midsegment $\overline{MP}$, which is parallel to $\overline{BH}$ and half as long as $\overline{BH}$. In addition, let $F$ be the intersection of $\overline{CM}$ and $\overline{BH}$.
We obtain the following diagram:
We can calculate that $PH=\frac{1}{2}$ and $HC=3$ By similar triangles $\triangle HFC \sim \triangle PMC$, we have $\frac{x}{PM}=\frac{HC}{PC}=\frac{3}{3+\frac{1}{2}}=\frac{6}{7}$. Also by similar triangles $\triangle PMA \sim \triangle HBA$, we have $\frac{PM}{x+y}=\frac{1}{2}$. Combining these, we get $\frac{x}{x+y}=\boxed{\textbf{(A) } \dfrac{3}{7}}$.
Note: Thanks MRENTHUSIASM for original diagram.
~gimmeaworkingusername | /* Made by MRENTHUSIASM */ size(250); pair A, B, C, H, M, F, P; A = (0,1); B = origin; C = (sqrt(3),0); H = foot(B,A,C); M = midpoint(A--B); F = intersectionpoint(B--H,C--M); P = midpoint(A--H); markscalefactor=0.008; draw(rightanglemark(A,B,C)^^rightanglemark(B,H,C),red+linewidth(1.25)); draw(A--B--C--cycle,linewidth(1.25)); draw(B--H^^C--M,linewidth(1.25)); draw(M--P,dashed+linewidth(1.25)); label("$1$",midpoint(A--M),1.5*W,red); label("$1$",midpoint(M--B),1.5*W,red); label("$2\sqrt{3}$",midpoint(B--C),1.5*S,red); label("$x$",midpoint(H--F),1.25*dir(-45),red); label("$y$",midpoint(F--B),1.5*dir(-10),red); label("$30^{\circ}$",C,7*dir(165),red); label("$60^{\circ}$",A,4*dir(-63),red); label("$\frac{1}{2}$",midpoint(A--P),1.5*midpoint(A--P),red); label("$\frac{1}{2}$",midpoint(P--H),1.5*midpoint(H--H),red); label("$3$",midpoint(H--C),1.5*dir(70),red); label("$A$",A,1.5*NW); label("$B$",B,1.5*SW); label("$C$",C,1.5*SE); label("$H$",H,1.5*NE); label("$M$",M,1.5*W); label("$P$",P,1.5*NE); label("$F$",F,2*dir(-75)); | [] |
971 | The altitude to the hypotenuse of a $30^\circ{-}60^\circ{-}90^\circ$ is divided into two segments of lengths $x<y$ by the median to the shortest side of the triangle. What is the ratio $\tfrac{x}{x+y}$?
$\textbf{(A) } \dfrac{3}{7} \qquad \textbf{(B) } \dfrac{\sqrt3}{4} \qquad \textbf{(C) } \dfrac{4}{9} \qquad \textbf{(D) } \dfrac{5}{11} \qquad \textbf{(E) } \dfrac{4\sqrt3}{15}$ | 2025 AMC 10B Problem 13 | This solution refers to the definitions and the diagram in Solution 1.
Let $M$ be the midpoint of $\overline{AC},$ so $DM=1.$ Let $G$ be the intersection of $\overline{BM}$ and $\overline{CE},$ so $G$ is the centroid of $\triangle ABC.$ In addition, we construct $\overline{DE},$ as shown below:
In any right triangle, since the median to the hypotenuse is half as long as the hypotenuse, we get $BM=\frac{1}{2}AC=2.$ Moreover, in any triangle, since the centroid divides each median into parts in the ratio $2:1,$ with the centroid being twice as close to the midpoint of a side as it is to the opposite vertex, we get $BG=\frac43$ and $GM=\frac23.$
Note that $\triangle ABM$ is an equilateral triangle with side-length $2.$ By the SAS similarity, we have $\triangle ABM \sim \triangle AED.$ Therefore, we conclude that $ED=1$ and $\overline{ED}\parallel\overline{BM}.$
By the AA Similarity, we have $\triangle DEF \sim \triangle BGF,$ from which $\frac xy = \frac{1}{4/3} = \frac34.$ It follows that \[\frac{x}{x+y}=\frac{3}{3+4}=\boxed{\textbf{(A) } \dfrac{3}{7}}.\]
~MRENTHUSIASM | /* Made by MRENTHUSIASM */ size(250); pair A, B, C, D, E, F, M, G; A = (0,1); B = origin; C = (sqrt(3),0); D = foot(B,A,C); E = midpoint(A--B); F = intersectionpoint(B--D,C--E); M = midpoint(A--C); G = centroid(A,B,C); markscalefactor=0.008; draw(rightanglemark(A,B,C)^^rightanglemark(B,D,C),red+linewidth(1.25)); draw(A--B--C--cycle,linewidth(1.25)); draw(B--D^^C--E,linewidth(1.25)); draw(E--D^^B--M,dashed+linewidth(1.25)); label("$1$",midpoint(A--E),1.5*W,red); label("$1$",midpoint(E--B),1.5*W,red); label("$2\sqrt{3}$",midpoint(B--C),1.5*S,red); label("$x$",midpoint(D--F),1.25*dir(-45),red); label("$y$",midpoint(F--B),1.5*dir(-10),red); label("$30^{\circ}$",C,7*dir(165),red); label("$60^{\circ}$",A,4*dir(-63),red); label("$1$",midpoint(A--D),1.5*midpoint(A--D),red); label("$1$",midpoint(D--M),1.5*midpoint(D--M),red); label("$A$",A,1.5*NW); label("$B$",B,1.5*SW); label("$C$",C,1.5*SE); label("$D$",D,1.5*D); label("$E$",E,1.5*W); label("$F$",F,2*dir(-75)); label("$M$",M,1.5*M); label("$G$",G,1.5*S); | [] |
971 | The altitude to the hypotenuse of a $30^\circ{-}60^\circ{-}90^\circ$ is divided into two segments of lengths $x<y$ by the median to the shortest side of the triangle. What is the ratio $\tfrac{x}{x+y}$?
$\textbf{(A) } \dfrac{3}{7} \qquad \textbf{(B) } \dfrac{\sqrt3}{4} \qquad \textbf{(C) } \dfrac{4}{9} \qquad \textbf{(D) } \dfrac{5}{11} \qquad \textbf{(E) } \dfrac{4\sqrt3}{15}$ | 2025 AMC 10B Problem 13 | This solution refers to the definitions and the diagram in Solution 1.
Since $E$ is the midpoint of $\overline{AB}$, we have $AE=1$, and because $\angle A=60^\circ$ and $AD=1$, $\triangle AED$ is equilateral. Its perpendicular bisector meets $\overline{AE}$ at $H$, giving $AH=HE=\tfrac12$ and $DH=\tfrac{\sqrt3}{2}$.
Extend $\overline{CE}$ past $E$ to $G$ so that $\overline{GD}$ is parallel to $\overline{BC}$ and is the perpendicular bisector of $\overline{AE}$, so $H$ lies on both $\overline{GD}$ and $\overline{AE}$:
By AA similarity, $\triangle BCE \sim \triangle HGE$ because $\angle BEC=\angle HEG$ and $\angle BCE=\angle HGE$ ($GD\parallel BC$). Thus $\frac{GH}{BC}=\frac{HE}{BE}$, and with $HE=\tfrac{1}{2}$, $BE=1$, and $BC=2\sqrt3$, we get $\frac{GH}{2\sqrt3}=\frac{\frac{1}{2}}{1}$ so $GH=\sqrt3$.
Also by AA similarity, $\triangle BCF \sim \triangle DGF$ because $\angle BFC=\angle DFG$ and $\angle BCF=\angle DGF$ ($GD\parallel BC$). Therefore $\frac{DF}{BF}=\frac{DG}{BC}$. Since $DG=DH+HG=\frac{\sqrt3}{2}+\sqrt3=\frac{3\sqrt3}{2}$ and $BC=2\sqrt3$, we obtain $\frac{DF}{BF}=\frac{x}{y}=\frac{\frac{3\sqrt3}{2}}{2\sqrt3}=\frac{3}{4}$, so $y=\frac43x$.
Finally, $\frac{x}{x+y}=\frac{x}{x+\frac{4}{3}x}=\frac{x}{\frac{7}{3}x}=\boxed{\textbf{(A) } \dfrac{3}{7}}$.
~Gymnastics15 (Mikenever Dai)
~MRENTHUSIASM (Original Diagram) | size(325); pair A, B, C, D, E, F, G, H; A = (0,2); B = origin; C = (2*sqrt(3),0); D = foot(B,A,C); E = midpoint(A--B); F = intersectionpoint(B--D,C--E); // G and H for the parallel/perpendicular construction G = (-sqrt(3), 1.5); H = (0, 1.5); markscalefactor=0.015; // Draw figure draw(rightanglemark(A,B,C)^^rightanglemark(B,D,C),red+linewidth(1.25)); draw(A--B--C--cycle,linewidth(1.25)); draw(B--D,linewidth(1.25)); draw(C--E,linewidth(1.25)); draw(A--E,linewidth(1.25)); draw(E--G,dashed+linewidth(1.25)); draw(G--D,dashed+linewidth(1.25)); draw(E--D,dashed+linewidth(1.25)); // equal tick marks add(pathticks(A--E, 1, .5, 0, 12, red+linewidth(1.25))); add(pathticks(E--B, 1, .5, 0, 12, red+linewidth(1.25))); // labels label("$x$",midpoint(D--F),1.25*dir(-45),red); label("$y$",midpoint(F--B),1.5*dir(-10),red); label("$30^{\circ}$",C,7*dir(165),red); label("$60^{\circ}$",A,4*dir(-63),red); label("$1$",midpoint(E--B),2.5*W,red); label("$2\sqrt{3}$",midpoint(B--C),1.5*S,red); label("$1$",midpoint(A--D),1.5*midpoint(A--D),red); label("$3$",midpoint(D--C),1.5*dir(70),red); // Point labels label("$A$",A,1.5*NW); label("$B$",B,1.5*SW); label("$C$",C,1.5*SE); label("$D$",D,1.5*NE); label("$E$",E,1.5*W); label("$F$",F,2*dir(-75)); label("$G$",G,1.5*NW); label("$H$",H,1.5*dir(135)); | [] |
971 | The altitude to the hypotenuse of a $30^\circ{-}60^\circ{-}90^\circ$ is divided into two segments of lengths $x<y$ by the median to the shortest side of the triangle. What is the ratio $\tfrac{x}{x+y}$?
$\textbf{(A) } \dfrac{3}{7} \qquad \textbf{(B) } \dfrac{\sqrt3}{4} \qquad \textbf{(C) } \dfrac{4}{9} \qquad \textbf{(D) } \dfrac{5}{11} \qquad \textbf{(E) } \dfrac{4\sqrt3}{15}$ | 2025 AMC 10B Problem 13 | WLOG $\overline{AB} = 2$ as the problem statement concerns only lengths ratios.
Draw $\overline{EM}$ connecting the median of $\triangle ABE$ to right angle $\angle E$, we know that by inscribing right $\triangle ABE$ in a circle and by Thales theorem, $\overline{ME} = \overline{MA} = \overline{MB}$ the circumradius.
$\overline{MB} = \overline{ME} \iff \triangle BEM \text{ is isosceles}$, and because $\angle B = 60^\circ$, then $\angle E = 60^\circ$, making $\triangle BEM$ equilateral.
Let $\overline{AD} = x$ and $\overline{DE} = y$. By $30^\circ - 60^\circ - 90^\circ$ special right triangle
\[\sqrt{3} = \overline{AE} = x + y\]
\[\overline{AM} = \sqrt{(2\sqrt{3})^2 + 1^2} = \sqrt{13}\]
\[\overline{CD} = \sqrt{y^2 + 3^2} = \sqrt{y^2 + 9} = \sqrt{x^2 - 2\sqrt{3}x + 12}\]
\[\overline{DM} = \sqrt{x^2 + 1^2 - 2 \cdot 1 \cdot x \cdot \cos{30^\circ} } = \sqrt{x^2 - \sqrt{3}x + 1}\]
\[\overline{CD} + \overline{DM} = \overline{AM}\]
\[\sqrt{x^2 - 2\sqrt{3}x + 12} + \sqrt{x^2 - \sqrt{3}x + 1} = \sqrt{13}\]
Algebra:
\begin{align*}
\sqrt{x^2 - 2\sqrt{3}x + 12} &= \sqrt{13} - \sqrt{x^2 - \sqrt{3}x + 1}\\
x^2 - 2\sqrt{3}x + 12 &= x^2 - \sqrt{3}x + 1 + 13 - 2\sqrt{13(x^2 - \sqrt{3}x + 1)}\\
-\sqrt{3}x - 2 &= -2\sqrt{13(x^2 - \sqrt{3}x + 1)}\\
3x^2 + 4\sqrt{3}x + 4 &= 4\cdot13(x^2 - \sqrt{3}x + 1)\\
49x^2 - 56\sqrt{3}x + 48 &= 0\\
(7x-4\sqrt{3})^2 &= 0
\end{align*}
$x = \frac{4\sqrt{3}}{7}$
$y = \frac{3\sqrt{3}}{7}$
$x$ is defined to be minimum of those lengths, so $x = \frac{3\sqrt{3}}{7}$.
Hence $\frac{x}{x+y} = \boxed{\textbf{(A) } \dfrac{3}{7}}$.
~clod | unitsize(0.3mm); /* Generated by Cloud's Excalidraw to Asymptote */ draw((-561, -520)--(-553, -515)--(-548, -523)--(-556, -528)--cycle, rgb(0.94, 0.54, 0)+linewidth(1)+solid);draw((-700, -590)--(-691, -590)--(-691, -599)--(-700, -599)--cycle, rgb(0.94, 0.54, 0)+linewidth(1)+solid);draw((-567, -510)--(-559, -505)--(-554, -513)--(-562, -518)--cycle, rgb(0.94, 0.54, 0)+linewidth(1)+solid);draw((-701, -260)--(-701, -600)--(-501, -600)--(-701, -260), rgb(0.11, 0.11, 0.11)+linewidth(2)+solid); draw((-701, -260)--(-601, -600), rgb(0.11, 0.11, 0.11)+linewidth(2)+solid); draw((-701, -600)--(-553, -515), rgb(0.11, 0.11, 0.11)+linewidth(2)+solid); draw((-648, -594)--(-648, -606), rgb(0.94, 0.54, 0)+linewidth(2)+solid); draw((-553, -594)--(-553, -606), rgb(0.94, 0.54, 0)+linewidth(2)+solid); draw((-602, -600)--(-552, -515), rgb(0.11, 0.11, 0.11)+linewidth(2)+solid); filldraw(shift(-701, -601) * rotate(0) * ellipse((0, 0), 4, 4), rgb(0.11, 0.11, 0.11), rgb(0, 0, 0)+linewidth(0)+solid); filldraw(shift(-601, -601) * rotate(0) * ellipse((0, 0), 4, 4), rgb(0.11, 0.11, 0.11), rgb(0, 0, 0)+linewidth(0)+solid); filldraw(shift(-502, -600) * rotate(0) * ellipse((0, 0), 4, 4), rgb(0.11, 0.11, 0.11), rgb(0, 0, 0)+linewidth(0)+solid); filldraw(shift(-701, -262) * rotate(0) * ellipse((0, 0), 4, 4), rgb(0.11, 0.11, 0.11), rgb(0, 0, 0)+linewidth(0)+solid); filldraw(shift(-552, -515) * rotate(0) * ellipse((0, 0), 4, 4), rgb(0.11, 0.11, 0.11), rgb(0, 0, 0)+linewidth(0)+solid); filldraw(shift(-616, -551) * rotate(0) * ellipse((0, 0), 4, 4), rgb(0.11, 0.11, 0.11), rgb(0, 0, 0)+linewidth(0)+solid); draw((-519, -598)--(-517, -590)--(-511, -586), rgb(0.87, 0.19, 0.19)+linewidth(2)+solid); draw((-699, -582)--(-691, -584)--(-687, -590), rgb(0.87, 0.19, 0.19)+linewidth(2)+solid); draw((-545, -530)--(-552, -532)--(-558, -530), rgb(0.87, 0.19, 0.19)+linewidth(2)+solid); draw((-593, -588)--(-587, -593)--(-586, -599), rgb(0.87, 0.19, 0.19)+linewidth(2)+solid); draw((-520, -555)--(-530, -561), rgb(0.94, 0.54, 0)+linewidth(2)+solid); draw((-583, -557)--(-573, -563), rgb(0.94, 0.54, 0)+linewidth(2)+solid); label("$B$", (-500, -616.5), fontsize(20pt) + rgb(0.11, 0.11, 0.11)+linewidth(1)+solid); label("$A$", (-702, -618.5), fontsize(20pt) + rgb(0.11, 0.11, 0.11)+linewidth(1)+solid); label("$C$", (-702, -247.5), fontsize(20pt) + rgb(0.11, 0.11, 0.11)+linewidth(1)+solid); label("$1$", (-649, -623), fontsize(16pt) + rgb(0.11, 0.11, 0.11)+linewidth(1)+solid); label("$1$", (-555, -622), fontsize(16pt) + rgb(0.11, 0.11, 0.11)+linewidth(1)+solid); label("$1$", (-507, -548), fontsize(16pt) + rgb(0.11, 0.11, 0.11)+linewidth(1)+solid); label("$2\sqrt{3}$", (-732.5, -457), fontsize(16pt) + rgb(0.11, 0.11, 0.11)+linewidth(1)+solid); label("$3$", (-596, -385), fontsize(16pt) + rgb(0.11, 0.11, 0.11)+linewidth(1)+solid); label("$D$", (-635, -544.5), fontsize(20pt) + rgb(0.11, 0.11, 0.11)+linewidth(1)+solid); label("$M$", (-602, -620.5), fontsize(20pt) + rgb(0.11, 0.11, 0.11)+linewidth(1)+solid); label("$E$", (-534, -502.5), fontsize(20pt) + rgb(0.11, 0.11, 0.11)+linewidth(1)+solid); | [] |
972 | A container has a $1\times 1$ square bottom, a $3\times 3$ open square top, and four congruent trapezoidal sides, as shown. Starting when the container is empty, a hose that runs water at a constant rate takes $35$ minutes to fill the container up to the midline of the trapezoids.
How many more minutes will it take to fill the remainder of the container?
$\text{(A) }70 \qquad \text{(B) }85 \qquad \text{(C) }90 \qquad \text{(D) }95 \qquad \text{(E) }105$ | 2025 AMC 10B Problem 19 | Extend the bottom of the container to form an upside down square pyramid. Then take a diagonal cross section as shown below.
You can notice that there are several similar triangles, so we write several equations.
\[\frac{x_3}{x_1+x_2+x_3}=\frac{\sqrt{2}}{3\sqrt{2}}\]
\[\frac{x_3}{x_2+x_3}=\frac{\sqrt{2}}{2\sqrt{2}}\]
Then solve the system to get: \[x_1=x_2=x_3=x\]
Write the volumes of the three pyramids with sides of the bases 1,2, and 3 respectively:
\[V_1=\frac{x}{3}\]
\[V_2=\frac{8x}{3}\]
\[V_3=9x\]
The volumes of the part of the container below the midline and above it are.
\[V_2-V_1=\frac{7x}{3}\] and \[V_3-V_2=\frac{19x}{3}\]
Then the time $t$ for the top to fill up is given by \[\frac{\frac{19x}{3}}{\frac{7x}{3}}=\frac{t}{35}\]
Therefore $t=\boxed{\mathbf{(D)}\ 95}$
-AARONPICK2 | size(220); // Outer (largest) upside-down isosceles triangle pair A = (0,-4); // bottom vertex pair B = (-3,0); // left base point pair C = (3,0); // right base point draw(A--B--C--cycle); // Middle triangle (scaled toward A) real k2 = 0.6; pair B2 = A + k2*(B - A); pair C2 = A + k2*(C - A); draw(A--B2--C2--cycle); // Smallest triangle (scaled toward A) real k3 = 0.3; pair B3 = A + k3*(B - A); pair C3 = A + k3*(C - A); draw(A--B3--C3--cycle); // --- Heights (altitudes) --- // Outer triangle height pair mid1 = (B + C)/2; draw(mid1--A); label("$x_1$", midpoint(mid1--A)+(0.25,1.25)); // centered on the segment label("$3\sqrt{2}$", midpoint(mid1--A)+(3.4,2)); // centered on the segment // Middle triangle height pair mid2 = (B2 + C2)/2; draw(mid2--A); label("$x_2$", midpoint(mid2--A)+(0.25,0.65)); // centered on the segment label("$2\sqrt{2}$", midpoint(mid2--A)+(2.3,1.2)); // centered on the segment // Smallest triangle height pair mid3 = (B3 + C3)/2; draw(mid3--A); label("$x_3$", midpoint(mid3--A)+(0.25,0.25)); // centered on the segment label("$\sqrt{2}$", midpoint(mid3--A)+(1.2,0.6)); // centered on the segment | [] |
973 | Four congruent semicircles are inscribed in a square of side length $1$ so that their diameters are on the sides of the square, one endpoint of each diameter is at a vertex of the square, and adjacent semicircles are tangent to each other. A small circle centered at the center of the square is tangent to each of the four semicircles, as shown below.
The diameter of the small circle can be written as $(\sqrt a+b)(\sqrt c+d)$, where $a$, $b$, $c$, and $d$ are integers. What is $a+b+c+d$?
$\textbf{(A) } 3 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 11$ | 2025 AMC 10B Problem 20 | Let the radius of the large semicircles be $r$, and let the diameter of the inner be $D$. Draw lines as follows:
By the Pythagorean Theorem, $r^2+(1-r)^2=(2r)^2$, so $2r^2+2r-1=0$. Hence $r=\frac{\sqrt{3}-1}{2}$ (we take the positive solution).
Now apply the Pythagorean Theorem on the quadrilateralβs altitude down to get $(1-2r)^2+1=(2r+D)^2$. Solving yields $2r+D=\sqrt{8-4\sqrt{3}}=\sqrt{6}-\sqrt{2}$. Then $D=\sqrt{6}-\sqrt{2}-\sqrt{3}+1=(\sqrt{3}-1)(\sqrt{2}-1)$. Hence the answer is $3-1+2-1=\boxed{\textbf{(A) } 3}$.
~ eevee9406 | // Block 1
import geometry;
size(6cm);
real r = (sqrt(3)-1)/2, R = (sqrt(2) - 1)*(sqrt(3)-1)/2;
draw((0,0) -- (1,0) -- (1, 1) -- (0,1) -- cycle);
draw(arc((r, 0), r, 0, 180));
draw(arc((1, r), r, 90, 270));
draw(arc((1 - r, 1), r, 180, 360));
draw(arc((0, 1 - r), r, 270, 450));
draw(circle((1/2,1/2),R));
draw((0.36602540378,0)--(0, 0.63397459622),dashed);
label(β$r$β,(0,0.45),W);
label(β$1-2r$β,(0,0.15),W);
label(β$r$β,(0.15,0),S);
label(β$r$β,(0.1,0.45),NE);
label(β$r$β,(0.275,0.15),NE);
label(β$r$β,(0.4,0.15),E);
label(β$r$β,(0.6,0.85),E);
label(β$D$β,(0.485,0.5),E);
draw((0.36602540378,0)--(0.63397459622,1),dashed);
label(β$r$β,(0.45,1),N);
label(β$1-2r$β,(0.15,1),N);
// Block 2
import geometry; size(6cm); real r = (sqrt(3)-1)/2, R = (sqrt(2) - 1)*(sqrt(3)-1)/2; draw((0,0) -- (1,0) -- (1, 1) -- (0,1) -- cycle); draw(arc((r, 0), r, 0, 180)); draw(arc((1, r), r, 90, 270)); draw(arc((1 - r, 1), r, 180, 360)); draw(arc((0, 1 - r), r, 270, 450)); draw(circle((1/2,1/2),R)); draw((0.36602540378,0)--(0, 0.63397459622),dashed); label(β$r$β,(0,0.45),W); label(β$1-2r$β,(0,0.15),W); label(β$r$β,(0.15,0),S); label(β$r$β,(0.1,0.45),NE); label(β$r$β,(0.275,0.15),NE); label(β$r$β,(0.4,0.15),E); label(β$r$β,(0.6,0.85),E); label(β$D$β,(0.485,0.5),E); draw((0.36602540378,0)--(0.63397459622,1),dashed); label(β$r$β,(0.45,1),N); label(β$1-2r$β,(0.15,1),N); | [] |
973 | Four congruent semicircles are inscribed in a square of side length $1$ so that their diameters are on the sides of the square, one endpoint of each diameter is at a vertex of the square, and adjacent semicircles are tangent to each other. A small circle centered at the center of the square is tangent to each of the four semicircles, as shown below.
The diameter of the small circle can be written as $(\sqrt a+b)(\sqrt c+d)$, where $a$, $b$, $c$, and $d$ are integers. What is $a+b+c+d$?
$\textbf{(A) } 3 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 11$ | 2025 AMC 10B Problem 20 | We have the diagram:
We label the vertices of the square $A$,$B$,$C$, and $D$, the centers of the semicircle $E$, $F$, $G$, and $H$, and the center of the small circle $O$. Lets call the radius of the big semicircle $x$ and the radius of the small circle $r$.
Looking at $\triangle HDG$, we can see it is a right triangle as $ABCD$ is a square, and we can also observe that since $DG$ is $x$ and the hypotenuse $GH$ is exactly twice at $2x$, $\triangle HDG$ is a $30-60-90^\circ$ triangle. That means $HD$ is equal to $\sqrt3 x$.
Upon inspection, we can also see that $AH$ is equal to $x$, and since $ABCD$ is a unit square with all sides equal to $1$, we can set the equation $HD+AH=1$, \[x+\sqrt3 x = 1\], so solving for $x$, we get that \[x=\frac{\sqrt{3}-1}{2}\]
Then, if we compose an auxiliary line from $O$ perpendicular to $HG$ at $J$, we can observe that $\triangle HOJ$ is an isosceles right triangle. That means $x+r=\sqrt2 x$ ($1-1-\sqrt2$ relations), which means \[r=(\sqrt2 -1)x\]
We're asked to find the diameter of the small circle, so we can multiply the equation by $2$ to get \[2r=2(\sqrt2 -1)x\]
Plugging x in, the $2$ multiplied and the $2$ in the denominator cancel out, and we get \[2r=2(\sqrt2 -1)(\frac{\sqrt{3}-1}{2})\] \[2r=(\sqrt2 -1)(\sqrt3 -1)\]
$a$, $b$, $c$, and $d$ are $2$, $-1$, $3$ respectively, and $-1$, so $2+(-1)+3+(-1)$ gets us $\boxed{\textbf{(A) } 3}$
-Neo | // Block 1
import geometry;
size(6cm);
real r = (sqrt(3)-1)/2, R = (sqrt(2) - 1)*(sqrt(3)-1)/2;
draw((0,0) -- (1,0) -- (1, 1) -- (0,1) -- cycle);
draw(arc((r, 0), r, 0, 180));
draw(arc((1, r), r, 90, 270));
draw(arc((1 - r, 1), r, 180, 360));
draw(arc((0, 1 - r), r, 270, 450));
draw(circle((1/2,1/2),R));
draw((0.36602540378,0)--(0, 0.63397459622),dashed);
draw((0.63397459622,1)--(0, 0.63397459622),dashed);
draw((0.63397459622,1)--(1, 0.36602540378),dashed);
draw((0.36602540378,0)--(1, 0.36602540378),dashed);
label(β$x$β,(0.15,0),S);
label(β$x$β,(0.1,0.45),SW);
label(β$x$β,(0.275,0.15),SW);
label(β$x$β,(0.22,0.64));
label(β$r$β,(0.45,0.55));
draw((0.5,0.5)--(0, 0.63397459622),dashed);
draw((0.36602540378,0)--(0.5,0.5),dashed);
draw((0.18301270189,0.31698729811)--(0.5,0.5),dashed);
label("$J$",(0.20301270189,0.29698729811),SW);
label(β$O$β,(0.485,0.5),E);
label("$A$",(0,1),NW);
label("$B$",(1,1),NE);
label("$D$",(0,0),SW);
label("$C$",(1,0),SE);
label("$G$",(0.36602540378,0),S);
label("$H$",(0, 0.63397459622),W);
label("$E$",(0.63397459622,1),N);
label("$F$",(1, 0.36602540378),E);
label(β$\sqrt3 x$β,(0,0.3),W);
// Block 2
import geometry; size(6cm); real r = (sqrt(3)-1)/2, R = (sqrt(2) - 1)*(sqrt(3)-1)/2; draw((0,0) -- (1,0) -- (1, 1) -- (0,1) -- cycle); draw(arc((r, 0), r, 0, 180)); draw(arc((1, r), r, 90, 270)); draw(arc((1 - r, 1), r, 180, 360)); draw(arc((0, 1 - r), r, 270, 450)); draw(circle((1/2,1/2),R)); draw((0.36602540378,0)--(0, 0.63397459622),dashed); draw((0.63397459622,1)--(0, 0.63397459622),dashed); draw((0.63397459622,1)--(1, 0.36602540378),dashed); draw((0.36602540378,0)--(1, 0.36602540378),dashed); label(β$x$β,(0.15,0),S); label(β$x$β,(0.1,0.45),SW); label(β$x$β,(0.275,0.15),SW); label(β$x$β,(0.22,0.64)); label(β$r$β,(0.45,0.55)); draw((0.5,0.5)--(0, 0.63397459622),dashed); draw((0.36602540378,0)--(0.5,0.5),dashed); draw((0.18301270189,0.31698729811)--(0.5,0.5),dashed); label("$J$",(0.20301270189,0.29698729811),SW); label(β$O$β,(0.485,0.5),E); label("$A$",(0,1),NW); label("$B$",(1,1),NE); label("$D$",(0,0),SW); label("$C$",(1,0),SE); label("$G$",(0.36602540378,0),S); label("$H$",(0, 0.63397459622),W); label("$E$",(0.63397459622,1),N); label("$F$",(1, 0.36602540378),E); label(β$\sqrt3 x$β,(0,0.3),W); | [] |
973 | Four congruent semicircles are inscribed in a square of side length $1$ so that their diameters are on the sides of the square, one endpoint of each diameter is at a vertex of the square, and adjacent semicircles are tangent to each other. A small circle centered at the center of the square is tangent to each of the four semicircles, as shown below.
The diameter of the small circle can be written as $(\sqrt a+b)(\sqrt c+d)$, where $a$, $b$, $c$, and $d$ are integers. What is $a+b+c+d$?
$\textbf{(A) } 3 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 11$ | 2025 AMC 10B Problem 20 | We realize that we can connect the radii of the semicircles to form an inner square (let $r$ denote the radius of the semicircles and $d$ denote their diameter)
To find the length of the inner square, we apply the Pythagorean Theorem
\[(1 - r)^{2}+r^{2}= (2r)^{2}\]
We now substitute $r = \frac{d}{2}$
\[(1 - \frac{d}{2})^2+(\frac{d}{2})^{2}= d^{2}\]
\[1-d + \frac{d^2}{4}+\frac{d^{2}}{4}= d^{2}\]
\[\frac{d^{2}}{2} -d+1= d^{2}\]
\[d^{2} -2d+2= 2d^{2}\]
\[-d^{2} -2d+2= 0\]
\[d^{2} +2d-2= 0\]
\[d = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\]
\[d = \frac{-2 \pm \sqrt{4-4(-2)}}{2}\]
\[d = \frac{-2 \pm \sqrt{12}}{2}\]
We must only consider the positive solution
\[d = \frac{-2 + \sqrt{12}}{2}\]
\[d = \frac{-2 + 2\sqrt{3}}{2}\]
\[d = -1 + \sqrt{3}\]
\[d = \sqrt{3} -1\]
The diagonal of the inner square measures $d\sqrt{2}$.
It can also be represented as $d+ d_{1}$ where $d_1$ denotes small circle's diameter. I demonstrate this below
\[2r+d_{1} = d+d_{1}\]
Therefore, we equate them, find $d_{1}$, and substitute $d= \sqrt{3}-1$
\[d+d_{1} = d\sqrt{2}\]
\[d_{1} = d\sqrt{2} - d\]
\[d_{1} = d(\sqrt{2} - 1)\]
\[d_{1} = (\sqrt{2}-1)(\sqrt{3}-1)\]
\[2+3-2 = 3 \implies \boxed{\textbf{(A) } 3}\]
~shockfront99 | // Block 1
import geometry;
size(6cm);
real r = (sqrt(3)-1)/2, R = (sqrt(2) - 1)*(sqrt(3)-1)/2;
draw((0,0) -- (1,0) -- (1, 1) -- (0,1) -- cycle);
draw((1,0.366025403784) -- (0.366025403784, 0));
draw((0.366025403784, 0) -- (0, 0.63397459621));
draw((1,0.366025403784) -- (0.63397459621, 1));
draw((0.63397459621, 1) -- (0, 0.63397459621));
draw(arc((r, 0), r, 0, 180));
draw(arc((1, r), r, 90, 270));
draw(arc((1 - r, 1), r, 180, 360));
draw(arc((0, 1 - r), r, 270, 450));
draw(circle((1/2,1/2),R));
draw(brace((0,0), (0, 0.63397459621)), black+0.4bp);
draw(brace((0.366025403784), (0,0)), black+0.4bp);
label("$1-r$", (-0.2, 0.3169872981));
label("$r$", (0.8, 0.28));
label("$r$", (0.18301270189, -0.08));
label("$r$", (0.55, 0.139));
// Block 2
import geometry;
size(6cm);
real r = (sqrt(3)-1)/2, R = (sqrt(2) - 1)*(sqrt(3)-1)/2;
draw((0,0) -- (1,0) -- (1, 1) -- (0,1) -- cycle);
draw((1,0.366025403784) -- (0.366025403784, 0));
draw((0.366025403784, 0) -- (0, 0.63397459621));
draw((1,0.366025403784) -- (0.63397459621, 1));
draw((0.63397459621, 1) -- (0, 0.63397459621));
draw((1,0.366025403784) -- (0, 0.63397459621));
draw(brace((1,0.366025403784), (0.649, 0.456)), black+0.4bp);
draw(brace((0.354, 0.54),(0, 0.63397459621)), black+0.4bp);
draw(brace((0.649, 0.456),(0.354, 0.54)), black+0.4bp);
draw(arc((r, 0), r, 0, 180));
draw(arc((1, r), r, 90, 270));
draw(arc((1 - r, 1), r, 180, 360));
draw(arc((0, 1 - r), r, 270, 450));
draw(circle((1/2,1/2),R));
label("$r$", (0.81, 0.338));
label("$r$", (0.164, 0.515));
label("$d_{1}$", (0.49, 0.42));
// Block 3
import geometry; size(6cm); real r = (sqrt(3)-1)/2, R = (sqrt(2) - 1)*(sqrt(3)-1)/2; draw((0,0) -- (1,0) -- (1, 1) -- (0,1) -- cycle); draw((1,0.366025403784) -- (0.366025403784, 0)); draw((0.366025403784, 0) -- (0, 0.63397459621)); draw((1,0.366025403784) -- (0.63397459621, 1)); draw((0.63397459621, 1) -- (0, 0.63397459621)); draw(arc((r, 0), r, 0, 180)); draw(arc((1, r), r, 90, 270)); draw(arc((1 - r, 1), r, 180, 360)); draw(arc((0, 1 - r), r, 270, 450)); draw(circle((1/2,1/2),R)); draw(brace((0,0), (0, 0.63397459621)), black+0.4bp); draw(brace((0.366025403784), (0,0)), black+0.4bp); label("$1-r$", (-0.2, 0.3169872981)); label("$r$", (0.8, 0.28)); label("$r$", (0.18301270189, -0.08)); label("$r$", (0.55, 0.139));
// Block 4
import geometry; size(6cm); real r = (sqrt(3)-1)/2, R = (sqrt(2) - 1)*(sqrt(3)-1)/2; draw((0,0) -- (1,0) -- (1, 1) -- (0,1) -- cycle); draw((1,0.366025403784) -- (0.366025403784, 0)); draw((0.366025403784, 0) -- (0, 0.63397459621)); draw((1,0.366025403784) -- (0.63397459621, 1)); draw((0.63397459621, 1) -- (0, 0.63397459621)); draw((1,0.366025403784) -- (0, 0.63397459621)); draw(brace((1,0.366025403784), (0.649, 0.456)), black+0.4bp); draw(brace((0.354, 0.54),(0, 0.63397459621)), black+0.4bp); draw(brace((0.649, 0.456),(0.354, 0.54)), black+0.4bp); draw(arc((r, 0), r, 0, 180)); draw(arc((1, r), r, 90, 270)); draw(arc((1 - r, 1), r, 180, 360)); draw(arc((0, 1 - r), r, 270, 450)); draw(circle((1/2,1/2),R)); label("$r$", (0.81, 0.338)); label("$r$", (0.164, 0.515)); label("$d_{1}$", (0.49, 0.42)); | [] |
974 | Square $ABCD$ has sides of length $4$. Points $P$ and $Q$ lie on $\overline{AD}$ and $\overline{CD}$, respectively, with $AP=\frac{8}{5}$ and $DQ=\frac{10}{3}$. A path begins along the segment from $P$ to $Q$ and continues by reflecting against the sides of $ABCD$ (with congruent incoming and outgoing angles). If the path hits a vertex of the square, it terminates there; otherwise it continues forever. At which vertex does the path terminate?
$\textbf{(A) } A \qquad \textbf{(B) } B \qquad \textbf{(C) } C \qquad \textbf{(D) } D \qquad \textbf{(E) } \text{The path continues forever.}$ | 2025 AMC 10B Problem 25 | The square can be placed with $A=(0,0),B=(4,0),C=(4,4),D=(0,4)$. Then \[P=(0,\frac{8}{5}),Q=(\frac{10}{3},4).\]
Straight line reflection is equivalent to passing through. The straight line from $P$ to $Q$ is
\[y = \frac{18}{25}x + \frac{8}{5}\]
\[25y = 18x + 40 \equiv 4 \pmod{18}\]
Notice that $25 \times 2 \equiv 7 \times 2 \equiv -4 \pmod{18}$, so $25 \times (-2) \equiv 4 \pmod{18}$, $y \equiv -2 \equiv 16 \pmod{18}$, then $x = 20$, the line passed through $(20, 16)$.
From $(0, \frac{8}{5})$ to $(20, 16)$, $x$ coordinate increases 20 whole units, $y$ coordinate increases 15 whole units. Excluding the case where $PQ$ passes through a lattice point at the final grid point, the segment $PQ$ crosses integer grid lines a total of $(20 - 1) + (15 - 1) = 33$ times. Each time the line crosses an integer grid line, it corresponds to one reflection; every four reflections, the path completes a full rotation within the square. $33 \equiv 1 \pmod{4}$,
after one reflection off side $CD$, the line $PQ$ will pass through point $\boxed{\textbf{(B) }B}$.
~reda_mandymath | // Block 1
unitsize(20);
pair _A, _B, _C, _D, _P, _Q;
_A = (0, 0);
_B = (0, 4);
_C = (4, 4);
_D = (4, 0);
_P = (0, 8/5);
_Q = (4, 10/3);
draw(_A -- _B -- _C -- _D -- _A ^^ _P -- _Q);
draw(extension(_P, _Q, _B, _C) -- _Q -- 2*_C - extension(_P, _Q, _B, _C), dashed);
label("$A$", _A, SW);
label("$B$", _B, NW);
label("$C$", _C, NE);
label("$D$", _D, SE);
label("$P$", _P, W);
label("$Q$", _Q, SE);
// Block 2
unitsize(20); pair _A, _B, _C, _D, _P, _Q; _A = (0, 0); _B = (0, 4); _C = (4, 4); _D = (4, 0); _P = (0, 8/5); _Q = (4, 10/3); draw(_A -- _B -- _C -- _D -- _A ^^ _P -- _Q); draw(extension(_P, _Q, _B, _C) -- _Q -- 2*_C - extension(_P, _Q, _B, _C), dashed); label("$A$", _A, SW); label("$B$", _B, NW); label("$C$", _C, NE); label("$D$", _D, SE); label("$P$", _P, W); label("$Q$", _Q, SE); | [] |
975 | The altitude to the hypotenuse of a $30^\circ{-}60^\circ{-}90^\circ$ is divided into two segments of lengths $x<y$ by the median to the shortest side of the triangle. What is the ratio $\tfrac{x}{x+y}$?
$\textbf{(A) } \dfrac{3}{7} \qquad \textbf{(B) } \dfrac{\sqrt3}{4} \qquad \textbf{(C) } \dfrac{4}{9} \qquad \textbf{(D) } \dfrac{5}{11} \qquad \textbf{(E) } \dfrac{4\sqrt3}{15}$ | 2025 AMC 12B Problem 10 | Without loss of generality, let $\triangle ABC$ have side-lengths $AB=2, BC=2\sqrt{3},$ and $AC=4.$ Let $D$ be the foot of the perpendicular from $B$ to $\overline{AC}, \ E$ be the midpoint of $\overline{AB},$ and $F$ be the intersection of $\overline{CE}$ and $\overline{BD}.$ Note that $\triangle ADB$ and $\triangle BDC$ are both $30^\circ{-}60^\circ{-}90^\circ$ triangles. From the side-length ratio, we get $AD=1$ and $DC=3.$
We obtain the following diagram:
From here, we will proceed with mass points. Throughout this solution, we will use $W_P$ to denote the weight of point $P.$
Let $W_C=1.$ Since $3AD=DC,$ it follows that $W_A=3$ and $W_D=W_C+W_A=4.$ Since $AE=EB$ and $W_A=3,$ it follows that $W_B=3.$
Now we focus on $\overline{BD}:$ Since $W_B=3$ and $W_D=4,$ we have $\frac{DF}{FB}=\frac xy=\frac34.$ Therefore, the answer is \[\frac{x}{x+y}=\frac{3}{3+4}=\boxed{\textbf{(A) } \dfrac{3}{7}}.\]
~lprado
~MRENTHUSIASM | // Block 1
/* Made by MRENTHUSIASM */
size(250);
pair A, B, C, D, E, F;
A = (0,1);
B = origin;
C = (sqrt(3),0);
D = foot(B,A,C);
E = midpoint(A--B);
F = intersectionpoint(B--D,C--E);
markscalefactor=0.008;
draw(rightanglemark(A,B,C)^^rightanglemark(B,D,C),red+linewidth(1.25));
draw(A--B--C--cycle,linewidth(1.25));
draw(B--D^^C--E,linewidth(1.25));
label("$1$",midpoint(A--E),1.5*W,red);
label("$1$",midpoint(E--B),1.5*W,red);
label("$2\sqrt{3}$",midpoint(B--C),1.5*S,red);
label("$x$",midpoint(D--F),1.25*dir(-45),red);
label("$y$",midpoint(F--B),1.5*dir(-10),red);
label("$30^{\circ}$",C,7*dir(165),red);
label("$60^{\circ}$",A,4*dir(-63),red);
label("$1$",midpoint(A--D),1.5*midpoint(A--D),red);
label("$3$",midpoint(D--C),1.5*dir(70),red);
label("$A$",A,1.5*NW);
label("$B$",B,1.5*SW);
label("$C$",C,1.5*SE);
label("$D$",D,1.5*NE);
label("$E$",E,1.5*W);
label("$F$",F,2*dir(-75));
// Block 2
/* Made by MRENTHUSIASM */ size(250); pair A, B, C, D, E, F; A = (0,1); B = origin; C = (sqrt(3),0); D = foot(B,A,C); E = midpoint(A--B); F = intersectionpoint(B--D,C--E); markscalefactor=0.008; draw(rightanglemark(A,B,C)^^rightanglemark(B,D,C),red+linewidth(1.25)); draw(A--B--C--cycle,linewidth(1.25)); draw(B--D^^C--E,linewidth(1.25)); label("$1$",midpoint(A--E),1.5*W,red); label("$1$",midpoint(E--B),1.5*W,red); label("$2\sqrt{3}$",midpoint(B--C),1.5*S,red); label("$x$",midpoint(D--F),1.25*dir(-45),red); label("$y$",midpoint(F--B),1.5*dir(-10),red); label("$30^{\circ}$",C,7*dir(165),red); label("$60^{\circ}$",A,4*dir(-63),red); label("$1$",midpoint(A--D),1.5*midpoint(A--D),red); label("$3$",midpoint(D--C),1.5*dir(70),red); label("$A$",A,1.5*NW); label("$B$",B,1.5*SW); label("$C$",C,1.5*SE); label("$D$",D,1.5*NE); label("$E$",E,1.5*W); label("$F$",F,2*dir(-75)); | [] |
975 | The altitude to the hypotenuse of a $30^\circ{-}60^\circ{-}90^\circ$ is divided into two segments of lengths $x<y$ by the median to the shortest side of the triangle. What is the ratio $\tfrac{x}{x+y}$?
$\textbf{(A) } \dfrac{3}{7} \qquad \textbf{(B) } \dfrac{\sqrt3}{4} \qquad \textbf{(C) } \dfrac{4}{9} \qquad \textbf{(D) } \dfrac{5}{11} \qquad \textbf{(E) } \dfrac{4\sqrt3}{15}$ | 2025 AMC 12B Problem 10 | This solution refers to the diagram in Solution 1, with the definitions slightly modified.
Let $AB=2$. Label the foot of the median $M$ and the foot of the altitude $H$. In $\triangle ABH$, we construct the midsegment $\overline{MP}$, which is parallel to $\overline{BH}$ and half as long as $\overline{BH}$. In addition, let $F$ be the intersection of $\overline{CM}$ and $\overline{BH}$.
We obtain the following diagram:
We can calculate that $PH=\frac{1}{2}$ and $HC=3$ By similar triangles $\triangle HFC \sim \triangle PMC$, we have $\frac{x}{PM}=\frac{HC}{PC}=\frac{3}{3+\frac{1}{2}}=\frac{6}{7}$. Also by similar triangles $\triangle PMA \sim \triangle HBA$, we have $\frac{PM}{x+y}=\frac{1}{2}$. Combining these, we get $\frac{x}{x+y}=\boxed{\textbf{(A) } \dfrac{3}{7}}$.
Note: Thanks MRENTHUSIASM for original diagram.
~gimmeaworkingusername | // Block 1
/* Made by MRENTHUSIASM */
size(250);
pair A, B, C, H, M, F, P;
A = (0,1);
B = origin;
C = (sqrt(3),0);
H = foot(B,A,C);
M = midpoint(A--B);
F = intersectionpoint(B--H,C--M);
P = midpoint(A--H);
markscalefactor=0.008;
draw(rightanglemark(A,B,C)^^rightanglemark(B,H,C),red+linewidth(1.25));
draw(A--B--C--cycle,linewidth(1.25));
draw(B--H^^C--M,linewidth(1.25));
draw(M--P,dashed+linewidth(1.25));
label("$1$",midpoint(A--M),1.5*W,red);
label("$1$",midpoint(M--B),1.5*W,red);
label("$2\sqrt{3}$",midpoint(B--C),1.5*S,red);
label("$x$",midpoint(H--F),1.25*dir(-45),red);
label("$y$",midpoint(F--B),1.5*dir(-10),red);
label("$30^{\circ}$",C,7*dir(165),red);
label("$60^{\circ}$",A,4*dir(-63),red);
label("$\frac{1}{2}$",midpoint(A--P),1.5*midpoint(A--P),red);
label("$\frac{1}{2}$",midpoint(P--H),1.5*midpoint(H--H),red);
label("$3$",midpoint(H--C),1.5*dir(70),red);
label("$A$",A,1.5*NW);
label("$B$",B,1.5*SW);
label("$C$",C,1.5*SE);
label("$H$",H,1.5*NE);
label("$M$",M,1.5*W);
label("$P$",P,1.5*NE);
label("$F$",F,2*dir(-75));
// Block 2
/* Made by MRENTHUSIASM */ size(250); pair A, B, C, H, M, F, P; A = (0,1); B = origin; C = (sqrt(3),0); H = foot(B,A,C); M = midpoint(A--B); F = intersectionpoint(B--H,C--M); P = midpoint(A--H); markscalefactor=0.008; draw(rightanglemark(A,B,C)^^rightanglemark(B,H,C),red+linewidth(1.25)); draw(A--B--C--cycle,linewidth(1.25)); draw(B--H^^C--M,linewidth(1.25)); draw(M--P,dashed+linewidth(1.25)); label("$1$",midpoint(A--M),1.5*W,red); label("$1$",midpoint(M--B),1.5*W,red); label("$2\sqrt{3}$",midpoint(B--C),1.5*S,red); label("$x$",midpoint(H--F),1.25*dir(-45),red); label("$y$",midpoint(F--B),1.5*dir(-10),red); label("$30^{\circ}$",C,7*dir(165),red); label("$60^{\circ}$",A,4*dir(-63),red); label("$\frac{1}{2}$",midpoint(A--P),1.5*midpoint(A--P),red); label("$\frac{1}{2}$",midpoint(P--H),1.5*midpoint(H--H),red); label("$3$",midpoint(H--C),1.5*dir(70),red); label("$A$",A,1.5*NW); label("$B$",B,1.5*SW); label("$C$",C,1.5*SE); label("$H$",H,1.5*NE); label("$M$",M,1.5*W); label("$P$",P,1.5*NE); label("$F$",F,2*dir(-75)); | [] |
975 | The altitude to the hypotenuse of a $30^\circ{-}60^\circ{-}90^\circ$ is divided into two segments of lengths $x<y$ by the median to the shortest side of the triangle. What is the ratio $\tfrac{x}{x+y}$?
$\textbf{(A) } \dfrac{3}{7} \qquad \textbf{(B) } \dfrac{\sqrt3}{4} \qquad \textbf{(C) } \dfrac{4}{9} \qquad \textbf{(D) } \dfrac{5}{11} \qquad \textbf{(E) } \dfrac{4\sqrt3}{15}$ | 2025 AMC 12B Problem 10 | This solution refers to the definitions and the diagram in Solution 1.
Let $M$ be the midpoint of $\overline{AC},$ so $DM=1.$ Let $G$ be the intersection of $\overline{BM}$ and $\overline{CE},$ so $G$ is the centroid of $\triangle ABC.$ In addition, we construct $\overline{DE},$ as shown below:
In any right triangle, since the median to the hypotenuse is half as long as the hypotenuse, we get $BM=\frac{1}{2}AC=2.$ Moreover, in any triangle, since the centroid divides each median into parts in the ratio $2:1,$ with the centroid being twice as close to the midpoint of a side as it is to the opposite vertex, we get $BG=\frac43$ and $GM=\frac23.$
Note that $\triangle ABM$ is an equilateral triangle with side-length $2.$ By the SAS similarity, we have $\triangle ABM \sim \triangle AED.$ Therefore, we conclude that $ED=1$ and $\overline{ED}\parallel\overline{BM}.$
By the AA Similarity, we have $\triangle DEF \sim \triangle BGF,$ from which $\frac xy = \frac{1}{4/3} = \frac34.$ It follows that \[\frac{x}{x+y}=\frac{3}{3+4}=\boxed{\textbf{(A) } \dfrac{3}{7}}.\]
~MRENTHUSIASM | // Block 1
/* Made by MRENTHUSIASM */
size(250);
pair A, B, C, D, E, F, M, G;
A = (0,1);
B = origin;
C = (sqrt(3),0);
D = foot(B,A,C);
E = midpoint(A--B);
F = intersectionpoint(B--D,C--E);
M = midpoint(A--C);
G = centroid(A,B,C);
markscalefactor=0.008;
draw(rightanglemark(A,B,C)^^rightanglemark(B,D,C),red+linewidth(1.25));
draw(A--B--C--cycle,linewidth(1.25));
draw(B--D^^C--E,linewidth(1.25));
draw(E--D^^B--M,dashed+linewidth(1.25));
label("$1$",midpoint(A--E),1.5*W,red);
label("$1$",midpoint(E--B),1.5*W,red);
label("$2\sqrt{3}$",midpoint(B--C),1.5*S,red);
label("$x$",midpoint(D--F),1.25*dir(-45),red);
label("$y$",midpoint(F--B),1.5*dir(-10),red);
label("$30^{\circ}$",C,7*dir(165),red);
label("$60^{\circ}$",A,4*dir(-63),red);
label("$1$",midpoint(A--D),1.5*midpoint(A--D),red);
label("$1$",midpoint(D--M),1.5*midpoint(D--M),red);
label("$A$",A,1.5*NW);
label("$B$",B,1.5*SW);
label("$C$",C,1.5*SE);
label("$D$",D,1.5*D);
label("$E$",E,1.5*W);
label("$F$",F,2*dir(-75));
label("$M$",M,1.5*M);
label("$G$",G,1.5*S);
// Block 2
/* Made by MRENTHUSIASM */ size(250); pair A, B, C, D, E, F, M, G; A = (0,1); B = origin; C = (sqrt(3),0); D = foot(B,A,C); E = midpoint(A--B); F = intersectionpoint(B--D,C--E); M = midpoint(A--C); G = centroid(A,B,C); markscalefactor=0.008; draw(rightanglemark(A,B,C)^^rightanglemark(B,D,C),red+linewidth(1.25)); draw(A--B--C--cycle,linewidth(1.25)); draw(B--D^^C--E,linewidth(1.25)); draw(E--D^^B--M,dashed+linewidth(1.25)); label("$1$",midpoint(A--E),1.5*W,red); label("$1$",midpoint(E--B),1.5*W,red); label("$2\sqrt{3}$",midpoint(B--C),1.5*S,red); label("$x$",midpoint(D--F),1.25*dir(-45),red); label("$y$",midpoint(F--B),1.5*dir(-10),red); label("$30^{\circ}$",C,7*dir(165),red); label("$60^{\circ}$",A,4*dir(-63),red); label("$1$",midpoint(A--D),1.5*midpoint(A--D),red); label("$1$",midpoint(D--M),1.5*midpoint(D--M),red); label("$A$",A,1.5*NW); label("$B$",B,1.5*SW); label("$C$",C,1.5*SE); label("$D$",D,1.5*D); label("$E$",E,1.5*W); label("$F$",F,2*dir(-75)); label("$M$",M,1.5*M); label("$G$",G,1.5*S); | [] |
975 | The altitude to the hypotenuse of a $30^\circ{-}60^\circ{-}90^\circ$ is divided into two segments of lengths $x<y$ by the median to the shortest side of the triangle. What is the ratio $\tfrac{x}{x+y}$?
$\textbf{(A) } \dfrac{3}{7} \qquad \textbf{(B) } \dfrac{\sqrt3}{4} \qquad \textbf{(C) } \dfrac{4}{9} \qquad \textbf{(D) } \dfrac{5}{11} \qquad \textbf{(E) } \dfrac{4\sqrt3}{15}$ | 2025 AMC 12B Problem 10 | This solution refers to the definitions and the diagram in Solution 1.
Since $E$ is the midpoint of $\overline{AB}$, we have $AE=1$, and because $\angle A=60^\circ$ and $AD=1$, $\triangle AED$ is equilateral. Its perpendicular bisector meets $\overline{AE}$ at $H$, giving $AH=HE=\tfrac12$ and $DH=\tfrac{\sqrt3}{2}$.
Extend $\overline{CE}$ past $E$ to $G$ so that $\overline{GD}$ is parallel to $\overline{BC}$ and is the perpendicular bisector of $\overline{AE}$, so $H$ lies on both $\overline{GD}$ and $\overline{AE}$:
By AA similarity, $\triangle BCE \sim \triangle HGE$ because $\angle BEC=\angle HEG$ and $\angle BCE=\angle HGE$ ($GD\parallel BC$). Thus $\frac{GH}{BC}=\frac{HE}{BE}$, and with $HE=\tfrac{1}{2}$, $BE=1$, and $BC=2\sqrt3$, we get $\frac{GH}{2\sqrt3}=\frac{\frac{1}{2}}{1}$ so $GH=\sqrt3$.
Also by AA similarity, $\triangle BCF \sim \triangle DGF$ because $\angle BFC=\angle DFG$ and $\angle BCF=\angle DGF$ ($GD\parallel BC$). Therefore $\frac{DF}{BF}=\frac{DG}{BC}$. Since $DG=DH+HG=\frac{\sqrt3}{2}+\sqrt3=\frac{3\sqrt3}{2}$ and $BC=2\sqrt3$, we obtain $\frac{DF}{BF}=\frac{x}{y}=\frac{\frac{3\sqrt3}{2}}{2\sqrt3}=\frac{3}{4}$, so $y=\frac43x$.
Finally, $\frac{x}{x+y}=\frac{x}{x+\frac{4}{3}x}=\frac{x}{\frac{7}{3}x}=\boxed{\textbf{(A) } \dfrac{3}{7}}$.
~Gymnastics15 (Mikenever Dai)
~MRENTHUSIASM (Original Diagram) | // Block 1
size(325);
pair A, B, C, D, E, F, G, H;
A = (0,2);
B = origin;
C = (2*sqrt(3),0);
D = foot(B,A,C);
E = midpoint(A--B);
F = intersectionpoint(B--D,C--E);
// G and H for the parallel/perpendicular construction
G = (-sqrt(3), 1.5);
H = (0, 1.5);
markscalefactor=0.015;
// Draw figure
draw(rightanglemark(A,B,C)^^rightanglemark(B,D,C),red+linewidth(1.25));
draw(A--B--C--cycle,linewidth(1.25));
draw(B--D,linewidth(1.25));
draw(C--E,linewidth(1.25));
draw(A--E,linewidth(1.25));
draw(E--G,dashed+linewidth(1.25));
draw(G--D,dashed+linewidth(1.25));
draw(E--D,dashed+linewidth(1.25));
// equal tick marks
add(pathticks(A--E, 1, .5, 0, 12, red+linewidth(1.25)));
add(pathticks(E--B, 1, .5, 0, 12, red+linewidth(1.25)));
// labels
label("$x$",midpoint(D--F),1.25*dir(-45),red);
label("$y$",midpoint(F--B),1.5*dir(-10),red);
label("$30^{\circ}$",C,7*dir(165),red);
label("$60^{\circ}$",A,4*dir(-63),red);
label("$1$",midpoint(E--B),2.5*W,red);
label("$2\sqrt{3}$",midpoint(B--C),1.5*S,red);
label("$1$",midpoint(A--D),1.5*midpoint(A--D),red);
label("$3$",midpoint(D--C),1.5*dir(70),red);
// Point labels
label("$A$",A,1.5*NW);
label("$B$",B,1.5*SW);
label("$C$",C,1.5*SE);
label("$D$",D,1.5*NE);
label("$E$",E,1.5*W);
label("$F$",F,2*dir(-75));
label("$G$",G,1.5*NW);
label("$H$",H,1.5*dir(135));
// Block 2
size(325); pair A, B, C, D, E, F, G, H; A = (0,2); B = origin; C = (2*sqrt(3),0); D = foot(B,A,C); E = midpoint(A--B); F = intersectionpoint(B--D,C--E); // G and H for the parallel/perpendicular construction G = (-sqrt(3), 1.5); H = (0, 1.5); markscalefactor=0.015; // Draw figure draw(rightanglemark(A,B,C)^^rightanglemark(B,D,C),red+linewidth(1.25)); draw(A--B--C--cycle,linewidth(1.25)); draw(B--D,linewidth(1.25)); draw(C--E,linewidth(1.25)); draw(A--E,linewidth(1.25)); draw(E--G,dashed+linewidth(1.25)); draw(G--D,dashed+linewidth(1.25)); draw(E--D,dashed+linewidth(1.25)); // equal tick marks add(pathticks(A--E, 1, .5, 0, 12, red+linewidth(1.25))); add(pathticks(E--B, 1, .5, 0, 12, red+linewidth(1.25))); // labels label("$x$",midpoint(D--F),1.25*dir(-45),red); label("$y$",midpoint(F--B),1.5*dir(-10),red); label("$30^{\circ}$",C,7*dir(165),red); label("$60^{\circ}$",A,4*dir(-63),red); label("$1$",midpoint(E--B),2.5*W,red); label("$2\sqrt{3}$",midpoint(B--C),1.5*S,red); label("$1$",midpoint(A--D),1.5*midpoint(A--D),red); label("$3$",midpoint(D--C),1.5*dir(70),red); // Point labels label("$A$",A,1.5*NW); label("$B$",B,1.5*SW); label("$C$",C,1.5*SE); label("$D$",D,1.5*NE); label("$E$",E,1.5*W); label("$F$",F,2*dir(-75)); label("$G$",G,1.5*NW); label("$H$",H,1.5*dir(135)); | [] |
975 | The altitude to the hypotenuse of a $30^\circ{-}60^\circ{-}90^\circ$ is divided into two segments of lengths $x<y$ by the median to the shortest side of the triangle. What is the ratio $\tfrac{x}{x+y}$?
$\textbf{(A) } \dfrac{3}{7} \qquad \textbf{(B) } \dfrac{\sqrt3}{4} \qquad \textbf{(C) } \dfrac{4}{9} \qquad \textbf{(D) } \dfrac{5}{11} \qquad \textbf{(E) } \dfrac{4\sqrt3}{15}$ | 2025 AMC 12B Problem 10 | WLOG $\overline{AB} = 2$ as the problem statement concerns only lengths ratios.
Draw $\overline{EM}$ connecting the median of $\triangle ABE$ to right angle $\angle E$, we know that by inscribing right $\triangle ABE$ in a circle and by Thales theorem, $\overline{ME} = \overline{MA} = \overline{MB}$ the circumradius.
$\overline{MB} = \overline{ME} \iff \triangle BEM \text{ is isosceles}$, and because $\angle B = 60^\circ$, then $\angle E = 60^\circ$, making $\triangle BEM$ equilateral.
Let $\overline{AD} = x$ and $\overline{DE} = y$. By $30^\circ - 60^\circ - 90^\circ$ special right triangle
\[\sqrt{3} = \overline{AE} = x + y\]
\[\overline{AM} = \sqrt{(2\sqrt{3})^2 + 1^2} = \sqrt{13}\]
\[\overline{CD} = \sqrt{y^2 + 3^2} = \sqrt{y^2 + 9} = \sqrt{x^2 - 2\sqrt{3}x + 12}\]
\[\overline{DM} = \sqrt{x^2 + 1^2 - 2 \cdot 1 \cdot x \cdot \cos{30^\circ} } = \sqrt{x^2 - \sqrt{3}x + 1}\]
\[\overline{CD} + \overline{DM} = \overline{AM}\]
\[\sqrt{x^2 - 2\sqrt{3}x + 12} + \sqrt{x^2 - \sqrt{3}x + 1} = \sqrt{13}\]
Algebra:
\begin{align*}
\sqrt{x^2 - 2\sqrt{3}x + 12} &= \sqrt{13} - \sqrt{x^2 - \sqrt{3}x + 1}\\
x^2 - 2\sqrt{3}x + 12 &= x^2 - \sqrt{3}x + 1 + 13 - 2\sqrt{13(x^2 - \sqrt{3}x + 1)}\\
-\sqrt{3}x - 2 &= -2\sqrt{13(x^2 - \sqrt{3}x + 1)}\\
3x^2 + 4\sqrt{3}x + 4 &= 4\cdot13(x^2 - \sqrt{3}x + 1)\\
49x^2 - 56\sqrt{3}x + 48 &= 0\\
(7x-4\sqrt{3})^2 &= 0
\end{align*}
$x = \frac{4\sqrt{3}}{7}$
$y = \frac{3\sqrt{3}}{7}$
$x$ is defined to be minimum of those lengths, so $x = \frac{3\sqrt{3}}{7}$.
Hence $\frac{x}{x+y} = \boxed{\textbf{(A) } \dfrac{3}{7}}$.
~clod | // Block 1
unitsize(0.3mm);
/* Generated by Cloud's Excalidraw to Asymptote */
draw((-561, -520)--(-553, -515)--(-548, -523)--(-556, -528)--cycle, rgb(0.94, 0.54, 0)+linewidth(1)+solid);draw((-700, -590)--(-691, -590)--(-691, -599)--(-700, -599)--cycle, rgb(0.94, 0.54, 0)+linewidth(1)+solid);draw((-567, -510)--(-559, -505)--(-554, -513)--(-562, -518)--cycle, rgb(0.94, 0.54, 0)+linewidth(1)+solid);draw((-701, -260)--(-701, -600)--(-501, -600)--(-701, -260), rgb(0.11, 0.11, 0.11)+linewidth(2)+solid);
draw((-701, -260)--(-601, -600), rgb(0.11, 0.11, 0.11)+linewidth(2)+solid);
draw((-701, -600)--(-553, -515), rgb(0.11, 0.11, 0.11)+linewidth(2)+solid);
draw((-648, -594)--(-648, -606), rgb(0.94, 0.54, 0)+linewidth(2)+solid);
draw((-553, -594)--(-553, -606), rgb(0.94, 0.54, 0)+linewidth(2)+solid);
draw((-602, -600)--(-552, -515), rgb(0.11, 0.11, 0.11)+linewidth(2)+solid);
filldraw(shift(-701, -601) * rotate(0) * ellipse((0, 0), 4, 4), rgb(0.11, 0.11, 0.11), rgb(0, 0, 0)+linewidth(0)+solid);
filldraw(shift(-601, -601) * rotate(0) * ellipse((0, 0), 4, 4), rgb(0.11, 0.11, 0.11), rgb(0, 0, 0)+linewidth(0)+solid);
filldraw(shift(-502, -600) * rotate(0) * ellipse((0, 0), 4, 4), rgb(0.11, 0.11, 0.11), rgb(0, 0, 0)+linewidth(0)+solid);
filldraw(shift(-701, -262) * rotate(0) * ellipse((0, 0), 4, 4), rgb(0.11, 0.11, 0.11), rgb(0, 0, 0)+linewidth(0)+solid);
filldraw(shift(-552, -515) * rotate(0) * ellipse((0, 0), 4, 4), rgb(0.11, 0.11, 0.11), rgb(0, 0, 0)+linewidth(0)+solid);
filldraw(shift(-616, -551) * rotate(0) * ellipse((0, 0), 4, 4), rgb(0.11, 0.11, 0.11), rgb(0, 0, 0)+linewidth(0)+solid);
draw((-519, -598)--(-517, -590)--(-511, -586), rgb(0.87, 0.19, 0.19)+linewidth(2)+solid);
draw((-699, -582)--(-691, -584)--(-687, -590), rgb(0.87, 0.19, 0.19)+linewidth(2)+solid);
draw((-545, -530)--(-552, -532)--(-558, -530), rgb(0.87, 0.19, 0.19)+linewidth(2)+solid);
draw((-593, -588)--(-587, -593)--(-586, -599), rgb(0.87, 0.19, 0.19)+linewidth(2)+solid);
draw((-520, -555)--(-530, -561), rgb(0.94, 0.54, 0)+linewidth(2)+solid);
draw((-583, -557)--(-573, -563), rgb(0.94, 0.54, 0)+linewidth(2)+solid);
label("$B$", (-500, -616.5), fontsize(20pt) + rgb(0.11, 0.11, 0.11)+linewidth(1)+solid);
label("$A$", (-702, -618.5), fontsize(20pt) + rgb(0.11, 0.11, 0.11)+linewidth(1)+solid);
label("$C$", (-702, -247.5), fontsize(20pt) + rgb(0.11, 0.11, 0.11)+linewidth(1)+solid);
label("$1$", (-649, -623), fontsize(16pt) + rgb(0.11, 0.11, 0.11)+linewidth(1)+solid);
label("$1$", (-555, -622), fontsize(16pt) + rgb(0.11, 0.11, 0.11)+linewidth(1)+solid);
label("$1$", (-507, -548), fontsize(16pt) + rgb(0.11, 0.11, 0.11)+linewidth(1)+solid);
label("$2\sqrt{3}$", (-732.5, -457), fontsize(16pt) + rgb(0.11, 0.11, 0.11)+linewidth(1)+solid);
label("$3$", (-596, -385), fontsize(16pt) + rgb(0.11, 0.11, 0.11)+linewidth(1)+solid);
label("$D$", (-635, -544.5), fontsize(20pt) + rgb(0.11, 0.11, 0.11)+linewidth(1)+solid);
label("$M$", (-602, -620.5), fontsize(20pt) + rgb(0.11, 0.11, 0.11)+linewidth(1)+solid);
label("$E$", (-534, -502.5), fontsize(20pt) + rgb(0.11, 0.11, 0.11)+linewidth(1)+solid);
// Block 2
unitsize(0.3mm); /* Generated by Cloud's Excalidraw to Asymptote */ draw((-561, -520)--(-553, -515)--(-548, -523)--(-556, -528)--cycle, rgb(0.94, 0.54, 0)+linewidth(1)+solid);draw((-700, -590)--(-691, -590)--(-691, -599)--(-700, -599)--cycle, rgb(0.94, 0.54, 0)+linewidth(1)+solid);draw((-567, -510)--(-559, -505)--(-554, -513)--(-562, -518)--cycle, rgb(0.94, 0.54, 0)+linewidth(1)+solid);draw((-701, -260)--(-701, -600)--(-501, -600)--(-701, -260), rgb(0.11, 0.11, 0.11)+linewidth(2)+solid); draw((-701, -260)--(-601, -600), rgb(0.11, 0.11, 0.11)+linewidth(2)+solid); draw((-701, -600)--(-553, -515), rgb(0.11, 0.11, 0.11)+linewidth(2)+solid); draw((-648, -594)--(-648, -606), rgb(0.94, 0.54, 0)+linewidth(2)+solid); draw((-553, -594)--(-553, -606), rgb(0.94, 0.54, 0)+linewidth(2)+solid); draw((-602, -600)--(-552, -515), rgb(0.11, 0.11, 0.11)+linewidth(2)+solid); filldraw(shift(-701, -601) * rotate(0) * ellipse((0, 0), 4, 4), rgb(0.11, 0.11, 0.11), rgb(0, 0, 0)+linewidth(0)+solid); filldraw(shift(-601, -601) * rotate(0) * ellipse((0, 0), 4, 4), rgb(0.11, 0.11, 0.11), rgb(0, 0, 0)+linewidth(0)+solid); filldraw(shift(-502, -600) * rotate(0) * ellipse((0, 0), 4, 4), rgb(0.11, 0.11, 0.11), rgb(0, 0, 0)+linewidth(0)+solid); filldraw(shift(-701, -262) * rotate(0) * ellipse((0, 0), 4, 4), rgb(0.11, 0.11, 0.11), rgb(0, 0, 0)+linewidth(0)+solid); filldraw(shift(-552, -515) * rotate(0) * ellipse((0, 0), 4, 4), rgb(0.11, 0.11, 0.11), rgb(0, 0, 0)+linewidth(0)+solid); filldraw(shift(-616, -551) * rotate(0) * ellipse((0, 0), 4, 4), rgb(0.11, 0.11, 0.11), rgb(0, 0, 0)+linewidth(0)+solid); draw((-519, -598)--(-517, -590)--(-511, -586), rgb(0.87, 0.19, 0.19)+linewidth(2)+solid); draw((-699, -582)--(-691, -584)--(-687, -590), rgb(0.87, 0.19, 0.19)+linewidth(2)+solid); draw((-545, -530)--(-552, -532)--(-558, -530), rgb(0.87, 0.19, 0.19)+linewidth(2)+solid); draw((-593, -588)--(-587, -593)--(-586, -599), rgb(0.87, 0.19, 0.19)+linewidth(2)+solid); draw((-520, -555)--(-530, -561), rgb(0.94, 0.54, 0)+linewidth(2)+solid); draw((-583, -557)--(-573, -563), rgb(0.94, 0.54, 0)+linewidth(2)+solid); label("$B$", (-500, -616.5), fontsize(20pt) + rgb(0.11, 0.11, 0.11)+linewidth(1)+solid); label("$A$", (-702, -618.5), fontsize(20pt) + rgb(0.11, 0.11, 0.11)+linewidth(1)+solid); label("$C$", (-702, -247.5), fontsize(20pt) + rgb(0.11, 0.11, 0.11)+linewidth(1)+solid); label("$1$", (-649, -623), fontsize(16pt) + rgb(0.11, 0.11, 0.11)+linewidth(1)+solid); label("$1$", (-555, -622), fontsize(16pt) + rgb(0.11, 0.11, 0.11)+linewidth(1)+solid); label("$1$", (-507, -548), fontsize(16pt) + rgb(0.11, 0.11, 0.11)+linewidth(1)+solid); label("$2\sqrt{3}$", (-732.5, -457), fontsize(16pt) + rgb(0.11, 0.11, 0.11)+linewidth(1)+solid); label("$3$", (-596, -385), fontsize(16pt) + rgb(0.11, 0.11, 0.11)+linewidth(1)+solid); label("$D$", (-635, -544.5), fontsize(20pt) + rgb(0.11, 0.11, 0.11)+linewidth(1)+solid); label("$M$", (-602, -620.5), fontsize(20pt) + rgb(0.11, 0.11, 0.11)+linewidth(1)+solid); label("$E$", (-534, -502.5), fontsize(20pt) + rgb(0.11, 0.11, 0.11)+linewidth(1)+solid); | [] |
976 | A circle has been divided into 6 sectors of different sizes. Then 2 of the sectors are painted red, 2 painted green, and 2 painted blue so that no two neighboring sectors are painted the same color. One such coloring is shown below.
How many different colorings are possible?
$\textbf{(A)}~12\qquad\textbf{(B)}~16\qquad\textbf{(C)}~18\qquad\textbf{(D)}~24\qquad\textbf{(E)}~28$ | 2025 AMC 12B Problem 13 | You can tackle the problem by forgetting about the different sizes at first. This gets you 4 combinations:
B-R-G-B-R-G
B-R-G-B-G-R
B-R-B-G-R-G
B-G-B-R-G-R
Now we can multiply by 6 since we have 6 different pieces. $6\cdot 4 =\boxed{\textbf{(D)}~24}$.
~Taha Jazaeri | // Block 1
size(300);
pair base = (0,0);
real r = 1;
// Define colors
pen B = blue;
pen R = red;
pen G = green;
// Patterns (each row = one circle)
// B,R,G,B,R,G etc.
pen[][] patterns = {
{B,R,G,B,R,G},
{B,R,G,B,G,R},
{B,R,B,G,R,G},
{B,G,B,R,G,R}
};
for (int c = 0; c < 4; ++c) {
pair O = base + (3*c, 0); // spacing circles horizontally
draw(circle(O, r)); // outline
for (int k = 0; k < 6; ++k) {
real th1 = k * 60;
real th2 = (k+1) * 60;
// Build sector path
path p = O -- arc(O, r, th1, th2) -- cycle;
// Fill according to pattern
filldraw(p, patterns[c][k], black);
}
}
// Block 2
size(300); pair base = (0,0); real r = 1; // Define colors pen B = blue; pen R = red; pen G = green; // Patterns (each row = one circle) // B,R,G,B,R,G etc. pen[][] patterns = { {B,R,G,B,R,G}, {B,R,G,B,G,R}, {B,R,B,G,R,G}, {B,G,B,R,G,R} }; for (int c = 0; c < 4; ++c) { pair O = base + (3*c, 0); // spacing circles horizontally draw(circle(O, r)); // outline for (int k = 0; k < 6; ++k) { real th1 = k * 60; real th2 = (k+1) * 60; // Build sector path path p = O -- arc(O, r, th1, th2) -- cycle; // Fill according to pattern filldraw(p, patterns[c][k], black); } } | [] |
977 | A container has a $1\times 1$ square bottom, a $3\times 3$ open square top, and four congruent trapezoidal sides, as shown. Starting when the container is empty, a hose that runs water at a constant rate takes $35$ minutes to fill the container up to the midline of the trapezoids.
How many more minutes will it take to fill the remainder of the container?
$\text{(A) }70 \qquad \text{(B) }85 \qquad \text{(C) }90 \qquad \text{(D) }95 \qquad \text{(E) }105$ | 2025 AMC 12B Problem 15 | Extend the bottom of the container to form an upside down square pyramid. Then take a diagonal cross section as shown below.
You can notice that there are several similar triangles, so we write several equations.
\[\frac{x_3}{x_1+x_2+x_3}=\frac{\sqrt{2}}{3\sqrt{2}}\]
\[\frac{x_3}{x_2+x_3}=\frac{\sqrt{2}}{2\sqrt{2}}\]
Then solve the system to get: \[x_1=x_2=x_3=x\]
Write the volumes of the three pyramids with sides of the bases 1,2, and 3 respectively:
\[V_1=\frac{x}{3}\]
\[V_2=\frac{8x}{3}\]
\[V_3=9x\]
The volumes of the part of the container below the midline and above it are.
\[V_2-V_1=\frac{7x}{3}\] and \[V_3-V_2=\frac{19x}{3}\]
Then the time $t$ for the top to fill up is given by \[\frac{\frac{19x}{3}}{\frac{7x}{3}}=\frac{t}{35}\]
Therefore $t=\boxed{\mathbf{(D)}\ 95}$
-AARONPICK2 | // Block 1
size(220);
// Outer (largest) upside-down isosceles triangle
pair A = (0,-4); // bottom vertex
pair B = (-3,0); // left base point
pair C = (3,0); // right base point
draw(A--B--C--cycle);
// Middle triangle (scaled toward A)
real k2 = 0.6;
pair B2 = A + k2*(B - A);
pair C2 = A + k2*(C - A);
draw(A--B2--C2--cycle);
// Smallest triangle (scaled toward A)
real k3 = 0.3;
pair B3 = A + k3*(B - A);
pair C3 = A + k3*(C - A);
draw(A--B3--C3--cycle);
// --- Heights (altitudes) ---
// Outer triangle height
pair mid1 = (B + C)/2;
draw(mid1--A);
label("$x_1$", midpoint(mid1--A)+(0.25,1.25)); // centered on the segment
label("$3\sqrt{2}$", midpoint(mid1--A)+(3.4,2)); // centered on the segment
// Middle triangle height
pair mid2 = (B2 + C2)/2;
draw(mid2--A);
label("$x_2$", midpoint(mid2--A)+(0.25,0.65)); // centered on the segment
label("$2\sqrt{2}$", midpoint(mid2--A)+(2.3,1.2)); // centered on the segment
// Smallest triangle height
pair mid3 = (B3 + C3)/2;
draw(mid3--A);
label("$x_3$", midpoint(mid3--A)+(0.25,0.25)); // centered on the segment
label("$\sqrt{2}$", midpoint(mid3--A)+(1.2,0.6)); // centered on the segment
// Block 2
size(220); // Outer (largest) upside-down isosceles triangle pair A = (0,-4); // bottom vertex pair B = (-3,0); // left base point pair C = (3,0); // right base point draw(A--B--C--cycle); // Middle triangle (scaled toward A) real k2 = 0.6; pair B2 = A + k2*(B - A); pair C2 = A + k2*(C - A); draw(A--B2--C2--cycle); // Smallest triangle (scaled toward A) real k3 = 0.3; pair B3 = A + k3*(B - A); pair C3 = A + k3*(C - A); draw(A--B3--C3--cycle); // --- Heights (altitudes) --- // Outer triangle height pair mid1 = (B + C)/2; draw(mid1--A); label("$x_1$", midpoint(mid1--A)+(0.25,1.25)); // centered on the segment label("$3\sqrt{2}$", midpoint(mid1--A)+(3.4,2)); // centered on the segment // Middle triangle height pair mid2 = (B2 + C2)/2; draw(mid2--A); label("$x_2$", midpoint(mid2--A)+(0.25,0.65)); // centered on the segment label("$2\sqrt{2}$", midpoint(mid2--A)+(2.3,1.2)); // centered on the segment // Smallest triangle height pair mid3 = (B3 + C3)/2; draw(mid3--A); label("$x_3$", midpoint(mid3--A)+(0.25,0.25)); // centered on the segment label("$\sqrt{2}$", midpoint(mid3--A)+(1.2,0.6)); // centered on the segment | [] |
978 | Two non-congruent triangles have the same area. Each triangle has sides of length $8$ and $9$, and the third side of each triangle has integer length. What is the sum of the lengths of the third sides?
$\textbf{(A) } 20 \qquad \textbf{(B) } 22 \qquad \textbf{(C) } 24 \qquad \textbf{(D) } 26 \qquad \textbf{(E) } 28$ | 2025 AMC 12B Problem 21 | ~Diagram by imosilver
Let the angles between the $8$ and $9$ sides be $\theta_1$ and $\theta_2$. Relating the two triangles' areas, we get
\[\frac{1}{2} \cdot 8 \cdot 9 \cdot \sin{\theta_1} = \frac{1}{2} \cdot 8 \cdot 9 \cdot \sin{\theta_2}\]
\[\sin{\theta_1} = \sin{\theta_2}\]
For this to be true, $\theta_1 + \theta_2 = 180.$
We will now apply Law of Cosines to get an expression for $s_1$ and $s_2$ in terms of $\theta_1$ and $\theta_2$.
\[s_1^2 = 8^2 + 9^2 - (2)(8)(9)\cos(\theta_1)\]
\[s_2^2 = 8^2 + 9^2 - (2)(8)(9)\cos(180-\theta_1).\]
Noting that $\cos{\theta_1} = -\cos{(180 - \theta_1)}$, we can add the two equations to get that
\[s_1^2 + s_2^2 = 290.\]
We now need two perfect squares that add to $290$. After some analysis, we find $121 + 169 = 290$. Therefore, $s_1 = 11$ and $s_2 = 13$, so our answer is $11 + 13 = \boxed{24}.$
Note that the $\cos{\theta_1} = 1/6.$ Also, note that $289 + 1 = 290$, but $17$ and $1$ are not valid side lengths due to the Triangle Inequality.
~lprado | // Block 1
size(250);
pair A = (0,0);
pair C1 = (-9,0);
pair C2 = (9,0);
pair B = (-4.0/3.0, 4*sqrt(35)/3);
pen pen1 = blue+1.2;
pen pen2 = red+1.2;
draw(B--A--C1--cycle, pen1);
draw(B--A--C2--cycle, pen2);
dot("$A$", A, S);
dot("$C_1$", C1, SW);
dot("$C_2$", C2, SE);
dot("$B$", B, N);
label("$8$", (A+B)/2, W);
label("$9$", (A+C1)/2, S);
label("$9$", (A+C2)/2, S);
// Block 2
size(250); pair A = (0,0); pair C1 = (-9,0); pair C2 = (9,0); pair B = (-4.0/3.0, 4*sqrt(35)/3); pen pen1 = blue+1.2; pen pen2 = red+1.2; draw(B--A--C1--cycle, pen1); draw(B--A--C2--cycle, pen2); dot("$A$", A, S); dot("$C_1$", C1, SW); dot("$C_2$", C2, SE); dot("$B$", B, N); label("$8$", (A+B)/2, W); label("$9$", (A+C1)/2, S); label("$9$", (A+C2)/2, S); | [] |
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