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

Revision as of 22:57, 6 April 2023

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)

$\triangle ADC$ 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, $[CDE]=\frac{1}{2}\cdot \sin 30\cdot CD^2=\frac{1}{4}\cdot (100-50\sqrt{3})=\frac{50-25\sqrt{3}}{2}\implies \boxed{\textbf{(D)}}$

Solution 5 (Basic Trigonometry)

Prerequisite knowledge for this solution: the side ratios of a 30-60-90, and 45-45-90 right triangle.

We let point C be the origin. Since $\overline{CD}$ and $\overline{CE}$ trisect $\angle ACB = 90^{\circ}$ , this means $m\angle CEB = 30^{\circ}$ and the equation of $\overline{CE}$ is $y=\frac{\sqrt{3}}{3}$ (you can figure out that the tangent of 30 degrees gives $\frac{\sqrt{3}}{3}$ ). Next, we can find A to be at $(0, 5)$ and B at $(5, 0)$ , so the equation of $\overline{AB}$ is $y=-x+5$ . So we have the system:

$\begin{cases}y=\frac{\sqrt{3}}{3}x\\y=-x+5\end{cases}$

By substituting values, we can arrive at $\frac{3+\sqrt{3}}{3}x=5$ , or $x=5\cdot\frac{3}{3+\sqrt{3}}=\frac{15}{3+\sqrt{3}}$ . We multiply $x=\frac{15}{3+\sqrt{3}}\cdot\frac{3-\sqrt{3}}{3-\sqrt{3}}=\frac{45-15\sqrt{3}}{6}=\frac{15-5\sqrt{3}}{2}$ .

Dropping an altitude from E onto $\overline{CB}$ , and calling the intersection point G, we find that $\triangle EGB$ is a 45-45-90 triangle with a leg of $\frac{15-5\sqrt{3}}{2}\cdot\frac{\sqrt{3}}{3}=\frac{15\sqrt{3}-15}{6}=\frac{5\sqrt{3}-5}{2}$ . Thus, $EB=\frac{5\sqrt{3}-5}{2}\sqrt{2}=\frac{5\sqrt{6}-5\sqrt{2}}{2}$ .

Dropping an altitude from C onto $\overline{AB}$ , and calling the intersection point H, we find that $CH=\frac{5\sqrt{2}}{2}=BH$ , and by the theorem of betweenness applied to H, E, and B, we get $HE=HB-EB=\frac{5\sqrt{2}}{2}-\frac{5\sqrt{6}-5\sqrt{2}}{2}=\frac{10\sqrt{2}-5\sqrt{6}}{2}$ .

We are almost done. By symmetry, $HD=HE$ , so to find the area of the triangle CED, we only need to multiply HE by CH, $\frac{10\sqrt{2}-5\sqrt{6}}{2}\cdot\frac{5\sqrt{2}}{2}=\frac{100-50\sqrt{3}}{4}=\frac{50-25\sqrt{3}}{2}$ . This is answer choice $\boxed{\textbf{(D) } \frac{50-25\sqrt{3}}{2}}$

~JH. L

Video Solution:

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

@@ Line 5: / Line 5: @@
 <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==
+==Solution 1 (No Trigonometry)==
+[[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>.
-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>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>.
+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 together, <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>.
 Solving gives <math>AF = DF = \frac{5\sqrt{3} - 5}{2}</math>.
@@ Line 25: / Line 27: @@
 <math>\frac{25}{2} = \frac{25\sqrt{3} - 25}{4} + \frac{25\sqrt{3} - 25}{4} + [CDE]</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>.
+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>
+===Note===
+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>
+==Solution 2 (Trigonometry)==
+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>
+==Solution 3 (Analytical Geometry)==
+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>
+==Solution 4 (Trigonometry)==
+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>.
+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>
+==Solution 5 (Basic Trigonometry)==
+Prerequisite knowledge for this solution: the side ratios of a 30-60-90, and 45-45-90 right triangle.
+We let point C be the origin. Since <math>\overline{CD}</math> and <math>\overline{CE}</math> trisect <math>\angle ACB = 90^{\circ}</math>, this means <math>m\angle CEB = 30^{\circ}</math> and the equation of <math>\overline{CE}</math> is <math>y=\frac{\sqrt{3}}{3}</math> (you can figure out that the tangent of 30 degrees gives <math>\frac{\sqrt{3}}{3}</math>). Next, we can find A to be at <math>(0, 5)</math> and B at <math>(5, 0)</math>, so the equation of <math>\overline{AB}</math> is <math>y=-x+5</math>. So we have the system:
+<cmath>\begin{cases}y=\frac{\sqrt{3}}{3}x\\y=-x+5\end{cases}</cmath>
+By substituting values, we can arrive at <math>\frac{3+\sqrt{3}}{3}x=5</math>, or <math>x=5\cdot\frac{3}{3+\sqrt{3}}=\frac{15}{3+\sqrt{3}}</math>. We multiply <math>x=\frac{15}{3+\sqrt{3}}\cdot\frac{3-\sqrt{3}}{3-\sqrt{3}}=\frac{45-15\sqrt{3}}{6}=\frac{15-5\sqrt{3}}{2}</math>.
+Dropping an altitude from E onto <math>\overline{CB}</math>, and calling the intersection point G, we find that <math>\triangle EGB</math> is a 45-45-90 triangle with a leg of <math>\frac{15-5\sqrt{3}}{2}\cdot\frac{\sqrt{3}}{3}=\frac{15\sqrt{3}-15}{6}=\frac{5\sqrt{3}-5}{2}</math>. Thus, <math>EB=\frac{5\sqrt{3}-5}{2}\sqrt{2}=\frac{5\sqrt{6}-5\sqrt{2}}{2}</math>.
+Dropping an altitude from C onto <math>\overline{AB}</math>, and calling the intersection point H, we find that <math>CH=\frac{5\sqrt{2}}{2}=BH</math>, and by the theorem of betweenness applied to H, E, and B, we get <math>HE=HB-EB=\frac{5\sqrt{2}}{2}-\frac{5\sqrt{6}-5\sqrt{2}}{2}=\frac{10\sqrt{2}-5\sqrt{6}}{2}</math>.
+We are almost done. By symmetry, <math>HD=HE</math>, so to find the area of the triangle CED, we only need to multiply HE by CH, <math>\frac{10\sqrt{2}-5\sqrt{6}}{2}\cdot\frac{5\sqrt{2}}{2}=\frac{100-50\sqrt{3}}{4}=\frac{50-25\sqrt{3}}{2}</math>. This is answer choice <math>\boxed{\textbf{(D) } \frac{50-25\sqrt{3}}{2}}</math>
+~JH. L
+==Video Solution:==
+https://www.youtube.com/watch?v=JWMIsCS0Ksk
+==See Also==
+{{AMC10 box|year=2015|ab=A|num-b=18|num-a=20}}
+{{MAA Notice}}
+[[Category: Introductory Geometry Problems]]

2015 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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