Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: In $\triangle ABC$ , the sides have integer lengths and $AB=AC$ . Circle $\omega$ has its center at the incenter of $\triangle ABC$ . An excircle of $\triangle ABC$ is a circle in the exterior of $\triangle ABC$ that is tangent to one side of the triangle and tangent to the extensions of the other two sides. Suppose that the excircle tangent to $\overline{BC}$ is internally tangent to $\omega$ , and the other two excircles are both externally tangent to $\omega$ . Find the minimum possible value of the perimeter of $\triangle ABC$

Solution: Let the tangent circle be $\omega$ . Some notation first: let $BC=a$ $AB=b$ $s$ be the semiperimeter, $\theta=\angle ABC$ , and $r$ be the inradius. Intuition tells us that the radius of $\omega$ is $r+\frac{2rs}{s-a}$ (using the exradius formula). However, the sum of the radius of $\omega$ and $\frac{rs}{s-b}$ is equivalent to the distance between the incenter and the the $B/C$ excenter. Denote the B excenter as $I_B$ and the incenter as $I$ . 
Lemma: $I_BI=\frac{2b*IB}{a}$ We draw the circumcircle of $\triangle ABC$ . Let the angle bisector of $\angle ABC$ hit the circumcircle at a second point $M$ . By the incenter-excenter lemma, $AM=CM=IM$ . Let this distance be $\alpha$ . Ptolemy's theorem on $ABCM$ gives us \[a\alpha+b\alpha=b(\alpha+IB)\to \alpha=\frac{b*IB}{a}\] Again, by the incenter-excenter lemma, $II_B=2IM$ so $II_b=\frac{2b*IB}{a}$ as desired.
Using this gives us the following equation: \[\frac{2b*IB}{a}=r+\frac{2rs}{s-a}+\frac{rs}{s-b}\] Motivated by the $s-a$ and $s-b$ , we make the following substitution: $x=s-a, y=s-b$ This changes things quite a bit. Here's what we can get from it: \[a=2y, b=x+y, s=x+2y\] It is known (easily proved with Heron's and $a=rs$ ) that \[r=\sqrt{\frac{(s-a)(s-b)(s-b)}{s}}=\sqrt{\frac{xy^2}{x+2y}}\] Using this, we can also find $IB$ : let the midpoint of $BC$ be $N$ . Using Pythagorean's Theorem on $\triangle INB$ \[IB^2=r^2+(\frac{a}{2})^2=\frac{xy^2}{x+2y}+y^2=\frac{2xy^2+2y^3}{x+2y}=\frac{2y^2(x+y)}{x+2y}\] We now look at the RHS of the main equation: \[r+\frac{2rs}{s-a}+\frac{rs}{s-b}=r(1+\frac{2(x+2y)}{x}+\frac{x+2y}{y})=r(\frac{x^2+5xy+4y^2}{xy})=\frac{r(x+4y)(x+y)}{xy}=\frac{2(x+y)IB}{2y}\] Cancelling some terms, we have \[\frac{r(x+4y)}{x}=IB\] Squaring, \[\frac{2y^2(x+y)}{x+2y}=\frac{(x+4y)^2*xy^2}{x^2(x+2y)}\to \frac{(x+4y)^2}{x}=2(x+y)\] Expanding and moving terms around gives \[(x-8y)(x+2y)=0\to x=8y\] Reverse substituting, \[s-a=8s-8b\to b=\frac{9}{2}a\] Clearly the smallest solution is $a=2$ and $b=9$ , so our answer is $2+9+9=\boxed{020}$ -franchester

Answer:  020
Problem: In $\triangle ABC$ , the sides have integer lengths and $AB=AC$ . Circle $\omega$ has its center at the incenter of $\triangle ABC$ . An excircle of $\triangle ABC$ is a circle in the exterior of $\triangle ABC$ that is tangent to one side of the triangle and tangent to the extensions of the other two sides. Suppose that the excircle tangent to $\overline{BC}$ is internally tangent to $\omega$ , and the other two excircles are both externally tangent to $\omega$ . Find the minimum possible value of the perimeter of $\triangle ABC$

Solution: Notice that the $A$ -excircle would have to be very small to fit the property that it is internally tangent to $\omega$ and the other two excircles are both externally tangent, given that circle $\omega$ 's centre is at the incenter of $\triangle ABC$ . If $BC=2$ , we see that $AB=AC$ must be somewhere in the $6$ to $13$ range. If we test $6$ by construction, we notice the $A$ -excircle is too big for it to be internally tangent to $\omega$ while the other two are externally tangent. This means we should test $8$ or $9$ next. I actually did this and found that $9$ worked, so the answer is $2+9+9=\boxed{20}$ . Note that $BC$ cannot be $1$ because then $AB=AC$ would have to be $4.5$ which is not an integer.

Answer:  20
Problem: In $\triangle ABC$ , the sides have integer lengths and $AB=AC$ . Circle $\omega$ has its center at the incenter of $\triangle ABC$ . An excircle of $\triangle ABC$ is a circle in the exterior of $\triangle ABC$ that is tangent to one side of the triangle and tangent to the extensions of the other two sides. Suppose that the excircle tangent to $\overline{BC}$ is internally tangent to $\omega$ , and the other two excircles are both externally tangent to $\omega$ . Find the minimum possible value of the perimeter of $\triangle ABC$

Solution: Let $M$ be the midpoint $BC, BM = a, AB= BC = b,$ $s = b+a$ be the semiperimeter, $r$ be the inradius. Let $I_A, I_B$ be excenters, $r_A, r_B$ be exradius, $R$ be radius $\omega.$ Then $R = r + 2 r_A,$ \[r = \sqrt{\frac{(s-2a)(s-b)^2}{s}} = a \sqrt{\frac{b-a}{b+a}},\] \[r \cdot s = r_A \cdot (s-2a) \implies r_A =  a \sqrt{\frac{b+a}{b–a}},\] \[r \cdot s = r_B \cdot (s-b) \implies r_B =  \sqrt{b^2 – a^2} = AM.\] \[II_B =R + r_B = r + 2r_A + r_B \implies\] \[II_B=   a \sqrt{\frac{b-a}{b+ a}}+ 2a  \sqrt{\frac{b+a}{b – a}} + \sqrt{b^2-a^2} = b\frac{3a+b}{\sqrt{b^2-a^2}}.\] \[\overline{AI_B}= \frac{\overline{A} \cdot 2a - \overline{B} \cdot b + \overline{C}\cdot b}{2a -b + b}-\overline{A} =b \frac{\overline{C}- \overline{B}}{2a},\] \[AI_B = b, \overline{AI_B}\perp AI \implies II_B = \sqrt{AI_B^2 + (AM – r)^2},\] \[II_B =  \sqrt{b^2 +\left(\sqrt{b^2-a^2}-  a \sqrt{\frac{b-a}{b+ a}}\right)^2} = b\sqrt{\frac{2b}{b+a}}\] Therefore we get problem’s condition in the form of \[b\frac{b +3a}{\sqrt{b^2-a^2}} = b \sqrt{\frac {2b}{b+a}} \implies  b + 3a = \sqrt{2b(b-a)} \implies (b-9a)(b+a) = 0 \implies b = 9a.\]
We use $a = 1$ an get $b = 9, 2s = 18+2 = \boxed{020}$

Answer:  020
Problem: Given $f(z) = z^2-19z$ , there are complex numbers $z$ with the property that $z$ $f(z)$ , and $f(f(z))$ are the vertices of a right triangle in the complex plane with a right angle at $f(z)$ . There are positive integers $m$ and $n$ such that one such value of $z$ is $m+\sqrt{n}+11i$ . Find $m+n$

Solution: Notice that we must have \[\frac{f(f(z))-f(z)}{f(z)-z}=-\frac{f(f(z))-f(z)}{z-f(z)}\in i\mathbb R .\] However, $f(t)-t=t(t-20)$ , so \begin{align*} \frac{f(f(z))-f(z)}{f(z)-z}&=\frac{(z^2-19z)(z^2-19z-20)}{z(z-20)}\\ &=\frac{z(z-19)(z-20)(z+1)}{z(z-20)}\\ &=(z-19)(z+1)\\ &=(z-9)^2-100. \end{align*} Then, the real part of $(z-9)^2$ is $100$ . Since $\text{Im}(z-9)=\text{Im}(z)=11$ , let $z-9=a+11i$ . Then, \[100=\text{Re}((a+11i)^2)=a^2-121\implies a=\pm\sqrt{221}.\] It follows that $z=9+\sqrt{221}+11i$ , and the requested sum is $9+221=\boxed{230}$

Answer:  230
Problem: Given $f(z) = z^2-19z$ , there are complex numbers $z$ with the property that $z$ $f(z)$ , and $f(f(z))$ are the vertices of a right triangle in the complex plane with a right angle at $f(z)$ . There are positive integers $m$ and $n$ such that one such value of $z$ is $m+\sqrt{n}+11i$ . Find $m+n$

Solution: We will use the fact that segments $AB$ and $BC$ are perpendicular in the complex plane if and only if $\frac{a-b}{b-c}\in i\mathbb{R}$ . To prove this, note that when dividing two complex numbers you subtract the angle of one from the other. Therefore, if the two complex numbers are perpendicular, the difference between their arguments will be 90 degrees, so subtracting the angles will yield an imaginary number with no real part(an argument of 90 degrees puts a complex number on the imaginary axis).
Now to apply this: \[\frac{f(z)-z}{f(f(z))-f(z)}\in i\mathbb{R}\] \[\frac{z^2-19z-z}{(z^2-19z)^2-19(z^2-19z)-(z^2-19z)}\] \[\frac{z^2-20z}{z^4-38z^3+341z^2+380z}\] \[\frac{z(z-20)}{z(z+1)(z-19)(z-20)}\] \[\frac{1}{(z+1)(z-19)}\in i\mathbb{R}\]
The factorization of the nasty denominator above is made easier with the intuition that $(z-20)$ must be a divisor for the problem to lead anywhere. Now we know $(z+1)(z-19)\in i\mathbb{R}$ so using the fact that the imaginary part of $z$ is $11i$ and calling the real part r,
\[(r+1+11i)(r-19+11i)\in i\mathbb{R}\] \[r^2-18r-140=0\]
solving the above quadratic yields $r=9+\sqrt{221}$ so our answer is $9+221=\boxed{230}$

Answer:  230
Problem: Given $f(z) = z^2-19z$ , there are complex numbers $z$ with the property that $z$ $f(z)$ , and $f(f(z))$ are the vertices of a right triangle in the complex plane with a right angle at $f(z)$ . There are positive integers $m$ and $n$ such that one such value of $z$ is $m+\sqrt{n}+11i$ . Find $m+n$

Solution: I would like to use a famous method, namely the coni method.
According to the question given, we can assume , $A= f(f(z)),B=f(z),C= z$ respectively.
WLOG, $Z_1= \frac{f(f(z))-f(z)}{z-f(z)}$ .
According to the question $\arg{Z_1}=\frac{\pi}{2}$
So, $\Re (Z_1)=0$
Now, $Z_1=\frac{z(z-19)(z+1)(z-20)}{-z(z-20)}$
$\implies Z_1= -(z^2-18z-19)$ .
WLOG, $z=a+11i$ .where $a=m+\sqrt{n}$
So, $\Re (Z_1)= -(a^2-18a-140)$ .
Solving, $a^2-18a-140=0$ .get ,
$a=9$ $\sqrt{221}$ .
So, possible value of $a=9+\sqrt{221}$
$m+n=\boxed{230}$ .
~ftheftics.

Answer:  230
Problem: Given $f(z) = z^2-19z$ , there are complex numbers $z$ with the property that $z$ $f(z)$ , and $f(f(z))$ are the vertices of a right triangle in the complex plane with a right angle at $f(z)$ . There are positive integers $m$ and $n$ such that one such value of $z$ is $m+\sqrt{n}+11i$ . Find $m+n$

Solution: The arguments of the two complex numbers differ by $90^\circ$ if the ratio of the numbers is a pure imaginary number. Thus three distinct complex numbers $A,\,B,$ and $C$ form a right triangle at $B$ if and only if $\tfrac{C-B}{B-A}$ has real part equal to $0.$ Hence \begin{align*} \frac{f(f(z))-f(z)}{f(z)-z}&=\frac{(z^2-19z)^2-19(z^2-19z)-(z^2-19z)}{(z^2-19z)-z}\\ &=\frac{(z^2-19z)(z^2-19z-19-1)}{z^2-20z}\\ &=\frac{z(z-19)(z+1)(z-20)}{z(z-20)} \\ &=z^2-18z-19 \end{align*} must have real part equal to $0.$ If $z=x+11i,$ the real part of $z^2-18z-19$ is $x^2-11^2-18x-19,$ which is $0$ when $x=9\pm\sqrt{221}.$ The requested sum is $9+221=\boxed{230}.$

Answer:  230
Problem: Given $f(z) = z^2-19z$ , there are complex numbers $z$ with the property that $z$ $f(z)$ , and $f(f(z))$ are the vertices of a right triangle in the complex plane with a right angle at $f(z)$ . There are positive integers $m$ and $n$ such that one such value of $z$ is $m+\sqrt{n}+11i$ . Find $m+n$

Solution: Firstly, the angle between the three complex numbers is equivalent to the angle between $f(z)-z,0,$ and $f(f(z))-f(z)$
Using $f(z)=z(z-19)$ to help expand,
$z-f(z)=20z-z^2$ and $f(f(z))-f(z)=(z^2-19z)(z^2-19z-19)-(z^2-19z)$
The second equation can be rewritten as $(z^2-19z)(z^2-19z-20)=z(z-19)(z-20)(z+1)$
Note that the angle between $(a+bi), 0,$ and $(a+bi)(c+di)$ is the same as the angle between $1, 0,$ and $c+di$ .
The proof of this is as follows:
Treating the complex numbers like vectors (e.g. $a+bi$ turns into $\left[\begin{array}{c}  a \\ b \end{array}\right]$ ) we have $cos \theta = \frac{a \cdot (ac-bd) + b \cdot (ad+bc)}{2\sqrt{a^2+b^2}\sqrt{a^2+b^2}\sqrt{c^2+d^2}}$
$cos \theta = \frac{a^2 c - abd + abd + b^2 c}{2(a^2+b^2)\sqrt{c^2+d^2}}$
$cos \theta = \frac{(a^2+b^2)c}{2(a^2+b^2)\sqrt{c^2+d^2}}=\frac{c}{2\sqrt{c^2+d^2}}$
Using the dot product formula for the cosine of the angle between the other two vectors (1, c + di) we get the same result. Thus, it is proven.
Now, we see that the three resulting complex numbers have the same angle as $-1,0,$ and $(z-19)(z+1)$ as a factor of $z(z-20)$ can be taken out of both expressions. Note that the value of $z$ clearly is not 0, nor is it 20 as it has a value of $11i$
Expanding $z^2-18z-19$ with $z=a+11i$ yields
$a^2-121+22ai-18a-198i-19=a^2-18a-140+22ai-198i$ . Turning these into vectors again, the resulting relation is
$cos(90^\circ)=0=\frac{-(a^2-18a-140)}{2\sqrt{(a^2-18a-140)^2 + (22a-198)^2} \cdot 1}$
In order for this thing to be 0, the numerator must be 0, so setting it equal to 0 yields
$a^2-18a-140=0 \Rightarrow a=9\pm \sqrt{81+140}$ . Neither of these make the denominator in the $cos(90^\circ)$ expression 0, so they are valid. The requested result is the positive root, so the answer is $\boxed{230}$

Answer:  230
Problem: Triangle $ABC$ has side lengths $AB=4$ $BC=5$ , and $CA=6$ . Points $D$ and $E$ are on ray $AB$ with $AB<AD<AE$ . The point $F \neq C$ is a point of intersection of the circumcircles of $\triangle ACD$ and $\triangle EBC$ satisfying $DF=2$ and $EF=7$ . Then $BE$ can be expressed as $\tfrac{a+b\sqrt{c}}{d}$ , where $a$ $b$ $c$ , and $d$ are positive integers such that $a$ and $d$ are relatively prime, and $c$ is not divisible by the square of any prime. Find $a+b+c+d$

Solution: [asy] unitsize(20); pair A, B, C, D, E, F, X, O1, O2; A = (0, 0); B = (4, 0); C = intersectionpoints(circle(A, 6), circle(B, 5))[0]; D = B + (5/4 * (1 + sqrt(2)), 0); E = D + (4 * sqrt(2), 0); F = intersectionpoints(circle(D, 2), circle(E, 7))[1]; X = extension(A, E, C, F); O1 = circumcenter(C, A, D); O2 = circumcenter(C, B, E);  filldraw(A--B--C--cycle, lightcyan, deepcyan); filldraw(D--E--F--cycle, lightmagenta, deepmagenta); draw(B--D, gray(0.6)); draw(C--F, gray(0.6)); draw(circumcircle(C, A, D), dashed); draw(circumcircle(C, B, E), dashed);  dot("$A$", A, dir(A-O1)); dot("$B$", B, dir(240)); dot("$C$", C, dir(120)); dot("$D$", D, dir(40)); dot("$E$", E, dir(E-O2)); dot("$F$", F, dir(270)); dot("$X$", X, dir(140));  label("$6$", (C+A)/2, dir(C-A)*I, deepcyan); label("$5$", (C+B)/2, dir(B-C)*I, deepcyan); label("$4$", (A+B)/2, dir(A-B)*I, deepcyan); label("$7$", (F+E)/2, dir(F-E)*I, deepmagenta); label("$2$", (F+D)/2, dir(D-F)*I, deepmagenta); label("$4\sqrt{2}$", (D+E)/2, dir(E-D)*I, deepmagenta); label("$a$", (B+X)/2, dir(B-X)*I, gray(0.3)); label("$a\sqrt{2}$", (D+X)/2, dir(D-X)*I, gray(0.3)); [/asy]
Notice that \[\angle DFE=\angle CFE-\angle CFD=\angle CBE-\angle CAD=180-B-A=C.\] By the Law of Cosines, \[\cos C=\frac{AC^2+BC^2-AB^2}{2\cdot AC\cdot BC}=\frac34.\] Then, \[DE^2=DF^2+EF^2-2\cdot DF\cdot EF\cos C=32\implies DE=4\sqrt2.\] Let $X=\overline{AB}\cap\overline{CF}$ $a=XB$ , and $b=XD$ . Then, \[XA\cdot XD=XC\cdot XF=XB\cdot XE\implies b(a+4)=a(b+4\sqrt2)\implies b=a\sqrt2.\] However, since $\triangle XFD\sim\triangle XAC$ $XF=\tfrac{4+a}3$ , but since $\triangle XFE\sim\triangle XBC$ \[\frac75=\frac{4+a}{3a}\implies a=\frac54\implies BE=a+a\sqrt2+4\sqrt2=\frac{5+21\sqrt2}4,\] and the requested sum is $5+21+2+4=\boxed{032}$

Answer:  032
Problem: Triangle $ABC$ has side lengths $AB=4$ $BC=5$ , and $CA=6$ . Points $D$ and $E$ are on ray $AB$ with $AB<AD<AE$ . The point $F \neq C$ is a point of intersection of the circumcircles of $\triangle ACD$ and $\triangle EBC$ satisfying $DF=2$ and $EF=7$ . Then $BE$ can be expressed as $\tfrac{a+b\sqrt{c}}{d}$ , where $a$ $b$ $c$ , and $d$ are positive integers such that $a$ and $d$ are relatively prime, and $c$ is not divisible by the square of any prime. Find $a+b+c+d$

Solution: Define $\omega_1$ to be the circumcircle of $\triangle ACD$ and $\omega_2$ to be the circumcircle of $\triangle EBC$
Because of exterior angles,
$\angle ACB = \angle CBE - \angle CAD$
But $\angle CBE = \angle CFE$ because $CBFE$ is cyclic. In addition, $\angle CAD = \angle CFD$ because $CAFD$ is cyclic. Therefore, $\angle ACB = \angle CFE - \angle CFD$ . But $\angle CFE - \angle CFD = \angle DFE$ , so $\angle ACB = \angle DFE$ . Using Law of Cosines on $\triangle ABC$ , we can figure out that $\cos(\angle ACB) = \frac{3}{4}$ . Since $\angle ACB = \angle DFE$ $\cos(\angle DFE) = \frac{3}{4}$ . We are given that $DF = 2$ and $FE = 7$ , so we can use Law of Cosines on $\triangle DEF$ to find that $DE = 4\sqrt{2}$
Let $G$ be the intersection of segment $\overline{AE}$ and $\overline{CF}$ . Using Power of a Point with respect to $G$ within $\omega_1$ , we find that $AG \cdot GD = CG \cdot GF$ . We can also apply Power of a Point with respect to $G$ within $\omega_2$ to find that $CG \cdot GF = BG \cdot GE$ . Therefore, $AG \cdot GD = BG \cdot GE$
$AG \cdot GD = BG \cdot GE$
$(AB + BG) \cdot GD = BG \cdot (GD + DE)$
$AB \cdot GD + BG \cdot GD = BG \cdot GD + BG \cdot DE$
$AB \cdot GD = BG \cdot DE$
$4 \cdot GD = BG \cdot 4\sqrt{2}$
$GD = BG \cdot \sqrt{2}$
Note that $\triangle GAC$ is similar to $\triangle GFD$ $GF = \frac{BG + 4}{3}$ . Also note that $\triangle GBC$ is similar to $\triangle GFE$ , which gives us $GF = \frac{7 \cdot BG}{5}$ . Solving this system of linear equations, we get $BG = \frac{5}{4}$ . Now, we can solve for $BE$ , which is equal to $BG(\sqrt{2} + 1) + 4\sqrt{2}$ . This simplifies to $\frac{5 + 21\sqrt{2}}{4}$ , which means our answer is $\boxed{032}$

Answer:  032
Problem: Triangle $ABC$ has side lengths $AB=4$ $BC=5$ , and $CA=6$ . Points $D$ and $E$ are on ray $AB$ with $AB<AD<AE$ . The point $F \neq C$ is a point of intersection of the circumcircles of $\triangle ACD$ and $\triangle EBC$ satisfying $DF=2$ and $EF=7$ . Then $BE$ can be expressed as $\tfrac{a+b\sqrt{c}}{d}$ , where $a$ $b$ $c$ , and $d$ are positive integers such that $a$ and $d$ are relatively prime, and $c$ is not divisible by the square of any prime. Find $a+b+c+d$

Solution: Construct $FC$ and let $FC\cap AE=K$ . Let $FK=x$ . Using $\triangle FKE\sim \triangle BKC$ \[BK=\frac{5}{7}x\] Using $\triangle FDK\sim ACK$ , it can be found that \[3x=AK=4+\frac{5}{7}x\to x=\frac{7}{4}\] This also means that $BK=\frac{21}{4}-4=\frac{5}{4}$ . It suffices to find $KE$ . It is easy to see the following: \[180-\angle ABC=\angle KBC=\angle KFE\] Using reverse Law of Cosines on $\triangle ABC$ $\cos{\angle ABC}=\frac{1}{8}\to \cos{180-\angle ABC}=\frac{-1}{8}$ . Using Law of Cosines on $\triangle EFK$ gives $KE=\frac{21\sqrt 2}{4}$ , so $BE=\frac{5+21\sqrt 2}{4}\to \boxed{032}$ .
-franchester

Answer:  032
Problem: Triangle $ABC$ has side lengths $AB=4$ $BC=5$ , and $CA=6$ . Points $D$ and $E$ are on ray $AB$ with $AB<AD<AE$ . The point $F \neq C$ is a point of intersection of the circumcircles of $\triangle ACD$ and $\triangle EBC$ satisfying $DF=2$ and $EF=7$ . Then $BE$ can be expressed as $\tfrac{a+b\sqrt{c}}{d}$ , where $a$ $b$ $c$ , and $d$ are positive integers such that $a$ and $d$ are relatively prime, and $c$ is not divisible by the square of any prime. Find $a+b+c+d$

Solution: Let $P=AE\cap CF$ . Let $CP=5x$ and $BP=5y$ ; from $\triangle{CBP}\sim\triangle{EFP}$ we have $EP=7x$ and $FP=7y$ . From $\triangle{CAP}\sim\triangle{DFP}$ we have $\frac{6}{4+5y}=\frac{2}{7y}$ giving $y=\frac{1}{4}$ . So $BP=\frac{5}{4}$ and $FP=\frac{7}{4}$ . These similar triangles also gives us $DP=\frac{5}{3}x$ so $DE=\frac{16}{3}x$ . Now, Stewart's Theorem on $\triangle{FEP}$ and cevian $FD$ tells us that \[\frac{560}{9}x^3+28x=\frac{49}{3}x+\frac{245}{3}x,\] so $x=\frac{3\sqrt{2}}{4}$ . Then $BE=\frac{5}{4}+7x=\frac{5+21\sqrt{2}}{4}$ so the answer is $\boxed{032}$ as desired. (Solution by Trumpeter, but not added to the Wiki by Trumpeter)

Answer:  032
Problem: Triangle $ABC$ has side lengths $AB=4$ $BC=5$ , and $CA=6$ . Points $D$ and $E$ are on ray $AB$ with $AB<AD<AE$ . The point $F \neq C$ is a point of intersection of the circumcircles of $\triangle ACD$ and $\triangle EBC$ satisfying $DF=2$ and $EF=7$ . Then $BE$ can be expressed as $\tfrac{a+b\sqrt{c}}{d}$ , where $a$ $b$ $c$ , and $d$ are positive integers such that $a$ and $d$ are relatively prime, and $c$ is not divisible by the square of any prime. Find $a+b+c+d$

Solution: Connect $CF$ meeting $AE$ at $J$ . We can observe that $\triangle{ACJ}\sim \triangle{FJD}$ Getting that $\frac{AJ}{FJ}=\frac{AC}{FD}=3$ . We can also observe that $\triangle{CBJ}\sim \triangle{EFJ}$ , getting that $\frac{CB}{EF}=\frac{BJ}{FJ}=\frac{CJ}{EJ}=\frac{5}{7}$
Assume that $BJ=5x;FJ=7x$ , since $\frac{AJ}{FJ}=3$ , we can get that $\frac{AJ}{FJ}=\frac{AB+BJ}{FJ}=\frac{4+5x}{7x}=3$ , getting that $x=\frac{1}{4}; BJ=\frac{5}{4}; FJ=\frac{7}{4}$
Using Power of Point, we can get that $BJ * EJ=CJ*FJ; DJ * AJ=CJ * FJ$ Assume that $DJ=5k,CJ=15k$ , getting that $JE=21k, DE=16k$
Now applying Law of Cosine on two triangles, $\triangle{ACJ};\triangle{FJE}$ separately, we can get two equations
$(1): (15k)^2+(\frac{21}{4})^2-2*15k *\frac{21}{4} * cos\angle{CJA}=36$
$(2):(21k)^2+(\frac{7}{4})^2-2*\frac{7}{4} * 21k*cos\angle{FJE}=49$
Since $\angle{CJA}=\angle{FJE}$ , we can use $15(2)-7(1)$ to eliminate the $cos$ term
Then we can get that $5040k^2=630$ , getting $k=\frac{\sqrt{2}}{4}$
$BE=21k=\frac{21\sqrt{2}}{4}; BJ=\frac{5}{4}$ , so the desired answer is $\frac{21\sqrt{2}+5}{4}$ , which leads to the answer $\boxed{032}$

Answer:  032
Problem: Triangle $ABC$ has side lengths $AB=4$ $BC=5$ , and $CA=6$ . Points $D$ and $E$ are on ray $AB$ with $AB<AD<AE$ . The point $F \neq C$ is a point of intersection of the circumcircles of $\triangle ACD$ and $\triangle EBC$ satisfying $DF=2$ and $EF=7$ . Then $BE$ can be expressed as $\tfrac{a+b\sqrt{c}}{d}$ , where $a$ $b$ $c$ , and $d$ are positive integers such that $a$ and $d$ are relatively prime, and $c$ is not divisible by the square of any prime. Find $a+b+c+d$

Solution: Nice problem!
First, let $AE$ and $CF$ intersect at $X$ . Our motivation here is to introduce cyclic quadrilaterals and find useful relationships in terms of angles. Observe that \[\angle DFE = \angle XFE - \angle XFD = \angle CBE - \angle CAB = 180 - \angle ABC - \angle CAB = \angle BAC\] By the so-called "Reverse Law of Cosines" on $\triangle ABC$ we have \[\cos(\angle BAC) = \frac{4^2 - 5^2 - 6^2}{-2 \cdot 5 \cdot 6} = \frac{3}{4}\] Applying on $\triangle DFE$ gives \[DE^2 = 2^2 + 7^2 - 2 \cdot 2 \cdot 7 \cos(\angle DFE)\] \[= 2^2 + 7^2 - 2 \cdot 2 \cdot 7 \cdot \frac{3}{4}\] \[=32\] So $DE = 4 \sqrt{2}$ , now by our cyclic quadrilaterals again, we are motivated by the multiple appearances of similar triangles throughout the figure. We want some that are related to $BX$ and $XD$ , which are crucial lengths in the problem. Suppose $BX = r, XD = s$ for simplicity. We have:
$\bullet~~~~\triangle AXC \sim \triangle FXD$ $\bullet~~~~\triangle BXC \sim \triangle FXE$
So \[\frac{AX}{FX} = \frac{XC}{XD} = \frac{AC}{FD} \implies \frac{4 + r}{FX} = \frac{XC}{s} = 3\] \[\frac{BX}{FX} = \frac{XC}{XE} = \frac{BC}{FE} \implies \frac{r}{FX} = \frac{XC}{s + 4 \sqrt{2}} = \frac{5}{7}\] \[\implies \frac{4 + r}{r} = \frac{s + 4 \sqrt{2}}{s} = \frac{21}{5}\] \[\implies r = \frac{5}{4}, s = \frac{5 \sqrt{2}}{4}\] So $BE = r + s + 4 \sqrt{2} = \frac{5 + 21 \sqrt{2}}{4}$ . The requested sum is $5 + 21 + 2 + 4 = \boxed{032}$

Answer:  032
Problem: Let $\overline{AB}$ be a chord of a circle $\omega$ , and let $P$ be a point on the chord $\overline{AB}$ . Circle $\omega_1$ passes through $A$ and $P$ and is internally tangent to $\omega$ . Circle $\omega_2$ passes through $B$ and $P$ and is internally tangent to $\omega$ . Circles $\omega_1$ and $\omega_2$ intersect at points $P$ and $Q$ . Line $PQ$ intersects $\omega$ at $X$ and $Y$ . Assume that $AP=5$ $PB=3$ $XY=11$ , and $PQ^2 = \frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: [asy] size(8cm); pair O, A, B, P, O1, O2, Q, X, Y; O=(0, 0); A=dir(140); B=dir(40); P=(3A+5B)/8; O1=extension((A+P)/2, (A+P)/2+(0, 1), A, O); O2=extension((B+P)/2, (B+P)/2+(0, 1), B, O); Q=intersectionpoints(circle(O1, length(A-O1)), circle(O2, length(B-O2)))[1]; X=intersectionpoint(P -- (P+(P-Q)*100), circle(O, 1)); Y=intersectionpoint(Q -- (Q+(Q-P)*100), circle(O, 1));  draw(circle(O, 1)); draw(circle(O1, length(A-O1))); draw(circle(O2, length(B-O2))); draw(A -- B); draw(X -- Y); draw(A -- O -- B); draw(O1 -- P -- O2);  dot("$O$", O, S); dot("$A$", A, A); dot("$B$", B, B); dot("$P$", P, dir(70)); dot("$Q$", Q, dir(200)); dot("$O_1$", O1, SW); dot("$O_2$", O2, SE); dot("$X$", X, X); dot("$Y$", Y, Y); [/asy] Let $O_1$ and $O_2$ be the centers of $\omega_1$ and $\omega_2$ , respectively. There is a homothety at $A$ sending $\omega$ to $\omega_1$ that sends $B$ to $P$ and $O$ to $O_1$ , so $\overline{OO_2}\parallel\overline{O_1P}$ . Similarly, $\overline{OO_1}\parallel\overline{O_2P}$ , so $OO_1PO_2$ is a parallelogram. Moreover, \[\angle O_1QO_2=\angle O_1PO_2=\angle O_1OO_2,\] whence $OO_1O_2Q$ is cyclic. However, \[OO_1=O_2P=O_2Q,\] so $OO_1O_2Q$ is an isosceles trapezoid. Since $\overline{O_1O_2}\perp\overline{XY}$ $\overline{OQ}\perp\overline{XY}$ , so $Q$ is the midpoint of $\overline{XY}$
By Power of a Point, $PX\cdot PY=PA\cdot PB=15$ . Since $PX+PY=XY=11$ and $XQ=11/2$ \[XP=\frac{11-\sqrt{61}}2\implies PQ=XQ-XP=\frac{\sqrt{61}}2\implies PQ^2=\frac{61}4,\] and the requested sum is $61+4=\boxed{065}$

Answer:  065