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

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(Solution 1 (No Trigonometry))
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<math> \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} </math>
 
<math> \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} </math>
  
==Solution==
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==Solution 1 (No Trigonometry)==
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[[File:2015AMC10AProblem19Picture.png]]
  
 
<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:a/sqrt{2}</math>, <math>DF = AF</math>.
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Because the side lengths of a <math>45-45-90</math> right triangle are in ratio <math>a:a:a\sqrt{2}</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>.
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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 + FC = 5</math>, <math>DF = \frac{5 - AF}{\sqrt{3}}</math>.
  
 
Setting the two equations for <math>DF</math> equal to each other, <math>AF = \frac{5 - AF}{\sqrt{3}}</math>.
 
Setting the two equations for <math>DF</math> equal to each other, <math>AF = \frac{5 - AF}{\sqrt{3}}</math>.
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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>
  
==Solution 2==
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===Note===
The area of <math>\triangle ABC</math> is <math>12.5</math>, and so the leg length of <math>45 - 45 - 90</math> <math>\triangle ABC</math> is <math>5.</math> Thus, the altitude to hypotenuse <math>AB</math>, <math>CF</math>, has length <math>\dfrac{5}{\sqrt{2}}</math> by <math>45 - 45 - 90</math> 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 <math>30 - 75 - 75</math> triangle. Thus, <math>DF = CF \tan 15^\circ = \dfrac{5}{\sqrt{2}} (2 - \sqrt{3})</math>, and so the area of <math>\triangle CDE</math> is <math>DF \cdot CF = \dfrac{25}{2} (2 - \sqrt{3})</math>. The answer is thus <math>\boxed{\textbf{(D) } \frac{50 - 25\sqrt{3}}{2}}</math>
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Another way to get <math>DF</math> is that you assume <math>AF=DF</math> to be equal to <math>a</math>, as previously mentioned, and <math>CF</math> is equal to <math>a\sqrt{3}</math>. <math>AF+DF=5=a+a\sqrt{3}</math>
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==Solution 2 (Trigonometry)==
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The area of <math>\triangle ABC</math> is <math>12.5</math>, and so the leg length of <math>45 - 45 - 90</math> <math>\triangle ABC</math> is <math>5.</math> Thus, the altitude to hypotenuse <math>AB</math>, <math>CF</math>, has length <math>\dfrac{5}{\sqrt{2}}</math> by <math>45 - 45 - 90</math> 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 <math>30 - 75 - 75</math> triangle. Thus, <math>DF = CF \tan 15^\circ = \dfrac{5}{\sqrt{2}} (2 - \sqrt{3})</math> by the Half-Angle formula, and so the area of <math>\triangle CDE</math> is <math>DF \cdot CF = \dfrac{25}{2} (2 - \sqrt{3})</math>. The answer is thus <math>\boxed{\textbf{(D) } \frac{50 - 25\sqrt{3}}{2}}</math>
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==Solution 3 (Analytical Geometry)==
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Because the area of triangle <math>ABC</math> is <math>12.5</math>, and the triangle is right and isosceles, we can quickly see that the leg length of the triangle <math>ABC</math> is 5. If we put the triangle on the coordinate plane, with vertex <math>C</math> at the origin, and the hypotenuse in the first quadrant, we can use slope-intercept form and tangents to get three lines that intersect at the origin, <math>D</math>, and <math>E</math>. Then, you can use the distance formula to get the length of <math>DE</math>. The height is just <math>\frac{5}{\sqrt{2}}</math>, so the area is just <math>DE \cdot \frac{5}{\sqrt{2}} \cdot \frac{1}{2}=\boxed{\textbf{(D) } \frac{50 - 25\sqrt{3}}{2}}</math>
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==Solution 4 (Trigonometry)==
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Just like with Solution 1, we drop a perpendicular from <math>D</math> onto <math>AC</math>, splitting it into a <math>30</math>-<math>60</math>-<math>90</math> triangle and a <math>45</math>-<math>45</math>-<math>90</math> triangle. We find that <math>AF=\frac{5\sqrt{3}-5}{2}</math>.
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Now, since <math>\triangle AEC\cong \triangle BDC</math> by ASA, <math>CE=CD</math>. Since, <math>DF=\frac{5\sqrt{3}-5}{2}</math>, <math>DC=2\cdot \frac{5\sqrt{3}-5}{2}=5\sqrt{3}-5</math>. By the sine area formula, <math>[CDE]=\frac{1}{2}\cdot \sin 30\cdot CD^2=\frac{1}{4}\cdot (100-50\sqrt{3})=\frac{50-25\sqrt{3}}{2}\implies \boxed{\textbf{(D)}}</math>
  
 
==See Also==
 
==See Also==
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Video Solution:
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https://www.youtube.com/watch?v=JWMIsCS0Ksk
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{{AMC10 box|year=2015|ab=A|num-b=18|num-a=20}}
 
{{AMC10 box|year=2015|ab=A|num-b=18|num-a=20}}
 
{{MAA Notice}}
 
{{MAA Notice}}
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[[Category: Introductory Geometry Problems]]

Revision as of 16:33, 6 December 2020

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 1 (No Trigonometry)

2015AMC10AProblem19Picture.png

$\triangle ADC$ 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 $a:a:a\sqrt{2}$, $DF = AF$.

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 $\boxed{\textbf{(D) }\frac{50 - 25\sqrt{3}}{2}}$

Note

Another way to get $DF$ is that you assume $AF=DF$ to be equal to $a$, as previously mentioned, and $CF$ is equal to $a\sqrt{3}$. $AF+DF=5=a+a\sqrt{3}$

Solution 2 (Trigonometry)

The area of $\triangle ABC$ is $12.5$, and so the leg length of $45 - 45 - 90$ $\triangle ABC$ 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, $DF = CF \tan 15^\circ = \dfrac{5}{\sqrt{2}} (2 - \sqrt{3})$ by the Half-Angle formula, and so the area of $\triangle CDE$ is $DF \cdot CF = \dfrac{25}{2} (2 - \sqrt{3})$. The answer is thus $\boxed{\textbf{(D) } \frac{50 - 25\sqrt{3}}{2}}$

Solution 3 (Analytical Geometry)

Because the area of triangle $ABC$ is $12.5$, and the triangle is right and isosceles, we can quickly see that the leg length of the triangle $ABC$ is 5. If we put the triangle on the coordinate plane, with vertex $C$ at the origin, and the hypotenuse in the first quadrant, we can use slope-intercept form and tangents to get three lines that intersect at the origin, $D$, and $E$. Then, you can use the distance formula to get the length of $DE$. The height is just $\frac{5}{\sqrt{2}}$, so the area is just $DE \cdot \frac{5}{\sqrt{2}} \cdot \frac{1}{2}=\boxed{\textbf{(D) } \frac{50 - 25\sqrt{3}}{2}}$


Solution 4 (Trigonometry)

Just like with Solution 1, we drop a perpendicular from $D$ onto $AC$, splitting it into a $30$-$60$-$90$ triangle and a $45$-$45$-$90$ triangle. We find that $AF=\frac{5\sqrt{3}-5}{2}$.

Now, since $\triangle AEC\cong \triangle BDC$ by ASA, $CE=CD$. Since, $DF=\frac{5\sqrt{3}-5}{2}$, $DC=2\cdot \frac{5\sqrt{3}-5}{2}=5\sqrt{3}-5$. By the sine area formula, $[CDE]=\frac{1}{2}\cdot \sin 30\cdot CD^2=\frac{1}{4}\cdot (100-50\sqrt{3})=\frac{50-25\sqrt{3}}{2}\implies \boxed{\textbf{(D)}}$

See Also

Video Solution:

https://www.youtube.com/watch?v=JWMIsCS0Ksk

2015 AMC 10A (ProblemsAnswer KeyResources)
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Problem 18
Followed by
Problem 20
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