Difference between revisions of "2015 AMC 10A Problems/Problem 19"

(Created page with "==Problem== The isosceles right triangle <math>ABC</math> has right angle at <math>C</math> and area <math>12.5</math>. The rays trisecting <math>\angle ACB</math> intersect <ma...")
 
(Solution)
Line 11: Line 11:
 
<math>\triangle ADC</math> can be split into a <math>45-45-90</math> right triangle and a <math>30-60-90</math> right triangle by dropping a perpendicular from <math>D</math> to side <math>AC</math>. Let <math>F</math> be where that perpendicular intersects <math>AC</math>.
 
<math>\triangle ADC</math> can be split into a <math>45-45-90</math> right triangle and a <math>30-60-90</math> right triangle by dropping a perpendicular from <math>D</math> to side <math>AC</math>. Let <math>F</math> be where that perpendicular intersects <math>AC</math>.
  
Because the side lengths of a <math>45-45-90</math> right triangle are in ratio <math>a:a:2a</math>, <math>DF = </math>AF<math>.
+
Because the side lengths of a <math>45-45-90</math> right triangle are in ratio <math>a:a:2a</math>, <math>DF = AF</math>.
  
Because the side lengths of a </math>30-60-90<math> right triangle are in ratio </math>a:a\sqrt{3}:2a<math> and </math>AF<math> + </math>FC = 5<math>, </math>DF = \frac{5 - AF}{\sqrt{3}}<math>.
+
Because the side lengths of a <math>30-60-90</math> right triangle are in ratio <math>a:a\sqrt{3}:2a</math> and <math>AF</math> + <math>FC = 5</math>, <math>DF = \frac{5 - AF}{\sqrt{3}}</math>.
  
Setting the two equations for </math>DF<math> equal together, </math>AF = \frac{5 - AF}{\sqrt{3}}<math>.
+
Setting the two equations for <math>DF</math> equal together, <math>AF = \frac{5 - AF}{\sqrt{3}}</math>.
  
Solving gives </math>AF = DF = \frac{5\sqrt{3} - 5}<math>.
+
Solving gives <math>AF = DF = \frac{5\sqrt{3} - 5}</math>.
  
The area of </math>\triangle ADC = \frac12 \cdot DF \cdot AC = \frac{25\sqrt{3} - 25}{2}<math>.
+
The area of <math>\triangle ADC = \frac12 \cdot DF \cdot AC = \frac{25\sqrt{3} - 25}{2}</math>.
  
</math>\triangle ADC<math> is congruent to </math>\triangle BEC<math>, so their areas are equal.
+
<math>\triangle ADC</math> is congruent to <math>\triangle BEC</math>, so their areas are equal.
  
A triangle's area can be written as the sum of the figures that make it up, so </math>[ABC] = [ADC] + [BEC] + [CDE]<math>.
+
A triangle's area can be written as the sum of the figures that make it up, so <math>[ABC] = [ADC] + [BEC] + [CDE]</math>.
  
</math>\frac{25}{2} = \frac{25\sqrt{3} - 25}{2} + \frac{25\sqrt{3} - 25}{2} + [CDE]<math>.
+
<math>\frac{25}{2} = \frac{25\sqrt{3} - 25}{2} + \frac{25\sqrt{3} - 25}{2} + [CDE]</math>.
  
Solving gives </math>[CDE] = \frac{50 - 25\sqrt{3}}{2}$, so the answer is \boxed{\textbf{(D) }\frac{50 - 25\sqrt{3}}{2}}.
+
Solving gives <math>[CDE] = \frac{50 - 25\sqrt{3}}{2}</math>, so the answer is \boxed{\textbf{(D) }\frac{50 - 25\sqrt{3}}{2}}.

Revision as of 17:46, 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

Let $D$ and $E$ be the points at which the angle trisectors intersect $AB$.

$\triangle ADC$ 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 together, $AF = \frac{5 - AF}{\sqrt{3}}$.

Solving gives $AF = DF = \frac{5\sqrt{3} - 5}$ (Error compiling LaTeX. ! Missing } inserted.).

The area of $\triangle ADC = \frac12 \cdot DF \cdot AC = \frac{25\sqrt{3} - 25}{2}$.

$\triangle ADC$ 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 $[ABC] = [ADC] + [BEC] + [CDE]$.

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

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

Invalid username
Login to AoPS