Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Let $ABCDEF$ be a regular hexagon . Let $G$ $H$ $I$ $J$ $K$ , and $L$ be the midpoints of sides $AB$ $BC$ $CD$ $DE$ $EF$ , and $AF$ , respectively. The segments $\overline{AH}$ $\overline{BI}$ $\overline{CJ}$ $\overline{DK}$ $\overline{EL}$ , and $\overline{FG}$ bound a smaller regular hexagon. Let the ratio of the area of the smaller hexagon to the area of $ABCDEF$ be expressed as a fraction $\frac {m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$

Solution: We can use coordinates. Let $O$ be at $(0,0)$ with $A$ at $(1,0)$
then $B$ is at $(\cos(60^\circ),\sin(60^\circ))=\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)$
$C$ is at $(\cos(120^\circ),\sin(120^\circ))=\left(-\frac{1}{2},\frac{\sqrt{3}}{2}\right)$
$D$ is at $(\cos(180^\circ),\sin(180^\circ))=(-1,0)$
\begin{align*}&H=\frac{B+C}{2}=\left(0,\frac{\sqrt{3}}{2}\right) \\ &I=\frac{C+D}{2}=\left(-\frac{3}{4},\frac{\sqrt{3}}{4}\right)\end{align*}
Line $AH$ has the slope of $-\frac{\sqrt{3}}{2}$ and the equation of $y=-\frac{\sqrt{3}}{2}(x-1)$
Line $BI$ has the slope of $\frac{\sqrt{3}}{5}$ and the equation $y-\frac{3}{2}=\frac{\sqrt{3}}{5}\left(x-\frac{1}{2}\right)$
Let's solve the system of equation to find $M$
\begin{align*}-\frac{\sqrt{3}}{2}(x-1)-\frac{3}{2}&=\frac{\sqrt{3}}{5}\left(x-\frac{1}{2}\right) \\ -5\sqrt{3}x&=2\sqrt{3}x-\sqrt{3} \\ x&=\frac{1}{7} \\ y&=-\frac{\sqrt{3}}{2}(x-1)=\frac{3\sqrt{3}}{7}\end{align*}
Finally,
\begin{align*}&\sqrt{x^2+y^2}=OM=\frac{1}{7}\sqrt{1^2+(3\sqrt{3})^2}=\frac{1}{7}\sqrt{28}=\frac{2}{\sqrt{7}} \\ &\frac{\text{Area of smaller hexagon}}{\text{Area of bigger hexagon}}=\left(\frac{OM}{OA}\right)^2=\left(\frac{2}{\sqrt{7}}\right)^2=\frac{4}{7}\end{align*}
Thus, the answer is $\boxed{011}$

Answer:  011
Problem: Find the number of second-degree polynomials $f(x)$ with integer coefficients and integer zeros for which $f(0)=2010$

Solution: Let $f(x) = a(x-r)(x-s)$ . Then $ars=2010=2\cdot3\cdot5\cdot67$ . First consider the case where $r$ and $s$ (and thus $a$ ) are positive. There are $3^4 = 81$ ways to split up the prime factors between $a$ $r$ , and $s$ . However, $r$ and $s$ are indistinguishable. In one case, $(a,r,s) = (2010,1,1)$ , we have $r=s$ . The other $80$ cases are double counting, so there are $40$
We must now consider the various cases of signs. For the $40$ cases where $|r|\neq |s|$ , there are a total of four possibilities, For the case $|r|=|s|=1$ , there are only three possibilities, $(r,s) = (1,1); (1,-1); (-1,-1)$ as $(-1,1)$ is not distinguishable from the second of those three.
You may ask: How can one of ${r, s}$ be positive and the other negative? $a$ will be negative as a result. That way, it's still $+2010$ that gets multiplied.
Thus the grand total is $4\cdot40 + 3 = \boxed{163}$

Answer:  163
Problem: Find the number of second-degree polynomials $f(x)$ with integer coefficients and integer zeros for which $f(0)=2010$

Solution: We use Burnside's Lemma . The set being acted upon is the set of integer triples $(a,r,s)$ such that $ars=2010$ . Because $r$ and $s$ are indistinguishable, the permutation group consists of the identity and the permutation that switches $r$ and $s$ . In cycle notation, the group consists of $(a)(r)(s)$ and $(a)(r \: s)$ . There are $4 \cdot 3^4$ fixed points of the first permutation (after distributing the primes among $a$ $r$ $s$ and then considering their signs. We have 4 ways since we can keep them all positive, first 2 negative, first and third negative, or last two negative) and $2$ fixed points of the second permutation ( $r=s=\pm 1$ ). By Burnside's Lemma, there are $\frac{1}{2} (4 \cdot 3^4+2)= \boxed{163}$ distinguishable triples $(a,r,s)$ .
Note: The permutation group is isomorphic to $\mathbb{Z}/2\mathbb{Z}$

Answer:  163
Problem: Find the number of second-degree polynomials $f(x)$ with integer coefficients and integer zeros for which $f(0)=2010$

Solution: The equation can be written in the form of $k(x-a)(x-b)$ , where $|k|$ is a divisor of $2010$ . Factor $2010=2\cdot 3\cdot 5\cdot 67$ . We divide into few cases to study.
Firstly, if $|k|=1$ , the equation can be $-(x-a)(x+b)$ or $(x-a)(x-b)$ or $(x+a)(x+b)$ , there are $2^4+2^4=32$ ways
If $|k|\in \{2,3,5,67\}$ . Take $|k|=2$ as an example, follow the procedure above, there are $2^3+2^3=16$ ways, and there are $\binom {4}{1}=4$ ways to choose $|k|$ , so there are $16\cdot 4=64$ ways
If $|k|$ is the product of two factors of $2010$ , there are $8$ ways for each. Thus there are $\binom {4}{2}\cdot 8=48$ ways
If $|k|$ is the product of three factors of $2010$ , there are $\binom {4}{3}\cdot 4=16$ ways
In the end, $|k|=2010$ , only $2010(x-1)(x-1); 2010(x+1)(x+1); -2010(x-1)(x+1)$ work, there are $3$ ways
Thus, $32+64+48+16+3=\boxed{163}$

Answer:  163
Problem: Define a T-grid to be a $3\times3$ matrix which satisfies the following two properties:
Find the number of distinct T-grids

Solution: The T-grid can be considered as a tic-tac-toe board: five $1$ 's (or X's) and four $0$ 's (or O's).
There are only $\dbinom{9}{5} = 126$ ways to fill the board with five $1$ 's and four $0$ 's. Now we just need to subtract the number of bad grids. Bad grids are ones with more than one person winning, or where someone has won twice.
Let three-in-a-row/column/diagonal be a "win" and let player $0$ be the one that fills in $0$ and player $1$ fills in $1$
Case $1$ Each player wins once.
If player X takes a diagonal, player Y cannot win. If either takes a row, all the columns are blocked, and visa versa. Therefore, they either both take a row or they both take a column.
Case $1$ $36$ cases
Case $2$ Player $1$ wins twice. (Player $0$ cannot win twice because he only has 4 moves.)
Case $2$ total: $22$
Thus, the answer is $126-22-36=\boxed{68}$

Answer:  68
Problem: Define a T-grid to be a $3\times3$ matrix which satisfies the following two properties:
Find the number of distinct T-grids

Solution: We can use generating functions to compute the case that no row or column is completely filled with $1$ 's. Given a row, let $a$ $b$ $c$ be the events that the first, second, third respective squares are $1$ 's. Then the generating function representing the possible events that exclude a row of $1,1,1$ or $0,0,0$ from occuring is \[ab+bc+ca+a+b+c.\] Therefore, the generating function representing the possible grids where no row is filled with $0$ 's and $1$ 's is \[P(a,b,c)=((ab+bc+ca)+(a+b+c))^3.\] We expand this by the Binomial Theorem to find \[P(a,b,c)=(ab+bc+ca)^3+3(ab+bc+ca)^2(a+b+c)+3(ab+bc+ca)(a+b+c)^2+(a+b+c)^3.\] Recall that our grid has five $1$ 's, hence we only want terms where the sum of the exponents is $5$ . This is given by \[3(ab+bc+ca)^2(a+b+c).\] When we expand this, we find \[3(2abc(a+b+c)+a^2b^2+b^2c^2+c^2a^2)(a+b+c).\] We also want to make sure that each of $a$ $b$ $c$ appears at least once (so there is no column filled with $0$ 's) and the power of each of $a$ $b$ $c$ is not greater than or equal to $3$ (so there is no column filled with $1$ 's). The sum of the coefficients of the above polynomial is clearly $81$ (using $a,b,c=1$ ), and the sum of the coefficients of the terms $a^3bc$ $ab^3c$ , and $abc^3$ is $6+6+6+3+3+3+3+3+3=36$ , hence the sum of the coefficients of the desired terms is $81-36=45$ . This counts the number of grids where no column or row is filled with $0$ 's or $1$ 's. However, we could potentially have both diagonals filled with $1$ 's, but this is the only exception to our $45$ possibilities, hence the number of $T$ -grids with no row or column filled with the same digit is $44$
On the other hand, if a row (column) is filled with $0$ 's, then by the Pigeonhole Principle, another row (column) must be filled with $1$ 's. Hence this is impossible, so all other	possible $T$ -grids have a row (column) filled with $1$ 's. If the top row is filled with $1$ 's, then we have two $1$ 's left to place. Clearly they cannot go in the same row, because then the other row is filled with $0$ 's. They also cannot appear in the same column. This leaves $3\cdot 2$ arrangements--3 choices for the location of the $1$ in the second row, and 2 choices for the location of the $1$ in the last row. However, two of these arrangements will fill a diagonal with $1$ 's. Hence there are only $4$ $T$ -grids where the top row is filled with $1$ 's. The same argument applies if any other row or column is filled with $1$ 's. Hence there are $4\cdot 6=24$ such $T$ -grids.
Thus the answer is $44+24=\boxed{68}$

Answer:  68
Problem: Two noncongruent integer-sided isosceles triangles have the same perimeter and the same area. The ratio of the lengths of the bases of the two triangles is $8: 7$ . Find the minimum possible value of their common perimeter.

Solution: Let $s$ be the semiperimeter of the two triangles. Also, let the base of the longer triangle be $16x$ and the base of the shorter triangle be $14x$ for some arbitrary factor $x$ . Then, the dimensions of the two triangles must be $s-8x,s-8x,16x$ and $s-7x,s-7x,14x$ . By Heron's Formula, we have
Since $15$ and $338$ are coprime, to minimize, we must have $s=338$ and $x=15$ . However, we want the minimum perimeter. This means that we must multiply our minimum semiperimeter by $2$ , which gives us a final answer of $\boxed{676}$

Answer:  676
Problem: Two noncongruent integer-sided isosceles triangles have the same perimeter and the same area. The ratio of the lengths of the bases of the two triangles is $8: 7$ . Find the minimum possible value of their common perimeter.

Solution: Let the first triangle have sides $16n,a,a$ , so the second has sides $14n,a+n,a+n$ . The height of the first triangle is $\frac{7}{8}$ the height of the second triangle. Therefore, we have \[a^2-64n^2=\frac{49}{64}((a+n)^2-49n^2).\] Multiplying this, we get \[64a^2-4096n^2=49a^2+98an-2352n^2,\] which simplifies to \[15a^2-98an-1744n^2=0.\] Solving this for $a$ , we get $a=n\cdot\frac{218}{15}$ , so $n=15$ and $a=218$ and the perimeter is $15\cdot16+218+218=\boxed{676}$

Answer:  676
Problem: The $52$ cards in a deck are numbered $1, 2, \cdots, 52$ . Alex, Blair, Corey, and Dylan each picks a card from the deck without replacement and with each card being equally likely to be picked, The two persons with lower numbered cards form a team, and the two persons with higher numbered cards form another team. Let $p(a)$ be the probability that Alex and Dylan are on the same team, given that Alex picks one of the cards $a$ and $a+9$ , and Dylan picks the other of these two cards. The minimum value of $p(a)$ for which $p(a)\ge\frac{1}{2}$ can be written as $\frac{m}{n}$ . where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: Once the two cards are drawn, there are $\dbinom{50}{2} = 1225$ ways for the other two people to draw. Alex and Dylan are the team with higher numbers if Blair and Corey both draw below $a$ , which occurs in $\dbinom{a-1}{2}$ ways. Alex and Dylan are the team with lower numbers if Blair and Corey both draw above $a+9$ , which occurs in $\dbinom{43-a}{2}$ ways. Thus, \[p(a)=\frac{\dbinom{43-a}{2}+\dbinom{a-1}{2}}{1225}.\] Simplifying, we get $p(a)=\frac{(43-a)(42-a)+(a-1)(a-2)}{2\cdot1225}$ , so we need $(43-a)(42-a)+(a-1)(a-2)\ge (1225)$ . If $a=22+b$ , then \begin{align*}(43-a)(42-a)+(a-1)(a-2)&=(21-b)(20-b)+(21+b)(20+b)=2b^2+2(21)(20)\ge (1225) \\ b^2\ge \frac{385}{2} &= 192.5 >13^2 \end{align*} So $b> 13$ or $b< -13$ , and $a=22+b<9$ or $a>35$ , so $a=8$ or $a=36$ . Thus, $p(8) = \frac{616}{1225} = \frac{88}{175}$ , and the answer is $88+175 = \boxed{263}$

Answer:  263
Problem: Triangle $ABC$ with right angle at $C$ $\angle BAC < 45^\circ$ and $AB = 4$ . Point $P$ on $\overline{AB}$ is chosen such that $\angle APC = 2\angle ACP$ and $CP = 1$ . The ratio $\frac{AP}{BP}$ can be represented in the form $p + q\sqrt{r}$ , where $p$ $q$ $r$ are positive integers and $r$ is not divisible by the square of any prime. Find $p+q+r$

Solution: Let $O$ be the circumcenter of $ABC$ and let the intersection of $CP$ with the circumcircle be $D$ . It now follows that $\angle{DOA} = 2\angle ACP = \angle{APC} = \angle{DPB}$ . Hence $ODP$ is isosceles and $OD = DP = 2$
Denote $E$ the projection of $O$ onto $CD$ . Now $CD = CP + DP = 3$ . By the Pythagorean Theorem $OE = \sqrt {2^2 - \frac {3^2}{2^2}} = \sqrt {\frac {7}{4}}$ . Now note that $EP = \frac {1}{2}$ . By the Pythagorean Theorem, $OP = \sqrt {\frac {7}{4} + \frac {1^2}{2^2}} = \sqrt {2}$ . Hence it now follows that,
\[\frac {AP}{BP} = \frac {AO + OP}{BO - OP} = \frac {2 + \sqrt {2}}{2 - \sqrt {2}} = 3 + 2\sqrt {2}\]
This gives that the answer is $\boxed{007}$

Answer:  007
Problem: Triangle $ABC$ with right angle at $C$ $\angle BAC < 45^\circ$ and $AB = 4$ . Point $P$ on $\overline{AB}$ is chosen such that $\angle APC = 2\angle ACP$ and $CP = 1$ . The ratio $\frac{AP}{BP}$ can be represented in the form $p + q\sqrt{r}$ , where $p$ $q$ $r$ are positive integers and $r$ is not divisible by the square of any prime. Find $p+q+r$

Solution: Let $\angle{ACP}$ be equal to $x$ . Then by Law of Sines, $PB = -\frac{\cos{x}}{\cos{3x}}$ and $AP = \frac{\sin{x}}{\sin{3x}}$ . We then obtain $\cos{3x} = 4\cos^3{x} - 3\cos{x}$ and $\sin{3x} = 3\sin{x} - 4\sin^3{x}$ . Solving, we determine that $\sin^2{x} = \frac{4 \pm \sqrt{2}}{8}$ . Plugging this in gives that $\frac{AP}{PB} = \frac{\sqrt{2}+1}{\sqrt{2}-1} = 3 + 2\sqrt{2}$ . The answer is $\boxed{007}$

Answer:  007
Problem: Triangle $ABC$ with right angle at $C$ $\angle BAC < 45^\circ$ and $AB = 4$ . Point $P$ on $\overline{AB}$ is chosen such that $\angle APC = 2\angle ACP$ and $CP = 1$ . The ratio $\frac{AP}{BP}$ can be represented in the form $p + q\sqrt{r}$ , where $p$ $q$ $r$ are positive integers and $r$ is not divisible by the square of any prime. Find $p+q+r$

Solution: Let $\alpha=\angle{ACP}$ $\beta=\angle{ABC}$ , and $x=BP$ . By Law of Sines,
$\frac{1}{\sin(\beta)}=\frac{x}{\sin(90-\alpha)}\implies \sin(\beta)=\frac{\cos(\alpha)}{x}$ (1), and
$\frac{4-x}{\sin(\alpha)}=\frac{4\sin(\beta)}{\sin(2\alpha)} \implies 4-x=\frac{2\sin(\beta)}{\cos(\alpha)}$ . (2)
Then, substituting (1) into (2), we get
$4-x=\frac{2}{x} \implies x^2-4x+2=0 \implies x=2-\sqrt{2} \implies \frac{4-x}{x}=\frac{2+\sqrt{2}}{2-\sqrt{2}}=3+2\sqrt{2}$
The answer is $\boxed{007}$ .
~Rowechen

Answer:  007
Problem: Triangle $ABC$ with right angle at $C$ $\angle BAC < 45^\circ$ and $AB = 4$ . Point $P$ on $\overline{AB}$ is chosen such that $\angle APC = 2\angle ACP$ and $CP = 1$ . The ratio $\frac{AP}{BP}$ can be represented in the form $p + q\sqrt{r}$ , where $p$ $q$ $r$ are positive integers and $r$ is not divisible by the square of any prime. Find $p+q+r$

Solution: Let $\angle{ACP}=x$ . Then, $\angle{APC}=2x$ and $\angle{A}=180-3x$ . Let the foot of the angle bisector of $\angle{APC}$ on side $AC$ be $D$ . Then,
$CD=DP$ and $\triangle{DAP}\sim{\triangle{APC}}$ due to the angles of these triangles.
Let $CD=a$ . By the Angle Bisector Theorem, $\frac{1}{a}=\frac{AP}{AD}$ , so $AD=a\cdot{AP}$ . Moreover, since $CD=DP=a$ , by similar triangle ratios, $\frac{AP}{a+a\cdot{AP}}=a$ . Therefore, $AP = \frac{a^2}{1-a^2}$
Construct the perpendicular from $D$ to $AP$ and denote it as $F$ . Denote the midpoint of $CP$ as $M$ . Since $PD$ is an angle bisector, $PF$ is congruent to $PM$ , so $PF=\frac{1}{2}$
Also, $\triangle{DFA}\sim{\triangle{BCA}}$ . Thus, $\frac{FA}{AC}=\frac{AD}{AB}\Longrightarrow\frac{\frac{a^2}{1-a^2}-\frac{1}{2}}{a+\frac{a^3}{1-a^2}}=\frac{\frac{a^3}{1-a^3}}{4}$ . After some major cancellation, we have $7a^4-8a^2+2=0$ , which is a quadratic in $a^2$ . Thus, $a^2 = \frac{4\pm\sqrt{2}}{7}$
Taking the negative root implies $AP<BP$ , contradiction. Thus, we take the positive root to find that $AP=2+\sqrt{2}$ . Thus, $BP=2-\sqrt{2}$ , and our desired ratio is $\frac{2+\sqrt{2}}{2-\sqrt{2}}\implies{3+2\sqrt{2}}$
The answer is $\boxed{007}$

Answer:  007
Problem: Triangle $ABC$ with right angle at $C$ $\angle BAC < 45^\circ$ and $AB = 4$ . Point $P$ on $\overline{AB}$ is chosen such that $\angle APC = 2\angle ACP$ and $CP = 1$ . The ratio $\frac{AP}{BP}$ can be represented in the form $p + q\sqrt{r}$ , where $p$ $q$ $r$ are positive integers and $r$ is not divisible by the square of any prime. Find $p+q+r$

Solution: Let $O$ be the circumcenter of $\triangle ABC$ . Since $\triangle ABC$ is a right triangle, $O$ will be on $\overline{AB}$ and $\overline{AO} \cong \overline{OB} \cong \overline{OC} = 2$ . Let $\overline{OP} = x$
Next, let's do some angle chasing. Label $\angle ACP = \theta^{\circ}$ , and $\angle APC = 2\theta^{\circ}$ . Thus, $\angle PAC = (180-3\theta)^{\circ}$ , and by isosceles triangles, $\angle ACO = (180-3\theta)^{\circ}$ . Then, by angle subtraction, $\angle OCP = (\theta - (180-3\theta))^{\circ} = (4\theta - 180)^{\circ}$
Using the Law of Sines: \[\frac{x}{\sin (4\theta-180)^{\circ}}=\frac{2}{\sin (2\theta)^{\circ}}\] Using trigonometric identies, $\sin (4\theta-180)^{\circ}=-\sin (4\theta)^{\circ}=-2\sin (2\theta)^{\circ}\cos (2\theta)^{\circ}$ . Plugging this back into the Law of Sines formula gives us: \[\frac{x}{-2\sin (2\theta)^{\circ}\cos (2\theta)^{\circ}}=\frac{2}{\sin (2\theta)^{\circ}}\]
\[-4\sin (2\theta)^{\circ}\cos (2\theta)^{\circ}=x\sin (2\theta)^{\circ}\] \[-4\cos (2\theta)^{\circ}=x\] \[\cos(2\theta)^{\circ}=\frac{-x}4\]
Next, using the Law of Cosines: \[2^2=1^2+x^2-2\cdot 1\cdot x\cdot \cos (2\theta)^{\circ}\] Substituting $\cos(2\theta)^{\circ}=\frac{-x}4$ gives us: \[2^2=1^2+x^2-2\cdot 1\cdot x\cdot \frac{-x}4\] \[4=1+x^2+\frac{x^2}{2}\]
Solving for x gives $x=\sqrt 2$
Finally: $\frac{\overline{AP}}{\overline{BP}}=\frac{\overline{AO}+\overline{OP}}{\overline{BO}-\overline{OP}}=\frac{2+\sqrt 2}{2-\sqrt 2}=3+2\sqrt2$ , which gives us an answer of $3+2+2=\boxed{007}$ . ~adyj

Answer:  007
Problem: In triangle $ABC$ $AC = 13$ $BC = 14$ ,  and $AB=15$ . Points $M$ and $D$ lie on $AC$ with $AM=MC$ and $\angle ABD = \angle DBC$ . Points $N$ and $E$ lie on $AB$ with $AN=NB$ and $\angle ACE = \angle ECB$ . Let $P$ be the point, other than $A$ , of intersection of the circumcircles of $\triangle AMN$ and $\triangle ADE$ . Ray $AP$ meets $BC$ at $Q$ . The ratio $\frac{BQ}{CQ}$ can be written in the form $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m-n$

Solution: Define the function $f:\mathbb{R}^{2}\rightarrow\mathbb{R}$ by \[f(X)=\text{Pow}_{(AMN)}(X)-\text{Pow}_{(ADE)}(X)\] for points $X$ in the plane. Then $f$ is linear, so $\frac{BQ}{CQ}=\frac{f(B)-f(Q)}{f(Q)-f(C)}$ . But $f(Q)=0$ since $Q$ lies on the radical axis of $(AMN)$ $(ADE)$ thus \[\frac{BQ}{CQ}=-\frac{f(B)}{f(C)}=-\frac{BN\cdot BA-BE\cdot BA}{CM\cdot CA-CD\cdot CA}\] Let $AC=b$ $BC=a$ and $AB=c$ . Note that $BN=\tfrac{c}{2}$ and $CM=\tfrac{b}{2}$ because they are midpoints, while $BE=\frac{ac}{a+b}$ and $CD=\frac{ab}{a+c}$ by Angle Bisector Theorem. Thus we can rewrite this expression as \begin{align*}&-\frac{\tfrac{c^{2}}{2}-\tfrac{ac^{2}}{a+b}}{\tfrac{b^{2}}{2}-\tfrac{ab^{2}}{a+c}} \\ =&-\left(\frac{c^{2}}{b^{2}}\right)\left(\frac{\tfrac{1}{2}-\tfrac{a}{a+b}}{\tfrac{1}{2}-\tfrac{a}{a+c}}\right) \\ =&\left(\frac{c^{2}}{b^{2}}\right)\left(\frac{a+c}{a+b}\right) \\ =&\left(\frac{225}{169}\right)\left(\frac{29}{27}\right) \\ =&~\frac{725}{507}\end{align*} so $m-n=\boxed{218}$

Answer:  218
Problem: In triangle $ABC$ $AC = 13$ $BC = 14$ ,  and $AB=15$ . Points $M$ and $D$ lie on $AC$ with $AM=MC$ and $\angle ABD = \angle DBC$ . Points $N$ and $E$ lie on $AB$ with $AN=NB$ and $\angle ACE = \angle ECB$ . Let $P$ be the point, other than $A$ , of intersection of the circumcircles of $\triangle AMN$ and $\triangle ADE$ . Ray $AP$ meets $BC$ at $Q$ . The ratio $\frac{BQ}{CQ}$ can be written in the form $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m-n$

Solution: Let $Y = MN \cap AQ$ $\frac {BQ}{QC} = \frac {NY}{MY}$ since $\triangle AMN \sim \triangle ACB$ . Since quadrilateral $AMPN$ is cyclic, $\triangle MYA \sim \triangle PYN$ and $\triangle MYP \sim \triangle AYN$ , yielding $\frac {YM}{YA} = \frac {MP}{AN}$ and $\frac {YA}{YN} = \frac {AM}{PN}$ . Multiplying these together yields * $\frac {YN}{YM} = \left(\frac {AN}{AM}\right) \left(\frac {PN}{PM}\right)$
$\frac {AN}{AM} = \frac {\frac {AB}{2}}{\frac {AC}{2}} = \frac {15}{13}$
Now we claim that $\triangle PMD \sim \triangle PNE$ . To prove this, we can use cyclic quadrilaterals.
From $AMPN$ $\angle PNY \cong \angle PAM$ and $\angle ANM \cong \angle APM$ . So, $m\angle PNA = m\angle PNY + m\angle ANM = m\angle PAM + m\angle APM = 180-m\angle PMA$ and $\angle PNA \cong \angle PMD$
From $ADPE$ $\angle PDE \cong \angle PAE$ and $\angle EDA \cong \angle EPA$ . Thus, $m\angle MDP = m\angle PDE + m\angle EDA =  m\angle PAE + m\angle EPA = 180-m\angle PEA$ and $\angle PDM \cong \angle PEN$
Thus, from AA similarity, $\triangle PMD \sim \triangle PNE$
Therefore, $\frac {PN}{PM} = \frac {NE}{MD}$ , which can easily be computed by the angle bisector theorem to be $\frac {145}{117}$ . It follows that * $\frac {BQ}{CQ} = \frac {15}{13} \cdot \frac {145}{117} = \frac {725}{507}$ , giving us an answer of $725 - 507 = \boxed{218}$

Answer:  218
Problem: In triangle $ABC$ $AC = 13$ $BC = 14$ ,  and $AB=15$ . Points $M$ and $D$ lie on $AC$ with $AM=MC$ and $\angle ABD = \angle DBC$ . Points $N$ and $E$ lie on $AB$ with $AN=NB$ and $\angle ACE = \angle ECB$ . Let $P$ be the point, other than $A$ , of intersection of the circumcircles of $\triangle AMN$ and $\triangle ADE$ . Ray $AP$ meets $BC$ at $Q$ . The ratio $\frac{BQ}{CQ}$ can be written in the form $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m-n$

Solution: This problem can be solved with barycentric coordinates. Let triangle $ABC$ be the reference triangle with $A=(1,0,0)$ $B=(0,1,0)$ , and $C=(0,0,1)$ . Thus, $N=(1:1:0)$ and $M=(1:0:1)$ . Using the Angle Bisector Theorem, we can deduce that $D=(14:0:15)$ and $E = (14:13:0)$ . Plugging the coordinates for triangles $ANM$ and $AED$ into the circle formula, we deduce that the equation for triangle $ANM$ is $-a^2yz-b^2zx-c^2xy+(\frac{c^2}{2}y+\frac{b^2}{2}z)(x+y+z)=0$ and the equation for triangle $AED$ is $-a^2yz-b^2zx-c^2xy+(\frac{14c^2}{27}y+\frac{14b^2}{29}z)(x+y+z)=0$ . Solving the system of equations, we get that $\frac{c^2y}{54}=\frac{b^2z}{58}$ . This equation determines the radical axis of circles $ANM$ and $AED$ , on which points $P$ and $Q$ lie. Thus, solving for $\frac{z}{y}$ gets the desired ratio of lengths, and $\frac{z}{y}=\frac{58c^2}{54b^2}$ and plugging in the lengths $b=13$ and $c=15$ gets $\frac{725}{507}$ . From this we get the desired answer of $725-507=\boxed{218}$ .
-wertguk

Answer:  218
Problem: Call a $3$ -digit number geometric if it has $3$ distinct digits which, when read from left to right, form a geometric sequence.  Find the difference between the largest and smallest geometric numbers.

Solution: Assume that the largest geometric number starts with a $9$ . We know that the common ratio must be a rational of the form $k/3$ for some integer $k$ , because a whole number should be attained for the 3rd term as well. When $k = 1$ , the number is $931$ . When $k = 2$ , the number is $964$ . When $k = 3$ , we get $999$ , but the integers must be distinct. By the same logic, the smallest geometric number is $124$ . The largest geometric number is $964$ and the smallest is $124$ . Thus the difference is $964 - 124 = \boxed{840}$

Answer:  840
Problem: Call a $3$ -digit number geometric if it has $3$ distinct digits which, when read from left to right, form a geometric sequence.  Find the difference between the largest and smallest geometric numbers.

Solution: Consider the three-digit number $abc$ . If its digits form a geometric progression, we must have that ${a \over b} = {b \over c}$ , that is, $b^2 = ac$
The minimum and maximum geometric numbers occur when $a$ is minimized and maximized, respectively. The minimum occurs when $a = 1$ ; letting $b = 2$ and $c = 4$ achieves this, so the smallest possible geometric number is 124.
For the maximum, we have that $b^2 = 9c$ $b$ is maximized when $9c$ is the greatest possible perfect square; this happens when $c = 4$ , yielding $b = 6$ . Thus, the largest possible geometric number is 964.
Our answer is thus $964 - 124 = \boxed{840}$

Answer:  840