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Difference between revisions of "Mock AIME 1 2010 Problems/Problem 13"

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== Solution ==
 
== Solution ==
−
<math>\boxed{372}</math>.
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<asy>
  
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import geometry;
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size(8cm);
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point B = origin;
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point A = (3,8);
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point C = (12,0);
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triangle t = triangle(A,B,C);
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circle c = circumcircle(t);
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point B1 = foot(t.VB);
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point C1 = foot(t.VC);
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point D = intersectionpoint(line(B1,C1), line(B,C));
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pair[] e = intersectionpoints(line(A,D), c);
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point E = e[0];
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// Triangle ABC and Circumcircle
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draw(t);
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draw(c);
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 +
// Altitudes
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draw(B--B1);
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draw(C--C1);
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// Segments AD,EB,BD, and B_1D
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draw(A--D);
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draw(E--B);
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draw(B--D);
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draw(B1--D);
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 +
// Point Labels
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dot(A);
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label("A",A,NW);
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dot(B);
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label("B",B,SSW);
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dot(C);
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label("C",C,SE);
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dot(B1);
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label("B$_1$",B1,NE);
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dot(C1);
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label("C$_1$",C1,NNW);
 +
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dot(D);
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label("D",D,SW);
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dot(E);
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label("E",E,NW);
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</asy>
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 +
Let <math>\measuredangle BED=\theta</math>. Because the problem gives us <math>ED^2+EB^2=3050</math>, we think to use the [[Law of Cosines]] in <math>\triangle BED</math>, which yields <math>BD^2=ED^2+EB^2-2ED\cdot EB\cos\theta</math>. Subtituting the values given by the problem, we get <math>28^2=3050-2ED\cdot EB\cos\theta</math>, which gives <math>ED\cdot EB=\tfrac{3050-784}{2\cos\theta}=\tfrac{1133}{\cos\theta}</math>.
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 +
To find another expression for <math>ED\cdot EB</math>, we think of the formula <math>[\triangle BED]=\tfrac12ED\cdot EB\sin\theta</math>. We know that the area of the triangle is <math>\tfrac12\cdot\tfrac{20\sqrt{159}}7\cdot28=40\sqrt{159}</math>. Substituting this in the previous equation for <math>[\triangle BED]</math>, we get that <math>40\sqrt{159}=\tfrac12ED\cdot EB\sin\theta</math>, so <math>ED\cdot EB=\tfrac{80\sqrt{159}}{\sin\theta}</math>.
 +
 +
Setting these two expressions for <math>ED\cdot EB</math> equal to each other reveals that <math>\tfrac{1133}{\cos\theta}=\tfrac{80\sqrt{159}}{\sin\theta}</math>, so <math>\tan\theta=\tfrac{80\sqrt{159}}{1133}</math> by the identity <math>\tan\theta=\tfrac{\sin\theta}{\cos\theta}.</math>
 +
 +
<math>\angle AEB</math> is [[supplementary]] to <math>\angle BED</math>, and <math>\angle ACB</math> is supplementary to <math>\angle AEB</math>, because <math>AEBC</math> is a [[cyclic quadrilateral]]. Thus, <math>\measuredangle BED=\measuredangle ACB=\theta</math>, so <math>\tan\measuredangle ACB=\tan\theta=\tfrac{80\sqrt{159}}{1133}</math>. Thus, <math>a+b+c=80+159+1133=1372</math>, so our answer is <math>\boxed{372}</math>.
  
 
== See Also ==
 
== See Also ==
 
{{Mock AIME box|year=2010|n=1|num-b=12|num-a=14}}
 
{{Mock AIME box|year=2010|n=1|num-b=12|num-a=14}}
 
[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]

Latest revision as of 16:54, 11 August 2024

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 $\boxed{372}$.

See Also

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