Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Let $\triangle ABC$ be a right triangle with right angle at $C.$ Let $D$ and $E$ be points on $\overline{AB}$ with $D$ between $A$ and $E$ such that $\overline{CD}$ and $\overline{CE}$ trisect $\angle C.$ If $\frac{DE}{BE} = \frac{8}{15},$ then $\tan B$ can be written as $\frac{m \sqrt{p}}{n},$ where $m$ and $n$ are relatively prime positive integers, and $p$ is a positive integer not divisible by the square of any prime. Find $m+n+p.$

Solution: (This solution avoids advanced trigonometry)
Let $X$ be the foot of the perpendicular from $D$ to $\overline{BC}$ , and let $Y$ be the foot of the perpendicular from $E$ to $\overline{BC}$
Now let $EY=x$ . Clearly, triangles $EYB$ and $DXB$ are similar with $\frac{BE}{BD}=\frac{15}{15+8}=\frac{15}{23}=\frac{EY}{DX}$ , so $DX=\frac{23}{15}x$
Since triangles $CDX$ and $CEY$ are 30-60-90 right triangles, we can easily find other lengths in terms of $x$ . For example, we see that $CY=x\sqrt{3}$ and $CX=\frac{\frac{23}{15}x}{\sqrt{3}}=\frac{23\sqrt{3}}{45}x$ . Therefore $XY=CY-CX=x\sqrt{3}-\frac{23\sqrt{3}}{45}x=\frac{22\sqrt{3}}{45}x$
Again using the fact that triangles $EYB$ and $DXB$ are similar, we see that $\frac{BX}{BY}=\frac{XY+BY}{BY}=\frac{XY}{BY}+1=\frac{23}{15}$ , so $BY=\frac{15}{8}XY=\frac{15}{8}*\frac{22\sqrt{3}}{45}=\frac{11\sqrt{3}}{2}$
Thus $\tan \angle B = \frac{x}{\frac{11\sqrt{3}}{12}x}=\frac{4\sqrt{3}}{11}$ , and our answer is $4+3+11=\boxed{018}$

Answer:  018
Problem: Let $\triangle ABC$ be a right triangle with right angle at $C.$ Let $D$ and $E$ be points on $\overline{AB}$ with $D$ between $A$ and $E$ such that $\overline{CD}$ and $\overline{CE}$ trisect $\angle C.$ If $\frac{DE}{BE} = \frac{8}{15},$ then $\tan B$ can be written as $\frac{m \sqrt{p}}{n},$ where $m$ and $n$ are relatively prime positive integers, and $p$ is a positive integer not divisible by the square of any prime. Find $m+n+p.$

Solution: (Another solution without trigonometry)
Extend $CD$ to point $F$ such that $\overline{AF} \parallel \overline{CB}$ . It is then clear that $\triangle AFD$ is similar to $\triangle BCD$
Let $AC=p$ $BC=q$ . Then $\tan \angle B = p/q$
With the Angle Bisector Theorem, we get that $CD=\frac{8}{15}q$ . From 30-60-90 $\triangle CAF$ , we get that $AF=\frac{1}{\sqrt{3}}p$ and $FD=FC-CD=\frac{2}{\sqrt{3}}p-\frac{8}{15}q$
From $\triangle AFD \sim \triangle BCD$ , we have that $\frac{FD}{CD}=\frac{FA}{CB}=\frac{\frac{2}{\sqrt{3}}p-\frac{8}{15}q}{\frac{8}{15}q}=\frac{\frac{1}{\sqrt{3}}p}{q}$ . Simplifying yields $\left(\frac{p}{q}\right)\left(\frac{2\sqrt{3}}{3}*\frac{15}{8}-\frac{\sqrt{3}}{3}\right)=1$ , and $\tan \angle B=\frac{p}{q}=\frac{4\sqrt{3}}{11}$ , so our answer is $4+3+11=\boxed{018}$

Answer:  018
Problem: Let $\triangle ABC$ be a right triangle with right angle at $C.$ Let $D$ and $E$ be points on $\overline{AB}$ with $D$ between $A$ and $E$ such that $\overline{CD}$ and $\overline{CE}$ trisect $\angle C.$ If $\frac{DE}{BE} = \frac{8}{15},$ then $\tan B$ can be written as $\frac{m \sqrt{p}}{n},$ where $m$ and $n$ are relatively prime positive integers, and $p$ is a positive integer not divisible by the square of any prime. Find $m+n+p.$

Solution: Let $CB = 1$ , and let the feet of the altitudes from $D$ and $E$ to $\overline{CB}$ be $D'$ and $E'$ , respectively. Also, let $DE = 8k$ and $EB = 15k$ . We see that $BD' = 15k\cos B$ and $BE' = 23k\cos B$ by right triangles $\triangle{BDD'}$ and $\triangle{BEE'}$ . From this we have that $D'E' = 8k\cos B$ . With the same triangles we have $DD' = 23k\sin B$ and $EE' = 15k\sin B$ . From 30-60-90 triangles $\triangle{CDD'}$ and $\triangle{CEE'}$ , we see that $CD' = \frac{23k\sqrt{3}\sin B}{3}$ and $CE' = 15k\sqrt{3}\sin B$ , so $D'E' = \frac{22k\sqrt{3}\sin B}{3}$ . From our two values of $D'E'$ we get: $8k\cos B = \frac{22k\sqrt{3}\sin B}{3}$
$\frac{\sin B}{\cos B} = \frac{8k}{\frac{22k\sqrt{3}}{3}} = \tan B$
$\tan B = \frac{8}{\frac{22\sqrt{3}}{3}} = \frac{24}{22\sqrt{3}} = \frac{8\sqrt{3}}{22} = \frac{4\sqrt{3}}{11}$ Our answer is then $4+3+11 = \boxed{018}$

Answer:  018
Problem: Three concentric circles have radii $3,$ $4,$ and $5.$ An equilateral triangle with one vertex on each circle has side length $s.$ The largest possible area of the triangle can be written as $a + \tfrac{b}{c} \sqrt{d},$ where $a,$ $b,$ $c,$ and $d$ are positive integers, $b$ and $c$ are relatively prime, and $d$ is not divisible by the square of any prime. Find $a+b+c+d.$

Solution: Reinterpret the problem in the following manner. Equilateral triangle $ABC$ has a point $X$ on the interior such that $AX = 5,$ $BX = 4,$ and $CX = 3.$ $60^\circ$ counter-clockwise rotation about vertex $A$ maps $X$ to $X'$ and $B$ to $C.$
Note that angle $XAX'$ is $60$ and $XA = X'A = 5$ which tells us that triangle $XAX'$ is equilateral and that $XX' = 5.$ We now notice that $XC = 3$ and $X'C = 4$ which tells us that angle $XCX'$ is $90$ because there is a $3$ $4$ $5$ Pythagorean triple. Now note that $\angle ABC + \angle ACB = 120^\circ$ and $\angle XCA + \angle XBA = 90^\circ,$ so $\angle XCB+\angle XBC = 30^\circ$ and $\angle BXC = 150^\circ.$ Applying the law of cosines on triangle $BXC$ yields
\[BC^2 = BX^2+CX^2 - 2 \cdot BX \cdot CX \cdot \cos150^\circ = 4^2+3^2-24 \cdot \frac{-\sqrt{3}}{2} = 25+12\sqrt{3}\]
and thus the area of $ABC$ equals \[\frac{\sqrt{3}}{4}\cdot BC^2 = 25\frac{\sqrt{3}}{4}+9.\]
so our final answer is $3+4+25+9 = \boxed{041}.$

Answer:  041
Problem: Three concentric circles have radii $3,$ $4,$ and $5.$ An equilateral triangle with one vertex on each circle has side length $s.$ The largest possible area of the triangle can be written as $a + \tfrac{b}{c} \sqrt{d},$ where $a,$ $b,$ $c,$ and $d$ are positive integers, $b$ and $c$ are relatively prime, and $d$ is not divisible by the square of any prime. Find $a+b+c+d.$

Solution: Let's call the circle center $X$ . It has a distance of 3, 4, 5 to an equilateral triangle $LMN$ . Consider $X$ ’s pedal triangle $ABC$ . Since $X$ ’s antipedal triangle is equilateral, $X$ must be the one of the isogonic centers of $\triangle{ABC}$ . We’ll take the one inside $ABC$ , i.e., the Fermat point, because it leads to larger $\triangle LMN$ . Now we construct the three equilateral triangles $ABD$ $ACE$ , and $BCF$ , the same way the Fermat point is constructed. Then we have $\angle DXE = \angle EXF = \angle FXE = 120$ . Since $AEMCX$ is concyclic with $XM$ =4 as diameter, we have $AC=4\sin(60)$ . Similarly, $AB=3\sin(60)$ , and $BC=5\sin(60)$ . So $\triangle ABC$ is a 3-4-5 right triangle with $\angle BAC=90$ . With some more angle chasing we get \[\angle MXC+\angle LXB = \angle MAC + \angle LAB = 180 – \angle BAC = 90\] \[\angle LXM = 360 – (\angle MXC + \angle LXB + \angle BXC) = 360 –(90+120)=150\] By Law of Cosines, we have \[LM^2 = 3^2+4^2-2*3*4\cos(150)=25+12\sqrt 3\] And the area follows. \[[LMN] = \frac{25}{4}\sqrt{3} + 9; \boxed{041}.\] By Mathdummy

Answer:  041
Problem: Three concentric circles have radii $3,$ $4,$ and $5.$ An equilateral triangle with one vertex on each circle has side length $s.$ The largest possible area of the triangle can be written as $a + \tfrac{b}{c} \sqrt{d},$ where $a,$ $b,$ $c,$ and $d$ are positive integers, $b$ and $c$ are relatively prime, and $d$ is not divisible by the square of any prime. Find $a+b+c+d.$

Solution: [asy] import cse5; size(350); defaultpen(linewidth(0.6)+fontsize(12));   pair O=origin; pair Op,Bp,Cp;  path c3 = CR(O,3), c4 = CR(O,4), c5 = CR(O,5);  var theta=55.75; pair A = 3*dir(theta), Ap = rotate(150,O)*A, F=IP(c4,O--2*Ap), C=rotate(60,A)*F, E=rotate(60,A)*C, B=IP(c5,O--E), D=foot(C,O,E);    filldraw(A--O--B--cycle, rgb(255,255,200)); draw(c3, cyan);  draw(c4, green);  draw(c5, purple); draw(A--F--C--A--E--C, red+0.8);  draw(D--B--C--D--O--C, purple+0.6); draw(A--O^^E--B, cyan+0.6); draw(A--B^^O--F, heavygreen+0.6); draw(A--D); dot("$A$",A,N); dot("$O$",O,dir(A-B)); dot("$F$",F,dir(F-E)); dot("$C$",C,SE); dot("$B$",B,dir(-45)); dot("$D$",D,dir(B-C)); dot("$E$",E,dir(E-F)); label("$s$",A--F,dir(A-E),red); label("$s$",F--C,dir(C-B),red); label("$s$",A--C,dir(C-E),red); label("$s$",A--E,dir(E-C),red); label("$s$",C--E,dir(C),red); label("$x$",B--E,down,royalblue); label("$x$",O--A,dir(B-A),royalblue); label("$y$",A--B,dir(F-A),heavygreen); label("$y$",O--F,dir(C),heavygreen); label("$z$",C--B,dir(E-A),purple); label("$z$",O--C,dir(F),purple); label("$m$",A--D,dir(0),blue+fontsize(9)); markscalefactor=0.03; draw(rightanglemark(C,D,B), gray); MA("\varphi",A,D,O,0.25,blue); [/asy] We have $x=3$ $y=4$ , and $z=5$ . Because $AD=m$ is the median of $\triangle AOB$ , by Stewart's Theorem we have \[m^2=\frac 12 (x^2+y^2)-\frac 14 \cdot z^2\quad \Rightarrow \quad m = \frac 52.\] Because $CD$ is the altitude of equilateral triangle $OBC$ , we have $CD=\frac{\sqrt{3}}2\cdot z$ . Then in $\triangle ADC$ , we have $\angle ADC=\varphi+90^\circ$ , so $\cos(\angle ADC)=-\sin\varphi$ , and the Law of Cosines gives \[s^2=m^2+\frac 34\cdot z^2 + mz\sqrt{3}\sin\varphi = 25 \left(1+\frac {\sqrt{3}}{2}\cdot\sin\varphi  \right)\] To calculate $\sin\phi$ we apply the Law of Cosines to $\triangle ADO$ to get \[mz\cos\varphi = m^2+\frac 14\cdot z^2 - x^2 \quad \Rightarrow \quad \cos\varphi = \frac 7{25}\quad \Rightarrow \quad \sin\varphi = \frac {24}{25}.\] Finally, we get $s^2=25+12\sqrt{3}$ and and thus the area of $\triangle ABC$ equals \[\frac{\sqrt{3}}{4}\cdot s^2 = 25\frac{\sqrt{3}}{4}+9.\] so our final answer is $3+4+25+9 = \boxed{041}.$

Answer:  041
Problem: Complex numbers $a,$ $b,$ and $c$ are zeros of a polynomial $P(z) = z^3 + qz + r,$ and $|a|^2 + |b|^2 + |c|^2 = 250.$ The points corresponding to $a,$ $b,$ and $c$ in the complex plane are the vertices of a right triangle with hypotenuse $h.$ Find $h^2.$

Solution: By Vieta's formula, the sum of the roots is equal to 0, or $a+b+c=0$ . Therefore, $\frac{(a+b+c)}{3}=0$ . Because the centroid of any triangle is the average of its vertices, the centroid of this triangle is the origin. Let one leg of the right triangle be $x$ and the other leg be $y$ . Without the loss of generality, let $\overline{ac}$ be the hypotenuse. The magnitudes of $a$ $b$ , and $c$ are just $\frac{2}{3}$ of the medians because the origin, or the centroid in this case, cuts the median in a ratio of $2:1$ . So, $|a|^2=\frac{4}{9}\cdot((\frac{x}{2})^2+y^2)=\frac{x^2}{9}+\frac{4y^2}{9}$ because $|a|$ is two thirds of the median from $a$ . Similarly, $|c|^2=\frac{4}{9}\cdot(x^2+(\frac{y}{2})^2)=\frac{4x^2}{9}+\frac{y^2}{9}$ . The median from $b$ is just half the hypotenuse because the median of any right triangle is just half the hypotenuse. So, $|b|^2=\frac{4}{9}\cdot\frac{x^2+y^2}{4}=\frac{x^2}{9}+\frac{y^2}{9}$ . Hence, $|a|^2+|b|^2+|c|^2=\frac{6x^2+6y^2}{9}=\frac{2x^2+2y^2}{3}=250$ . Therefore, $h^2=x^2+y^2=\frac{3}{2}\cdot250=\boxed{375}$

Answer:  375
Problem: Complex numbers $a,$ $b,$ and $c$ are zeros of a polynomial $P(z) = z^3 + qz + r,$ and $|a|^2 + |b|^2 + |c|^2 = 250.$ The points corresponding to $a,$ $b,$ and $c$ in the complex plane are the vertices of a right triangle with hypotenuse $h.$ Find $h^2.$

Solution: Assume $q$ and $r$ are real, so at least one of $a,$ $b,$ and $c$ must be real, with the remaining roots being pairs of complex conjugates. Without loss of generality, we assume $a$ is real and $b$ and $c$ are $x + yi$ and $x - yi$ respectively. By symmetry, the triangle described by $a,$ $b,$ and $c$ must be isosceles and is thus an isosceles right triangle with hypotenuse $\overline{bc}.$ Now since $P(z)$ has no $z^2$ term, we must have $a+b+c = a + (x + yi) + (x - yi) = 0$ and thus $a = -2x.$ Also, since the length of the altitude from the right angle of an isosceles triangle is half the length of the hypotenuse, $a-x=y$ and thus $y=-3x.$ We can then solve for $x$
\begin{align*} |a|^2 + |b|^2 + |c|^2 &= 250\\ |-2x|^2 + |x-3xi|^2 + |x+3xi|^2 &= 250\\ 4x^2 + (x^2 + 9x^2) + (x^2 + 9x^2) &= 250\\ x^2 &= \frac{250}{24} \end{align*}
Now $h$ is the distance between $b$ and $c,$ so $h = 2y = -6x$ and thus $h^2 = 36x^2 = 36 \cdot \frac{250}{24} = \boxed{375.}$

Answer:  375.
Problem: Complex numbers $a,$ $b,$ and $c$ are zeros of a polynomial $P(z) = z^3 + qz + r,$ and $|a|^2 + |b|^2 + |c|^2 = 250.$ The points corresponding to $a,$ $b,$ and $c$ in the complex plane are the vertices of a right triangle with hypotenuse $h.$ Find $h^2.$

Solution: Let the roots $a$ $b$ , and $c$ each be represented by complex numbers $m + ni$ $p + qi$ , and $r + ti$ . 
By Vieta's formulas, their sum is 0. Breaking into real and imaginary components, we get:
$m + p + r = 0$
$n + q + t = 0$
And, we know that the sum of the squares of the magnitudes of each is 250, so
$m^2 + n^2 + p^2 + q^2 + r^2 + t^2 = 250$
Given the complex plane, we set each of these complex numbers to points: $(m, n)$ $(p, q)$ $(r, t)$ . 
WLOG let $(r, t)$ be the vertex opposite the hypotenuse.
If the three points form a right triangle, the vectors from $(r, t)$ to $(m, m)$ and $(p, q)$ 's dot product is 0. $mp + r^2 - r(m + p) + nq + t^2 - t(n + q) = 0$
Substituting $m + p + r = 0$ and likewise, simplifying: $mp + 2r^2 + nq + 2t^2 = 0$
Rearranging we get: $r^2 + t^2 = -\frac{mp + nq}{2}$
The answer is the distance from $(m, n)$ to $(p, q)$ $m^2 + n^2 + p^2 + q^2 - 2(mp + nq)$ .
Substituting the equation equal to 250,
$= 250 - r^2 - t^2 - 2(mp + nq)$ $= 250 + \frac{mp + nq}{2} - 2(mp + nq)$ $= 250 - \frac{3}{2} \cdot (mp + nq)$
Taking our original equations summing to 0, and squaring each we get:
$n + q = -t$ $m + p = -r$
$n^2 + 2nq + q^2 = t^2$ $m^2 + 2mp + p^2 = r^2$
Adding, we get:
$m^2 + n^2 + p^2 + q^2 + 2(mp + nq) = r^2 + t^2$
Substituting again we obtain:
$250 - r^2 - t^2 + 2(mp + nq) = r^2 + t^2$ $2(r^2 + t^2) = 250 + 2(mp + nq)$ $r^2 + t^2 = 125 + (mp + nq)$
Substituting the equivalence of $r^2 + t^2$
$-\frac{mp + nq}{2} = 125 + (mp + nq)$
Solving for $mp + nq$ , we find it equal to $-\frac{250}{3}$
Substituting this value into our answer expression, we get:
$250 - \frac{3}{2} \cdot (-\frac{250}{3})$ , 
Answer = $\boxed{375}$

Answer:  375
Problem: Complex numbers $a,$ $b,$ and $c$ are zeros of a polynomial $P(z) = z^3 + qz + r,$ and $|a|^2 + |b|^2 + |c|^2 = 250.$ The points corresponding to $a,$ $b,$ and $c$ in the complex plane are the vertices of a right triangle with hypotenuse $h.$ Find $h^2.$

Solution: As noted in the previous solutions, $a+b+c = 0$ .
Let $a = a_1+a_2 i$ $b = b_1+b_2 i$ $c = c_1+c_2 i$ and we have $a_1 + b_1 + c_1 = a_2 + b_2 + c_2 = 0$ . 
Then the given $|a|^2 + |b|^2 + |c|^2 = 250$ translates to $\sum_{} ( {a_1}^2 + {a_2}^2 ) = 250.$ Note that in a right triangle, the sum of the squares of the three sides is equal to two times the square of the hypotenuse, by the pythagorean theorem.
Thus, we have \[2h^2 = (a_1 - b_1)^2 + (a_2 - b_2)^2 + (b_1 - c_1)^2 + (b_2 - c_2)^2 + (a_1 - c_1)^2 + (a_2 - c_2)^2\] \[= 2 \left( \sum_{} ( {a_1}^2 + {a_2}^2 )  \right) - 2 \left( \sum_{cyc} a_1 b_1 + \sum_{cyc} a_2 b_2 \right)\] \[= 500 - \left( (a_1 + b_1 + c_1)^2 + (a_2 + b_2 + c_2)^2 -  \sum_{cyc} ( {a_1}^2 + {a_2}^2 )     \right)\] \[= 500 - (0^2 + 0^2 - 250)\] so $h^2 = \boxed{375}$ and we may conclude.
~ rzlng

Answer:  375
Problem: Complex numbers $a,$ $b,$ and $c$ are zeros of a polynomial $P(z) = z^3 + qz + r,$ and $|a|^2 + |b|^2 + |c|^2 = 250.$ The points corresponding to $a,$ $b,$ and $c$ in the complex plane are the vertices of a right triangle with hypotenuse $h.$ Find $h^2.$

Solution: First, note that the roots of this cubic will be $a, b$ and $-(a+b)$ due to Vieta's, which means that the sum of the roots are 0.
Now, we use some god level wishful thinking. It would really be nice if one of these roots was on the real axis, and it was a right isosceles triangle, because that would be very convenient and easy to work with. The neat part is that it actually works
Set one of the roots as $r$ , where $r$ is any real number. WLOG, assume that this is the right angle. With symmetry to respect of the x axis (because symmetry makes the imaginary parts of the other 2 roots cancel out, besides the fact that complex conjugate root theorem forces it). This way, we can set the other 2 roots as $\frac{r}{2}+ni$ and $\frac{r}{2}-ni$ , making the roots add up to 0.
Now, as we want the roots to satisfy the original condition (right triangle) we are going to have to set an equation to find $n$ out. We use the fact that it is an isosceles right triangle to find that $\frac{3r}{2}=n$ , which means that the 2 other roots are now $\frac{r}{2}+\frac{3r}{2}i$ and $\frac{r}{2}-\frac{3r}{2}i$
Now we use the fact that $|a|^2+|b|^2+|c|^2=250$ . Clearly one of these is $r$ away from the origin, so that gets $r^2$ , and then we get $2*\frac{r}{2}^2+\frac{3r}{2}^2$ which gets us $5r^2$ , getting $r^2+5r^2=250$ , so $r=\sqrt{\frac{250}{6}}$ . So the final answer comes out to \[(\frac{250}{6}*\frac{9}{4}*2)^2=\boxed{375}\]

Answer:  375
Problem: Complex numbers $a,$ $b,$ and $c$ are zeros of a polynomial $P(z) = z^3 + qz + r,$ and $|a|^2 + |b|^2 + |c|^2 = 250.$ The points corresponding to $a,$ $b,$ and $c$ in the complex plane are the vertices of a right triangle with hypotenuse $h.$ Find $h^2.$

Solution: As shown in the other solutions, $a+b+c = 0$
Without loss of generality, let $b$ be the complex number opposite the hypotenuse.
Note that there is an isomorphism between $\mathbb{C}$ under $+$ and $\mathbb{R}^2$ under $+$
Let $\Vec{a}$ $\Vec{b}$ , and $\Vec{c}$ be the corresponding vectors to $a$ $b$ , and $c$
Thus $\Vec{a} + \Vec{b} + \Vec{c} = \Vec{0}$
$\Rightarrow 0 = \Vec{0}\cdot \Vec{0} = (\Vec{a} + \Vec{b} + \Vec{c})\cdot (\Vec{a} + \Vec{b} + \Vec{c}) = \Vec{a}\cdot \Vec{a} + \Vec{b}\cdot \Vec{b} + \Vec{c}\cdot \Vec{c} + 2(\Vec{a}\cdot \Vec{b} + \Vec{a}\cdot \Vec{c} + \Vec{b}\cdot \Vec{c})$
Now $|a|^2 + |b|^2 + |c|^2 = 250$ implies that $\lVert \Vec{a}\rVert^2 + \lVert \Vec{b}\rVert^2 + \lVert \Vec{c}\rVert^2 = 250$
$\Rightarrow \Vec{a}\cdot \Vec{a} + \Vec{b}\cdot \Vec{b} + \Vec{c}\cdot \Vec{c} = \lVert \Vec{a}\rVert^2 + \lVert \Vec{b}\rVert^2 + \lVert \Vec{c}\rVert^2 = 250$
Also note that because there is a right angle at $b$ $\Vec{a} - \Vec{b}$ and $\Vec{c} - \Vec{b}$ are perpendicular.
$\Rightarrow (\Vec{a} - \Vec{b})\cdot (\Vec{c} - \Vec{b}) = 0$
$\Rightarrow 0 = (\Vec{a} - \Vec{b})\cdot (\Vec{c} - \Vec{b}) = \Vec{a}\cdot \Vec{c} + \Vec{b} \cdot \Vec{b} - \Vec{a} \cdot \Vec{b} - \Vec{b} \cdot \Vec{c}$
Note that $h^2 = |a-c|^2$
$\Rightarrow h^2 = \lVert \Vec{a} - \Vec{c} \rVert^2 = (\Vec{a} - \Vec{c})\cdot (\Vec{a} - \Vec{c}) = \Vec{a} \cdot \Vec{a} + \Vec{c}\cdot \Vec{c} - \Vec{a}\cdot \Vec{c} - \Vec{a}\cdot \Vec{c} = \Vec{a} \cdot \Vec{a} + \Vec{c}\cdot \Vec{c} - 2\Vec{a}\cdot \Vec{c}$
$\Rightarrow 0 = \Vec{a}\cdot \Vec{a} + \Vec{b}\cdot \Vec{b} + \Vec{c}\cdot \Vec{c} + 2(\Vec{a}\cdot \Vec{b} + \Vec{a}\cdot \Vec{c} + \Vec{b}\cdot \Vec{c}) = 250 + 2(\Vec{a}\cdot \Vec{b} + \Vec{a}\cdot \Vec{c} + \Vec{b}\cdot \Vec{c})$
$\Rightarrow -250 = 2(\Vec{a}\cdot \Vec{b} + \Vec{a}\cdot \Vec{c} + \Vec{b}\cdot \Vec{c})$
$\Rightarrow -125 = \Vec{a}\cdot \Vec{b} + \Vec{a}\cdot \Vec{c} + \Vec{b}\cdot \Vec{c}$
$\Rightarrow -125 = -125 + 0 = (\Vec{a}\cdot \Vec{b} + \Vec{a}\cdot \Vec{c} + \Vec{b}\cdot \Vec{c}) + (\Vec{a}\cdot \Vec{c} + \Vec{b} \cdot \Vec{b} - \Vec{a} \cdot \Vec{b} - \Vec{b} \cdot \Vec{c}) = 2\Vec{a}\cdot \Vec{c} + \Vec{b} \cdot \Vec{b}$
$\Rightarrow 125 = - \Vec{b} \cdot \Vec{b} - 2\Vec{a}\cdot \Vec{c}$
$\Rightarrow 375 = 250 + 125 = \Vec{a}\cdot \Vec{a} + \Vec{b}\cdot \Vec{b} + \Vec{c}\cdot \Vec{c} - \Vec{b} \cdot \Vec{b} - 2\Vec{a}\cdot \Vec{c} = \Vec{a}\cdot \Vec{a} + \Vec{c}\cdot \Vec{c} - 2\Vec{a}\cdot \Vec{c} = h^2$
$\Rightarrow h^2 = \boxed{375}$

Answer:  375
Problem: Complex numbers $a,$ $b,$ and $c$ are zeros of a polynomial $P(z) = z^3 + qz + r,$ and $|a|^2 + |b|^2 + |c|^2 = 250.$ The points corresponding to $a,$ $b,$ and $c$ in the complex plane are the vertices of a right triangle with hypotenuse $h.$ Find $h^2.$

Solution: Note that the roots of the polynomial $(a,b,c)$ must sum to $0$ due to the $z^2$ coefficient equaling $0$ because of Vieta's Formulas. This tells us that $a+b+c = 0$ and $\overline{a} + \overline{b} + \overline{c} = 0$ so $(a+b+c)(\overline{a+b+c}) = |a|^2 + |b|^2 + |c|^2 + a\overline{b} + \overline{a}b + b\overline{c} + \overline{b}c + c\overline{a} + \overline{c}a = 250 + a\overline{b} + \overline{a}b + b\overline{c} + \overline{b}c + c\overline{a} + \overline{c}a =  0$ so $a\overline{b} + \overline{a}b + b\overline{c} + \overline{b}c + c\overline{a} + \overline{c}a = -250.$
Note that terms similar to $\overline{a}b + \overline{b}a$ appear in $|a-b|^2$ so we decide to sum $|a-b|^2 + |b-c|^2 + |c-a|^2$ out of intuition. Note that this corresponds to the sum of the squares of the sidelengths of the right triangle and if WLOG the $|c-a|^2$ side is the hypotenuse then our sum is equal to $2|c-a|^2 = 2h^2.$
\[2|a|^2 + 2|b|^2 + 2|c|^2 -a\overline{b} - \overline{a}b - b\overline{c} - \overline{b}c - c\overline{a} - \overline{c}a = 500 + 250 = 2h^2\]
As a result we know that $h^2 = \boxed{375}$ ~SailS

Answer:  375
Problem: There are $n$ mathematicians seated around a circular table with $n$ seats numbered $1,$ $2,$ $3,$ $...,$ $n$ in clockwise order. After a break they again sit around the table. The mathematicians note that there is a positive integer $a$ such that
Find the number of possible values of $n$ with $1 < n < 1000.$

Solution: It is a well-known fact that the set $0, a, 2a, ... (n-1)a$ forms a complete set of residues if and only if $a$ is relatively prime to $n$
Thus, we have $a$ is relatively prime to $n$ . In addition, for any seats $p$ and $q$ , we must have $ap - aq$ not be equivalent to either $p - q$ or $q - p$ modulo $n$ to satisfy our conditions. These simplify to $(a-1)p \not\equiv (a-1)q$ and $(a+1)p \not\equiv (a+1)q$ modulo $n$ , so multiplication by both $a-1$ and $a+1$ must form a complete set of residues mod $n$ as well.
Thus, we have $a-1$ $a$ , and $a+1$ are relatively prime to $n$ . We must find all $n$ for which such an $a$ exists. $n$ obviously cannot be a multiple of $2$ or $3$ , but for any other $n$ , we can set $a = n-2$ , and then $a-1 = n-3$ and $a+1 = n-1$ . All three of these will be relatively prime to $n$ , since two numbers $x$ and $y$ are relatively prime if and only if $x-y$ is relatively prime to $x$ . In this case, $1$ $2$ , and $3$ are all relatively prime to $n$ , so $a = n-2$ works.
Now we simply count all $n$ that are not multiples of $2$ or $3$ , which is easy using inclusion-exclusion. We get a final answer of $998 - (499 + 333 - 166) = \boxed{332}$

Answer:  332
Problem: Find the number of ordered pairs of positive integer solutions $(m, n)$ to the equation $20m + 12n = 2012$

Solution: Solving for $m$ gives us $m = \frac{503-3n}{5},$ so in order for $m$ to be an integer, we must have $3n \equiv 503 \mod 5 \longrightarrow n \equiv 1 \mod 5.$ The smallest possible value of $n$ is obviously $1,$ and the greatest is $\frac{503 - 5}{3} = 166,$ so the total number of solutions is $\frac{166-1}{5}+1 = \boxed{34}$

Answer:  34
Problem: Find the number of ordered pairs of positive integer solutions $(m, n)$ to the equation $20m + 12n = 2012$

Solution: Dividing by $4$ gives us $5m + 3n = 503$ . Thus, we have $503-5m\equiv 0 \pmod 3$ since $n$ is an integer. Rearranging then gives $m\equiv 1\pmod 3.$ Since $503-5m>0,$ we know that $m<503/5.$ Because $m$ is an integer, we can rewrite this as $m\le 100.$ Therefore, $m$ ranges from \[0\cdot 3+1\quad \text{to} \quad 33\cdot 3+1,\] giving $\boxed{034}$ values. 
~vaporwave

Answer:  034
Problem: Find the number of ordered pairs of positive integer solutions $(m, n)$ to the equation $20m + 12n = 2012$

Solution: Because the x-intercept of the equation is $\frac{2012}{20}$ , and the y-intercept is $\frac{2012}{12}$ , the slope is $\frac{\frac{-2012}{12}}{\frac{2012}{20}} = \frac{-5}{3}$ . Now, notice the first obvious solution: (100,1). From it, we derive all the other solutions by applying the slope in reverse, i.e: $(100,1), (97,6), (94,11)...$ Because the solutions are only positive, we can generate only 33 more solutions, so in total we have $33+1=\boxed{34}$ solutions.

Answer:  34