Difference between revisions of "2015 AMC 10A Problems/Problem 19"
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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== | + | ==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>. | <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: | + | 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 | + | 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 | + | 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>. | Solving gives <math>AF = DF = \frac{5\sqrt{3} - 5}{2}</math>. | ||
− | The area of <math>\triangle ADC = \frac12 \cdot DF \cdot AC = \frac{25\sqrt{3} - 25}{ | + | The area of <math>\triangle ADC = \frac12 \cdot DF \cdot AC = \frac{25\sqrt{3} - 25}{4}</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. | ||
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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}{ | + | <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> | ||
+ | |||
+ | ==See Also== | ||
+ | Video Solution: | ||
+ | |||
+ | https://www.youtube.com/watch?v=JWMIsCS0Ksk | ||
+ | |||
+ | {{AMC10 box|year=2015|ab=A|num-b=18|num-a=20}} | ||
+ | {{MAA Notice}} | ||
+ | |||
+ | [[Category: Introductory Geometry Problems]] |
Latest revision as of 15:33, 6 December 2020
Contents
Problem
The isosceles right triangle has right angle at and area . The rays trisecting intersect at and . What is the area of ?
Solution 1 (No Trigonometry)
can be split into a right triangle and a right triangle by dropping a perpendicular from to side . Let be where that perpendicular intersects .
Because the side lengths of a right triangle are in ratio , .
Because the side lengths of a right triangle are in ratio and , .
Setting the two equations for equal to each other, .
Solving gives .
The area of .
is congruent to , so their areas are equal.
A triangle's area can be written as the sum of the figures that make it up, so .
.
Solving gives , so the answer is
Note
Another way to get is that you assume to be equal to , as previously mentioned, and is equal to .
Solution 2 (Trigonometry)
The area of is , and so the leg length of is Thus, the altitude to hypotenuse , , has length by right triangles. Now, it is clear that , and so by the Exterior Angle Theorem, is an isosceles triangle. Thus, by the Half-Angle formula, and so the area of is . The answer is thus
Solution 3 (Analytical Geometry)
Because the area of triangle is , and the triangle is right and isosceles, we can quickly see that the leg length of the triangle is 5. If we put the triangle on the coordinate plane, with vertex 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, , and . Then, you can use the distance formula to get the length of . The height is just , so the area is just
Solution 4 (Trigonometry)
Just like with Solution 1, we drop a perpendicular from onto , splitting it into a -- triangle and a -- triangle. We find that .
Now, since by ASA, . Since, , . By the sine area formula,
See Also
Video Solution:
https://www.youtube.com/watch?v=JWMIsCS0Ksk
2015 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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