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Difference between revisions of "2015 AMC 10A Problems/Problem 19"
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Solving gives <math>[CDE] = \frac{50 - 25\sqrt{3}}{2}</math>, so the answer is <math>\boxed{\textbf{(D) }\frac{50 - 25\sqrt{3}}{2}}</math>.
 
Solving gives <math>[CDE] = \frac{50 - 25\sqrt{3}}{2}</math>, so the answer is <math>\boxed{\textbf{(D) }\frac{50 - 25\sqrt{3}}{2}}</math>.
  
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==Solution==
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==Solution 2==
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The area of <math>ABC</math> is 12.5, and so the leg length of 45-45-90 <math>ABC</math> is 5. Thus, the altitude to hypotenuse <math>AB</math>, <math>CF</math>, has length <math>\dfrac{5}{\sqrt{2}}</math> by 45-45-90 right triangles. Now, it is clear that <math>\angle{ACD} = \angle{BCE} = 30^\circ</math>, and so by the Exterior Angle Theorem, <math>\triangle{CDE}</math> is an isosceles 30-75-75 triangle. Thus, <math>DF = CF tan 15^\circ = \dfrac{5}{\sqrt{2}} (2 - \sqrt{3})</math>, and so the area of <math>CDE</math> is <math>DF \cdot CF = \dfrac{25}{2} (2 - \sqrt{3})</math>. The answer is <math>\textbf{D}</math>.
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The area of <math>ABC</math> is 12.5, and so the leg length of 45-45-90 <math>ABC</math> is 5. Thus, the altitude to hypotenuse <math>AB</math>, <math>CF</math>, has length <math>\dfrac{5}{\sqrt{2}}</math> by 45-45-90 right triangles. Now, it is clear that <math>\angle{ACD} = \angle{BCE} = 30^\circ</math>, and so by the Exterior Angle Theorem, <math>\triangle{CDE}</math> is an isosceles 30-75-75 triangle. Thus, <math>DF = CF tan 15^\circ = \dfrac{5}{\sqrt{2}} (2 - \sqrt{3})</math>, and so the area of <math>CDE</math> is <math>DF \cdot CF = \dfrac{25}{2} (2 - \sqrt{3})</math>. The answer is <math>\textbf{(D)}</math>.

Revision as of 18:45, 4 February 2015

Problem

The isosceles right triangle $ABC$ has right angle at $C$ and area $12.5$. The rays trisecting $\angle ACB$ intersect $AB$ at $D$ and $E$. What is the area of $\bigtriangleup CDE$?

$\textbf{(A) }\dfrac{5\sqrt{2}}{3}\qquad\textbf{(B) }\dfrac{50\sqrt{3}-75}{4}\qquad\textbf{(C) }\dfrac{15\sqrt{3}}{8}\qquad\textbf{(D) }\dfrac{50-25\sqrt{3}}{2}\qquad\textbf{(E) }\dfrac{25}{6}$

Solution

$\triangle ADC$ can be split into a $45-45-90$ right triangle and a $30-60-90$ right triangle by dropping a perpendicular from $D$ to side $AC$. Let $F$ be where that perpendicular intersects $AC$.

Because the side lengths of a $45-45-90$ right triangle are in ratio $a:a:2a$, $DF = AF$.

Because the side lengths of a $30-60-90$ right triangle are in ratio $a:a\sqrt{3}:2a$ and $AF$ + $FC = 5$, $DF = \frac{5 - AF}{\sqrt{3}}$.

Setting the two equations for $DF$ equal to each other, $AF = \frac{5 - AF}{\sqrt{3}}$.

Solving gives $AF = DF = \frac{5\sqrt{3} - 5}{2}$.

The area of $\triangle ADC = \frac12 \cdot DF \cdot AC = \frac{25\sqrt{3} - 25}{4}$.

$\triangle ADC$ is congruent to $\triangle BEC$, so their areas are equal.

A triangle's area can be written as the sum of the figures that make it up, so $[ABC] = [ADC] + [BEC] + [CDE]$.

$\frac{25}{2} = \frac{25\sqrt{3} - 25}{4} + \frac{25\sqrt{3} - 25}{4} + [CDE]$.

Solving gives $[CDE] = \frac{50 - 25\sqrt{3}}{2}$, so the answer is $\boxed{\textbf{(D) }\frac{50 - 25\sqrt{3}}{2}}$.

Solution 2

The area of $ABC$ is 12.5, and so the leg length of 45-45-90 $ABC$ is 5. Thus, the altitude to hypotenuse $AB$, $CF$, has length $\dfrac{5}{\sqrt{2}}$ by 45-45-90 right triangles. Now, it is clear that $\angle{ACD} = \angle{BCE} = 30^\circ$, and so by the Exterior Angle Theorem, $\triangle{CDE}$ is an isosceles 30-75-75 triangle. Thus, $DF = CF tan 15^\circ = \dfrac{5}{\sqrt{2}} (2 - \sqrt{3})$, and so the area of $CDE$ is $DF \cdot CF = \dfrac{25}{2} (2 - \sqrt{3})$. The answer is $\textbf{(D)}$.

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