Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Let $A_1 A_2 A_3 A_4 A_5 A_6 A_7 A_8$ be a regular octagon.  Let $M_1$ $M_3$ $M_5$ , and $M_7$ be the midpoints of sides $\overline{A_1 A_2}$ $\overline{A_3 A_4}$ $\overline{A_5 A_6}$ , and $\overline{A_7 A_8}$ , respectively.  For $i = 1, 3, 5, 7$ , ray $R_i$ is constructed from $M_i$ towards the interior of the octagon such that $R_1 \perp R_3$ $R_3 \perp R_5$ $R_5 \perp R_7$ , and $R_7 \perp R_1$ .  Pairs of rays $R_1$ and $R_3$ $R_3$ and $R_5$ $R_5$ and $R_7$ , and $R_7$ and $R_1$ meet at $B_1$ $B_3$ $B_5$ $B_7$ respectively.  If $B_1 B_3 = A_1 A_2$ , then $\cos 2 \angle A_3 M_3 B_1$ can be written in the form $m - \sqrt{n}$ , where $m$ and $n$ are positive integers.  Find $m + n$

Solution: Let $\theta=\angle M_1 M_3 B_1$ . Thus we have that $\cos 2 \angle A_3 M_3 B_1=\cos \left(2\theta + \frac{\pi}{2} \right)=-\sin2\theta$
Since $A_1 A_2 A_3 A_4 A_5 A_6 A_7 A_8$ is a regular octagon and $B_1 B_3 = A_1 A_2$ , let $k=A_1 A_2 = A_2 A_3 = B_1 B_3$
Extend $\overline{A_1 A_2}$ and $\overline{A_3 A_4}$ until they intersect. Denote their intersection as $I_1$ . Through similar triangles & the $45-45-90$ triangles formed, we find that $M_1 M_3=\frac{k}{2}(2+\sqrt2)$
We also have that $\triangle M_7 B_7 M_1 =\triangle M_1 B_1 M_3$ through ASA congruence ( $\angle B_7 M_7 M_1 =\angle B_1 M_1 M_3$ $M_7 M_1 = M_1 M_3$ $\angle B_7 M_1 M_7 =\angle B_1 M_3 M_1$ ). Therefore, we may let $n=M_1 B_7 = M_3 B_1$
Thus, we have that $\sin\theta=\frac{n-k}{\frac{k}{2}(2+\sqrt2)}$ and that $\cos\theta=\frac{n}{\frac{k}{2}(2+\sqrt2)}$ . Therefore $\cos\theta-\sin\theta=\frac{k}{\frac{k}{2}(2+\sqrt2)}=\frac{2}{2+\sqrt2}=2-\sqrt2$
Squaring gives that $\sin^2\theta - 2\sin\theta\cos\theta + \cos^2\theta = 6-4\sqrt2$ and consequently that $-2\sin\theta\cos\theta = 5-4\sqrt2 = -\sin2\theta$ through the identities $\sin^2\theta + \cos^2\theta = 1$ and $\sin2\theta = 2\sin\theta\cos\theta$
Thus we have that $\cos 2 \angle A_3 M_3 B_1=5-4\sqrt2=5-\sqrt{32}$ . Therefore $m+n=5+32=\boxed{037}$

Answer:  037
Problem: Let $A_1 A_2 A_3 A_4 A_5 A_6 A_7 A_8$ be a regular octagon.  Let $M_1$ $M_3$ $M_5$ , and $M_7$ be the midpoints of sides $\overline{A_1 A_2}$ $\overline{A_3 A_4}$ $\overline{A_5 A_6}$ , and $\overline{A_7 A_8}$ , respectively.  For $i = 1, 3, 5, 7$ , ray $R_i$ is constructed from $M_i$ towards the interior of the octagon such that $R_1 \perp R_3$ $R_3 \perp R_5$ $R_5 \perp R_7$ , and $R_7 \perp R_1$ .  Pairs of rays $R_1$ and $R_3$ $R_3$ and $R_5$ $R_5$ and $R_7$ , and $R_7$ and $R_1$ meet at $B_1$ $B_3$ $B_5$ $B_7$ respectively.  If $B_1 B_3 = A_1 A_2$ , then $\cos 2 \angle A_3 M_3 B_1$ can be written in the form $m - \sqrt{n}$ , where $m$ and $n$ are positive integers.  Find $m + n$

Solution: Let $A_1A_2 = 2$ .  Then $B_1$ and $B_3$ are the projections of $M_1$ and $M_5$ onto the line $B_1B_3$ , so $2=B_1B_3=-M_1M_5\cos x$ , where $x = \angle A_3M_3B_1$ .  Then since $M_1M_5 = 2+2\sqrt{2}, \cos x = \dfrac{-2}{2+2\sqrt{2}}= 1-\sqrt{2}$ $\cos 2x = 2\cos^2 x -1 = 5 - 4\sqrt{2} = 5-\sqrt{32}$ , and $m+n=\boxed{037}$

Answer:  037
Problem: Let $A_1 A_2 A_3 A_4 A_5 A_6 A_7 A_8$ be a regular octagon.  Let $M_1$ $M_3$ $M_5$ , and $M_7$ be the midpoints of sides $\overline{A_1 A_2}$ $\overline{A_3 A_4}$ $\overline{A_5 A_6}$ , and $\overline{A_7 A_8}$ , respectively.  For $i = 1, 3, 5, 7$ , ray $R_i$ is constructed from $M_i$ towards the interior of the octagon such that $R_1 \perp R_3$ $R_3 \perp R_5$ $R_5 \perp R_7$ , and $R_7 \perp R_1$ .  Pairs of rays $R_1$ and $R_3$ $R_3$ and $R_5$ $R_5$ and $R_7$ , and $R_7$ and $R_1$ meet at $B_1$ $B_3$ $B_5$ $B_7$ respectively.  If $B_1 B_3 = A_1 A_2$ , then $\cos 2 \angle A_3 M_3 B_1$ can be written in the form $m - \sqrt{n}$ , where $m$ and $n$ are positive integers.  Find $m + n$

Solution: Notice that $R_3$ and $R_7$ are parallel ( $B_1B_3B_5B_7$ is a square by symmetry and since the rays are perpendicular) and $B_1B_3=B_3B_5=s=$ the distance between the parallel rays. If the regular hexagon as a side length of $s$ , then $M_3M_7$ has a length of $s+s\sqrt{2}$ . Let $X$ be on $R_3$ such that $M_7X$ is perpendicular to $M_3X$ , and $\phi=\angle M_7M_3X$ . The distance between $R_3$ and $R_7$ is $s=M_7X$ , so $\sin\phi=\frac{s}{s+s\sqrt{2}}=\frac{1}{1+\sqrt{2}}$
Since we are considering a regular hexagon, $M_3$ is directly opposite to $M_7$ and $\angle A_3M_3B_1=90 ^\circ +\phi$ . All that's left is to calculate $\cos 2\angle A_3M_3B_1=\cos^2(90^\circ+\phi)-\sin^2(90^\circ+\phi)=\sin^2\phi-\cos^2\phi$ . By drawing a right triangle or using the Pythagorean identity, $\cos^2\phi=\frac{2+2\sqrt2}{3+2\sqrt2}$ and $\cos 2\angle A_3M_3B_1=\frac{-1-2\sqrt2}{3+2\sqrt2}=5-4\sqrt2=5-\sqrt{32}$ , so $m+n=\boxed{037}$

Answer:  037
Problem: Let $A_1 A_2 A_3 A_4 A_5 A_6 A_7 A_8$ be a regular octagon.  Let $M_1$ $M_3$ $M_5$ , and $M_7$ be the midpoints of sides $\overline{A_1 A_2}$ $\overline{A_3 A_4}$ $\overline{A_5 A_6}$ , and $\overline{A_7 A_8}$ , respectively.  For $i = 1, 3, 5, 7$ , ray $R_i$ is constructed from $M_i$ towards the interior of the octagon such that $R_1 \perp R_3$ $R_3 \perp R_5$ $R_5 \perp R_7$ , and $R_7 \perp R_1$ .  Pairs of rays $R_1$ and $R_3$ $R_3$ and $R_5$ $R_5$ and $R_7$ , and $R_7$ and $R_1$ meet at $B_1$ $B_3$ $B_5$ $B_7$ respectively.  If $B_1 B_3 = A_1 A_2$ , then $\cos 2 \angle A_3 M_3 B_1$ can be written in the form $m - \sqrt{n}$ , where $m$ and $n$ are positive integers.  Find $m + n$

Solution: Assume that $A_1A_2=1.$ Denote the center $O$ , and the midpoint of $B_1$ and $B_3$ as $B_2$ . Then we have that \[\cos\angle A_3M_3B_1=\cos(\angle A_3M_3O+\angle OM_3B_1)=-\sin(\angle OM_3B_1)=-\frac{OB_2}{OM_3}=-\frac{1/2}{1/2+\sqrt2/2}=-\frac{1}{\sqrt2+1}=1-\sqrt2.\] Thus, by the cosine double-angle theorem, \[\cos2\angle A_3M_3B_1=2(1-\sqrt2)^2-1=5-\sqrt{32},\] so $m+n=\boxed{037}$

Answer:  037
Problem: For some integer $m$ , the polynomial $x^3 - 2011x + m$ has the three integer roots $a$ $b$ , and $c$ .  Find $|a| + |b| + |c|$

Solution: From Vieta's formulas, we know that $a+b+c = 0$ , and $ab+bc+ac = -2011$ . Thus $a = -(b+c)$ . All three of $a$ $b$ , and $c$ are non-zero: say, if $a=0$ , then $b=-c=\pm\sqrt{2011}$ (which is not an integer). $\textsc{wlog}$ , let $|a| \ge |b| \ge |c|$ . If $a > 0$ , then $b,c < 0$ and if $a < 0$ , then $b,c > 0$ . We have \[-2011=ab+bc+ac = a(b+c)+bc = -a^2+bc\] Thus $a^2 = 2011 + bc$ . We know that $b$ $c$ have the same sign. So $|a| \ge 45 = \lceil \sqrt{2011} \rceil$
Also, if we fix $a$ $b+c$ is fixed, so $bc$ is maximized when $b = c$ . Hence, \[2011 = a^2 - bc > \tfrac{3}{4}a^2 \qquad \Longrightarrow \qquad a ^2 < \tfrac{4}{3}\cdot 2011 = 2681+\tfrac{1}{3}\] So $|a| \le 51$ . Thus we have bounded $a$ as $45\le |a| \le 51$ , i.e. $45\le |b+c| \le 51$ since $a=-(b+c)$ . Let's analyze $bc=(b+c)^2-2011$ . Here is a table:
We can tell we don't need to bother with $45$
$105 = (3)(5)(7)$ , So $46$ won't work. $198/47 > 4$
$198$ is not divisible by $5$ $198/6 = 33$ , which is too small to get $47$
$293/48 > 6$ $293$ is not divisible by $7$ or $8$ or $9$ , we can clearly tell that $10$ is too much.
Hence, $|a| = 49$ $a^2 -2011 = 390$ $b = 39$ $c = 10$
Answer: $\boxed{098}$

Answer:  098
Problem: For some integer $m$ , the polynomial $x^3 - 2011x + m$ has the three integer roots $a$ $b$ , and $c$ .  Find $|a| + |b| + |c|$

Solution: Starting off like the previous solution, we know that $a + b + c = 0$ , and $ab + bc + ac = -2011$
Therefore, $c = -b-a$
Substituting, $ab + b(-b-a) + a(-b-a) = ab-b^2-ab-ab-a^2 = -2011$
Factoring the perfect square, we get: $ab-(b+a)^2=-2011$ or $(b+a)^2-ab=2011$
Therefore, a sum ( $a+b$ ) squared minus a product ( $ab$ ) gives $2011$ ..
We can guess and check different $a+b$ ’s starting with $45$ since $44^2 < 2011$
$45^2 = 2025$ therefore $ab = 2025-2011 = 14$
Since no factors of $14$ can sum to $45$ $1+14$ being the largest sum), a + b cannot equal $45$
$46^2 = 2116$ making $ab = 105 = 3 * 5 * 7$
$5 * 7 + 3 < 46$ and $3 * 5 * 7 > 46$ so $46$ cannot work either.
We can continue to do this until we reach $49$
$49^2 =  2401$ making $ab = 390 = 2 * 3 * 5* 13$
$3 * 13 + 2* 5 = 49$ , so one root is $10$ and another is $39$ . The roots sum to zero, so the last root must be $-49$
$|-49|+10+39 = \boxed{098}$

Answer:  098
Problem: For some integer $m$ , the polynomial $x^3 - 2011x + m$ has the three integer roots $a$ $b$ , and $c$ .  Find $|a| + |b| + |c|$

Solution: Let us first note the obvious that is derived from Vieta's formulas: $a+b+c=0, ab+bc+ac=-2011$ . Now, due to the first equation, let us say that $a+b=-c$ , meaning that $a,b>0$ and $c<0$ . Now, since both $a$ and $b$ are greater than 0, their absolute values are both equal to $a$ and $b$ , respectively. Since $c$ is less than 0, it equals $-a-b$ . Therefore, $|c|=|-a-b|=a+b$ , meaning $|a|+|b|+|c|=2(a+b)$ . We now apply Newton's sums to get that $a^2+b^2+ab=2011$ ,or $(a+b)^2-ab=2011$ . Solving, we find that $49^2-390$ satisfies this, meaning $a+b=49$ , so $2(a+b)=\boxed{098}$

Answer:  098
Problem: For some integer $m$ , the polynomial $x^3 - 2011x + m$ has the three integer roots $a$ $b$ , and $c$ .  Find $|a| + |b| + |c|$

Solution: We have
As a result, we have
$a+b+c=0$
$ab+bc+ac=-2011$
$abc=-m$
So, $a=-b-c$
As a result, $ab+bc+ac=(-b-c)b+(-b-c)c+bc=-b^2-c^2-bc=-2011$
Solve $b=\frac {-c+\sqrt{c^2-4(c^2-2011)}}{2}$ and $\Delta =8044-3c^2=k^2$ , where $k$ is an integer
Cause $89<\sqrt{8044}<90$
So, after we tried for $2$ times, we get $k=88$ and $c=10$
then $b=39$ $a=-b-c=-49$
As a result, $|a|+|b|+|c|=10+39+49=\boxed{098}$

Answer:  098
Problem: For some integer $m$ , the polynomial $x^3 - 2011x + m$ has the three integer roots $a$ $b$ , and $c$ .  Find $|a| + |b| + |c|$

Solution: First, derive the equations $a=-b-c$ and $ab+bc+ca=-2011\implies b^2+bc+c^2=2011$ . Since the product is negative, $a$ is negative, and $b$ and $c$ positive. Now, a simple mod 3 testing of all cases shows that $b\equiv \{1,2\} \pmod{3}$ , and $c$ has the repective value. We can choose $b$ not congruent to 0, make sure you see why. Now, we bash on values of $b$ , testing the quadratic function to see if $c$ is positive. You can also use a delta argument like solution 4, but this is simpler. We get that for $b=10$ $c=39, -49$ . Choosing $c$ positive we get $a=-49$ , so $|a|+|b|+|c|=10+29+39=\boxed{098}$ ~firebolt360

Answer:  098
Problem: Gary purchased a large beverage, but only drank $m/n$ of it, where $m$ and $n$ are relatively prime positive integers. If he had purchased half as much and drunk twice as much, he would have wasted only $2/9$ as much beverage. Find $m+n$

Solution: Let $x$ be the fraction consumed, then $(1-x)$ is the fraction wasted. We have $\frac{1}{2} - 2x =\frac{2}{9} (1-x)$ , or $9 - 36x = 4 - 4x$ , or $32x = 5$ or $x = 5/32$ . Therefore, $m + n = 5 + 32 = \boxed{037}$

Answer:  037
Problem: The degree measures of the angles in a convex 18-sided polygon form an increasing arithmetic sequence with integer values. Find the degree measure of the smallest angle

Solution: Another way to solve this problem would be to use exterior angles. Exterior angles of any polygon add up to $360^{\circ}$ . Since there are $18$ exterior angles in an 18-gon, the average measure of an exterior angles is $\frac{360}{18}=20^\circ$ . We know from the problem that since the exterior angles must be in an arithmetic sequence, the median and average of them is $20$ . Since there are even number of exterior angles, the middle two must be $19^\circ$ and $21^\circ$ , and the difference between terms must be $2$ . Check to make sure the smallest exterior angle is greater than $0$ $19-2(8)=19-16=3^\circ$ . It is, so the greatest exterior angle is $21+2(8)=21+16=37^\circ$ and the smallest interior angle is $180-37=\boxed{143}$

Answer:  143
Problem: The degree measures of the angles in a convex 18-sided polygon form an increasing arithmetic sequence with integer values. Find the degree measure of the smallest angle

Solution: The sum of the angles in a 18-gon is $(18-2) \cdot 180^\circ = 2880 ^\circ.$ Because the angles are in an arithmetic sequence, we can also write the sum of the angles as $a+(a+d)+(a+2d)+\dots+(a+17d)=18a+153d,$ where $a$ is the smallest angle and $d$ is the common difference. Since these two are equal, we know that $18a+153d = 2880 ^\circ,$ or $2a+17d = 320^\circ.$ The smallest value of $d$ that satisfies this is $d=2,$ so $a=143.$ Other values of $d$ and $a$ satisfy that equation, but if we tried any of them the last angle would be greater than $180,$ so the only value of $a$ that works is $a=\boxed{143}$

Answer:  143
Problem: The degree measures of the angles in a convex 18-sided polygon form an increasing arithmetic sequence with integer values. Find the degree measure of the smallest angle

Solution: Each individual angle in a $18$ -gon is $\frac {(18-2) \cdot 180^\circ}{18} = 160^\circ$ . Since no angle in a convex polygon can be larger than $180^\circ$ , the smallest angle possible is in the set $159, 161, 157, 163, 155, 165, 153, 167, 151, 169, 149, 171, 147, 173, 145, 175, 143, 177$
Our smallest possible angle is $\boxed{143}$

Answer:  143
Problem: In triangle $ABC$ $AB=20$ and $AC=11$ . The angle bisector of $\angle A$ intersects $BC$ at point $D$ , and point $M$ is the midpoint of $AD$ . Let $P$ be the point of the intersection of $AC$ and $BM$ . The ratio of $CP$ to $PA$ can be expressed in the form $\dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: [asy] pointpen = black; pathpen = linewidth(0.7);  pair A = (0,0), C= (11,0), B=IP(CR(A,20),CR(C,18)), D = IP(B--C,CR(B,20/31*abs(B-C))), M = (A+D)/2, P = IP(M--2*M-B, A--C), D2 = IP(D--D+P-B, A--C);  D(MP("A",D(A))--MP("B",D(B),N)--MP("C",D(C))--cycle); D(A--MP("D",D(D),NE)--MP("D'",D(D2))); D(B--MP("P",D(P))); D(MP("M",M,NW)); MP("20",(B+D)/2,ENE); MP("11",(C+D)/2,ENE);  [/asy] Let $D'$ be on $\overline{AC}$ such that $BP \parallel DD'$ . It follows that $\triangle BPC \sim \triangle DD'C$ , so \[\frac{PC}{D'C} = 1 + \frac{BD}{DC} = 1 + \frac{AB}{AC} = \frac{31}{11}\] by the Angle Bisector Theorem . Similarly, we see by the Midline Theorem that $AP = PD'$ . Thus, \[\frac{CP}{PA} = \frac{1}{\frac{PD'}{PC}} = \frac{1}{1 - \frac{D'C}{PC}} = \frac{31}{20},\] and $m+n = \boxed{51}$

Answer:  51
Problem: In triangle $ABC$ $AB=20$ and $AC=11$ . The angle bisector of $\angle A$ intersects $BC$ at point $D$ , and point $M$ is the midpoint of $AD$ . Let $P$ be the point of the intersection of $AC$ and $BM$ . The ratio of $CP$ to $PA$ can be expressed in the form $\dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: Assign mass points as follows: by Angle-Bisector Theorem, $BD / DC = 20/11$ , so we assign $m(B) = 11, m(C) = 20, m(D) = 31$ . Since $AM = MD$ , then $m(A) = 31$ , and $\frac{CP}{PA} = \frac{m(A) }{ m(C)} = \frac{31}{20}$ , so $m+n = \boxed{51}$

Answer:  51
Problem: In triangle $ABC$ $AB=20$ and $AC=11$ . The angle bisector of $\angle A$ intersects $BC$ at point $D$ , and point $M$ is the midpoint of $AD$ . Let $P$ be the point of the intersection of $AC$ and $BM$ . The ratio of $CP$ to $PA$ can be expressed in the form $\dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: By Menelaus' Theorem on $\triangle ACD$ with transversal $PB$ \[1 = \frac{CP}{PA} \cdot \frac{AM}{MD} \cdot \frac{DB}{CB} = \frac{CP}{PA} \cdot \left(\frac{1}{1+\frac{AC}{AB}}\right) \quad \Longrightarrow \quad \frac{CP}{PA} = \frac{31}{20}.\] So $m+n = \boxed{051}$

Answer:  051
Problem: In triangle $ABC$ $AB=20$ and $AC=11$ . The angle bisector of $\angle A$ intersects $BC$ at point $D$ , and point $M$ is the midpoint of $AD$ . Let $P$ be the point of the intersection of $AC$ and $BM$ . The ratio of $CP$ to $PA$ can be expressed in the form $\dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: We will use barycentric coordinates. Let $A = (1, 0, 0)$ $B = (0, 1, 0)$ $C = (0, 0, 1)$ . By the Angle Bisector Theorem $D = [0:11:20] = \left(0, \frac{11}{31}, \frac{20}{31}\right)$ . Since $M$ is the midpoint of $AD$ $M = \frac{A + D}{2} = \left(\frac{1}{2}, \frac{11}{62}, \frac{10}{31}\right)$ . Therefore, the equation for line BM is $20x = 31z$ . Let $P = (x, 0, 1-x)$ . Using the equation for $BM$ , we get \[20x = 31(1-x)\] \[x = \frac{31}{51}\] Therefore, $\frac{CP}{PA} = \frac{1-x}{x} = \frac{31}{20}$ so the answer is $\boxed{051}$

Answer:  051
Problem: In triangle $ABC$ $AB=20$ and $AC=11$ . The angle bisector of $\angle A$ intersects $BC$ at point $D$ , and point $M$ is the midpoint of $AD$ . Let $P$ be the point of the intersection of $AC$ and $BM$ . The ratio of $CP$ to $PA$ can be expressed in the form $\dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: Let $DC=x$ . Then by the Angle Bisector Theorem, $BD=\frac{20}{11}x$ . By the Ratio Lemma, we have that $\frac{PC}{AP}=\frac{\frac{31}{11}x\sin\angle PBC}{20\sin\angle ABP}.$ Notice that $[\triangle BAM]=[\triangle BMD]$ since their bases have the same length and they share a height. By the sin area formula, we have that \[\frac{1}{2}\cdot20\cdot BM\cdot \sin\angle ABP=\frac{1}{2}\cdot \frac{20}{11}x\cdot BM\cdot\sin\angle PBC.\] Simplifying, we get that $\frac{\sin\angle PBC}{\sin\angle ABP}=\frac{11}{x}.$ Plugging this into what we got from the Ratio Lemma, we have that $\frac{PC}{AP}=\frac{31}{20}\implies\boxed{051.}$

Answer:  051.
Problem: In triangle $ABC$ $AB=20$ and $AC=11$ . The angle bisector of $\angle A$ intersects $BC$ at point $D$ , and point $M$ is the midpoint of $AD$ . Let $P$ be the point of the intersection of $AC$ and $BM$ . The ratio of $CP$ to $PA$ can be expressed in the form $\dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: First, we will find $\frac{MP}{BP}$ . By Menelaus on $\triangle BDM$ and the line $AC$ , we have \[\frac{BC}{CD}\cdot\frac{DA}{AM}\cdot\frac{MP}{PB}=1\implies \frac{62MP}{11BP}=1\implies \frac{MP}{BP}=\frac{11}{62}.\] This implies that $\frac{MB}{BP}=1-\frac{MP}{BP}=\frac{51}{62}$ . Then, by Menelaus on $\triangle AMP$ and line $BC$ , we have \[\frac{AD}{DM}\cdot\frac{MB}{BP}\cdot\frac{PC}{CA}=1\implies \frac{PC}{CA}=\frac{31}{51}.\] Therefore, $\frac{PC}{AP}=\frac{31}{51-31}=\frac{31}{20}.$ The answer is $\boxed{051}$ . -brainiacmaniac31

Answer:  051
Problem: In triangle $ABC$ $AB=20$ and $AC=11$ . The angle bisector of $\angle A$ intersects $BC$ at point $D$ , and point $M$ is the midpoint of $AD$ . Let $P$ be the point of the intersection of $AC$ and $BM$ . The ratio of $CP$ to $PA$ can be expressed in the form $\dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: Assume $ABC$ is a right triangle at $A$ . Line $AD = x$ and $BC = \tfrac{-11}{20}x + 11$ . These two lines intersect at $D$ which have coordinates $(\frac{220}{31},\frac{220}{31})$ and thus $M$ has coordinates $(\frac{110}{31},\frac{110}{31})$ . Thus, the line $BM = \tfrac{11}{51} \cdot (20-x)$ . When $x = 0$ $P$ has $y$ coordinate equal to $\frac{11\cdot20}{51} \frac{AP + CP}{AP} = 1 + \frac{CP}{AP}$ $\tfrac{51}{20} = 1 + \frac{CP}{AP},$ which equals ${\tfrac{31}{20}},$ giving an answer of $\boxed{51}.$

Answer:  51
Problem: In triangle $ABC$ $AB=20$ and $AC=11$ . The angle bisector of $\angle A$ intersects $BC$ at point $D$ , and point $M$ is the midpoint of $AD$ . Let $P$ be the point of the intersection of $AC$ and $BM$ . The ratio of $CP$ to $PA$ can be expressed in the form $\dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: We start by using Menelaus' theorem on $\triangle ABD$ and $EC$ .
So, we see that $\frac{BC}{DC}\cdot\frac{DM}{AM}\cdot\frac{AE}{EB}=1$ .
By Angle Bisector theorem, $\frac{BC}{DC}=\frac{31}{11}$ , and therefore after plugging in our values we get $\frac{AE}{EB}=\frac{11}{31}$ .
Then, by Ceva's on the whole figure, we have $\frac{CP}{PA}\cdot\frac{AE}{EB}\cdot\frac{BD}{DC}=1$ .
Plugging in our values, we get $\frac{CP}{PA}=\frac{31}{20}$ , giving an answer of $\boxed{51}$ .
~ESAOPS

Answer:  51
Problem: The sum of the first $2011$ terms of a geometric sequence is $200$ . The sum of the first $4022$ terms is $380$ . Find the sum of the first $6033$ terms.

Solution: Since the sum of the first $2011$ terms is $200$ , and the sum of the first $4022$ terms is $380$ , the sum of the second $2011$ terms is $180$ .
This is decreasing from the first 2011, so the common ratio is less than one.
Because it is a geometric sequence and the sum of the first 2011 terms is $200$ , second $2011$ is $180$ , the ratio of the second $2011$ terms to the first $2011$ terms is $\frac{9}{10}$ . Following the same pattern, the sum of the third $2011$ terms is $\frac{9}{10}*180 = 162$
Thus, $200+180+162=542$ , so the sum of the first $6033$ terms is $\boxed{542}$

Answer:  542
Problem: The sum of the first $2011$ terms of a geometric sequence is $200$ . The sum of the first $4022$ terms is $380$ . Find the sum of the first $6033$ terms.

Solution: Solution by e_power_pi_times_i
The sum of the first $2011$ terms can be written as $\dfrac{a_1(1-k^{2011})}{1-k}$ , and the first $4022$ terms can be written as $\dfrac{a_1(1-k^{4022})}{1-k}$ . Dividing these equations, we get $\dfrac{1-k^{2011}}{1-k^{4022}} = \dfrac{10}{19}$ . Noticing that $k^{4022}$ is just the square of $k^{2011}$ , we substitute $x = k^{2011}$ , so $\dfrac{1}{x+1} = \dfrac{10}{19}$ . That means that $k^{2011} = \dfrac{9}{10}$ . Since the sum of the first $6033$ terms can be written as $\dfrac{a_1(1-k^{6033})}{1-k}$ , dividing gives $\dfrac{1-k^{2011}}{1-k^{6033}}$ . Since $k^{6033} = \dfrac{729}{1000}$ , plugging all the values in gives $\boxed{542}$

Answer:  542
Problem: The sum of the first $2011$ terms of a geometric sequence is $200$ . The sum of the first $4022$ terms is $380$ . Find the sum of the first $6033$ terms.

Solution: The sum of the first 2011 terms of the sequence is expressible as $a_1 + a_1r + a_1r^2 + a_1r^3$ .... until $a_1r^{2010}$ . The sum of the 2011 terms following the first 2011 is expressible as $a_1r^{2011} + a_1r^{2012} + a_1r^{2013}$ .... until $a_1r^{4021}$ . Notice that the latter sum of terms can be expressed as $(r^{2011})(a_1 + a_1r + a_1r^2 + a_1r^3...a_1r^{2010})$ . We also know that the latter sum of terms can be obtained by subtracting 200 from 380, which then means that $r^{2011} = 9/10$ . The terms from 4023 to 6033 can be expressed as $(r^{4022})(a_1 + a_1r + a_1r^2 + a_1r^3...a_1r^{2010})$ , which is equivalent to $((9/10)^2)(200) = 162$ . Adding 380 and 162 gives the answer of $\boxed{542}$

Answer:  542
Problem: Define an ordered quadruple of integers $(a, b, c, d)$ as interesting if $1 \le a<b<c<d \le 10$ , and $a+d>b+c$ . How many interesting ordered quadruples are there?

Solution: We first start out when the value of $a=1$
Doing casework, we discover that $d=5,6,7,8,9,10$ . We quickly find a pattern.
Now, doing this for the rest of the values of $a$ and $d$ , we see that the answer is simply:
$(1)+(2)+(1+3)+(2+4)+(1+3+5)+(2+4+6)+(1)+(2)+(1+3)+(2+4)$ $+(1+3+5)+(1)+(2)+(1+3)+(2+4)+(1)+(2)+(1+3)+(1)+(2)+(1)=\boxed{080}$

Answer:  080
Problem: Define an ordered quadruple of integers $(a, b, c, d)$ as interesting if $1 \le a<b<c<d \le 10$ , and $a+d>b+c$ . How many interesting ordered quadruples are there?

Solution: Notice that if $a+d>b+c$ , then $(11-a)+(11-d)<(11-b)+(11-c)$ , so there is a bijection between the number of ordered quadruples with $a+d>b+c$ and the number of ordered quadruples with $a+d<b+c$
Quick counting gives that the number of ordered quadruples with $a+d=b+c$ is 50. To count this, consider our numbers $1, 2, 3, 4, 5, 6, 7, 8, 9, 10$ . Notice that if, for example, $a+d=b+c=8$ , that the average of $a,d$ and $b,c$ must both be $4$ . In this way, there is a symmetry for this case, centered at $4$ . If instead, say, $a+d=b+c=7$ , an odd number, then there is symmetry with $(a,d);(b,c)$ about $3.5$ . Further, the number of cases for each of these centers of symmetry correspond to a triangular number. Eg centered at $2.5,3,8,8.5$ , there is $1$ case for each and so on, until centered at $5.5$ , there are $10$ possible cases. Adding these all, we have $2(1+3+6)+10=50$
Thus the answer is $\frac{\binom{10}{4}-50}{2} = \boxed{080}.$

Answer:  080