Mock AIME 1 2010 Problems/Problem 13

Problem

Suppose $\triangle ABC$ is inscribed in circle $\Gamma$ . $B_1$ and $C_1$ are the feet of the altitude from $B$ to $CA$ and $C$ to $AB$ , respectively. Let $D$ be the intersection of lines $\overline{B_1 C_1}$ and $\overline{BC}$ , let $E$ be the point of intersection of $\Gamma$ and line $\overline{DA}$ distinct from $A$ , and let $F$ be the foot of the perpendicular from $E$ to $BD$ . Given that $BD = 28$ , $EF = \frac{20 \sqrt{159}}{7}$ , and $ED^2 + EB^2 = 3050$ , and that $\tan m \angle ACB$ can be expressed in the form $\frac{a \sqrt{b}}{c}$ , where $a$ and $c$ are relatively prime positive integers and $b$ is an integer not divisible by the square of any prime, find the last three digits of $a + b + c$ .

Solution

$[asy] import geometry; size(8cm); point B = origin; point A = (3,8); point C = (12,0); triangle t = triangle(A,B,C); circle c = circumcircle(t); point B1 = foot(t.VB); point C1 = foot(t.VC); point D = intersectionpoint(line(B1,C1), line(B,C)); pair[] e = intersectionpoints(line(A,D), c); point E = e[0]; // Triangle ABC and Circumcircle draw(t); draw(c); // Altitudes draw(B--B1); draw(C--C1); // Segments AD,EB,BD, and B_1D draw(A--D); draw(E--B); draw(B--D); draw(B1--D); // Point Labels dot(A); label("A",A,NW); dot(B); label("B",B,SSW); dot(C); label("C",C,SE); dot(B1); label("B$_1$",B1,NE); dot(C1); label("C$_1$",C1,NNW); dot(D); label("D",D,SW); dot(E); label("E",E,NW); [/asy]$

Let $\measuredangle BED=\theta$ . Because the problem gives us $ED^2+EB^2=3050$ , we think to use the Law of Cosines in $\triangle BED$ , which yields $BD^2=ED^2+EB^2-2ED\cdot EB\cos\theta$ . Subtituting the values given by the problem, we get $28^2=3050-2ED\cdot EB\cos\theta$ , which gives $ED\cdot EB=\tfrac{3050-784}{2\cos\theta}=\tfrac{1133}{\cos\theta}$ .

To find another expression for $ED\cdot EB$ , we think of the formula $[\triangle BED]=\tfrac12ED\cdot EB\sin\theta$ . We know that the area of the triangle is $\tfrac12\cdot\tfrac{20\sqrt{159}}7\cdot28=40\sqrt{159}$ . Substituting this in the previous equation for $[\triangle BED]$ , we get that $40\sqrt{159}=\tfrac12ED\cdot EB\sin\theta$ , so $ED\cdot EB=\tfrac{80\sqrt{159}}{\sin\theta}$ .

Setting these two expressions for $ED\cdot EB$ equal to each other reveals that $\tfrac{1133}{\cos\theta}=\tfrac{80\sqrt{159}}{\sin\theta}$ , so $\tan\theta=\tfrac{80\sqrt{159}}{1133}$ by the identity $\tan\theta=\tfrac{\sin\theta}{\cos\theta}.$

$\angle AEB$ is supplementary to $\angle BED$ , and $\angle ACB$ is supplementary to $\angle AEB$ , because $AEBC$ is a cyclic quadrilateral. Thus, $\measuredangle BED=\measuredangle ACB=\theta$ , so $\tan\measuredangle ACB=\tan\theta=\tfrac{80\sqrt{159}}{1133}$ . Thus, $a+b+c=80+159+1133=1372$ , so our answer is $\boxed{372}$ .

Mock AIME 1 2010 (Problems, Source)
Preceded by Problem 12	Followed by Problem 14
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Mock AIME 1 2010 Problems/Problem 13

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