Difference between revisions of "2019 AMC 10B Problems/Problem 15"
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+ | ==Problem== | ||
+ | |||
+ | Right triangles <math>T_1</math> and <math>T_2</math>, have areas of 1 and 2, respectively. A side of <math>T_1</math> is congruent to a side of <math>T_2</math>, and a different side of <math>T_1</math> is congruent to a different side of <math>T_2</math>. What is the square of the product of the lengths of the other (third) side of <math>T_1</math> and <math>T_2</math>? | ||
+ | |||
+ | <math>\textbf{(A) }\frac{28}{3}\qquad\textbf{(B) }10\qquad\textbf{(C) }\frac{32}{3}\qquad\textbf{(D) }\frac{34}{3}\qquad\textbf{(E) }12</math> | ||
+ | |||
+ | ==Solution 1== | ||
+ | |||
+ | First of all, let the two sides which are congruent be <math>x</math> and <math>y</math>, where <math>y > x</math>. The only way that the conditions of the problem can be satisfied is if <math>x</math> is the shorter leg of <math>T_{2}</math> and the longer leg of <math>T_{1}</math>, and <math>y</math> is the longer leg of <math>T_{2}</math> and the hypotenuse of <math>T_{1}</math>. | ||
+ | |||
+ | Notice that this means the value we are looking for is the square of <math>\sqrt{x^{2}+y^{2}} \cdot \sqrt{y^{2}-x^{2}} = \sqrt{y^{4}-x^{4}}</math>, which is just <math>y^{4}-x^{4}</math>. | ||
+ | |||
+ | The area conditions give us two equations: <math>\frac{xy}{2} = 2</math> and <math>\frac{x\sqrt{y^{2}-x^{2}}}{2} = 1</math>. | ||
+ | |||
+ | This means that <math>y = \frac{4}{x}</math> and that <math>\frac{4}{x^{2}} = y^{2} - x^{2}</math>. | ||
+ | |||
+ | Taking the second equation, we get <math>x^{2}y^{2} - x^{4} = 4</math>, so since <math>xy = 4</math>, <math>x^{4} = 12</math>. | ||
+ | |||
+ | Since <math>y = \frac{4}{x}</math>, we get <math>y^{4} = \frac{256}{12} = \frac{64}{3}</math>. | ||
+ | |||
+ | The value we are looking for is just <math>y^{4}-x^{4} = \frac{64-36}{3} = \frac{28}{3}</math> so the answer is <math>\boxed{\textbf{(A) }\frac{28}{3}}</math>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | Like in Solution 1, we have <math>\frac{xy}{2} = 2</math> and <math>\frac{x\sqrt{y^{2}-x^{2}}}{2} = 1</math>. | ||
+ | |||
+ | Squaring both equations yields <math>x^2y^2=16</math> and <math>x^2(y^2-x^2)=4</math>. | ||
+ | |||
+ | Let <math>a = x^2</math> and <math>b = y^2</math>. Then <math>b = \frac{16}{a}</math>, and <math>a\left(\frac{16}{a}-a\right)=4 \implies 16 - a^2 = 4 \implies a = 2\sqrt3</math>, so <math>b = \frac{16}{2\sqrt3} = \frac{8\sqrt3}{3}</math>. | ||
+ | |||
+ | We are looking for the value of <math>y^4 - x^4 = b^2 - a^2</math>, so the answer is <math>\frac{64}{3} - 12 = \boxed{\textbf{(A) }\frac{28}{3}}</math>. | ||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | Firstly, let the right triangles be <math>\triangle ABC</math> and <math>\triangle EDF</math>, with <math>\triangle ABC</math> being the smaller triangle. As in Solution 1, let <math>\overline{AB} = \overline{EF} = x</math> and <math>\overline{BC} = \overline{DF} = y</math>. Additionally, let <math>\overline{AC} = z</math> and <math>\overline{DE} = w</math>. | ||
+ | |||
+ | We are given that <math>[ABC] = 1</math> and <math>[EDF] = 2</math>, so using <math>\text{area} = \frac{bh}{2}</math>, we have <math>\frac{xy}{2} = 1</math> and <math>\frac{xw}{2} = 2</math>. Dividing the two equations, we get <math>\frac{xy}{xw}</math> = <math>\frac{y}{w} = 2</math>, so <math>y = 2w</math>. | ||
+ | |||
+ | Thus <math>\triangle EDF</math> is a <math>30^{\circ}-60^{\circ}-90^{\circ}</math> right triangle, meaning that <math>x = w\sqrt{3}</math>. Now by the Pythagorean Theorem in <math>\triangle ABC</math>, <math>\left(w\sqrt{3}\right)^2 + \left(2w\right)^2 = z^2 \Rightarrow 3w^2 + 4w^2 = z^2 \Rightarrow 7w^2 = z^2 \Rightarrow w\sqrt{7} = z</math>. | ||
+ | |||
+ | The problem requires the square of the product of the third side lengths of each triangle, which is <math>(wz)^2</math>. By substitution, we see that <math>wz</math> = <math>\left(w\right)\left(w\sqrt{7}\right) = w^2\sqrt{7}</math>. We also know <math>\frac{xw}{2} = 1 \Rightarrow\frac{(w)\left(w\sqrt{3}\right)}{2} =1 \Rightarrow (w)\left(w\sqrt{3}\right) = 2 \Rightarrow w^2\sqrt{3} = 2 \Rightarrow w^2 = \frac{2\sqrt{3}}{3}</math>. | ||
+ | |||
+ | Since we want <math>\left(w^2\sqrt{7}\right)^2</math>, multiplying both sides by <math>\sqrt{7}</math> gets us <math>w^2\sqrt{7} = \frac{2\sqrt{21}}{3}</math>. Now squaring gives <math>\left(\frac{2\sqrt{21}}{3}\right)^2 = \frac{4*21}{9} = \boxed{\textbf{(A) }\frac{28}{3}}</math>. | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/mXvetCMMzpU | ||
+ | |||
+ | ~IceMatrix | ||
+ | |||
+ | ==See Also== | ||
+ | {{AMC10 box|year=2019|ab=B|num-b=14|num-a=16}} | ||
+ | {{MAA Notice}} |
Latest revision as of 04:13, 28 January 2020
Problem
Right triangles and , have areas of 1 and 2, respectively. A side of is congruent to a side of , and a different side of is congruent to a different side of . What is the square of the product of the lengths of the other (third) side of and ?
Solution 1
First of all, let the two sides which are congruent be and , where . The only way that the conditions of the problem can be satisfied is if is the shorter leg of and the longer leg of , and is the longer leg of and the hypotenuse of .
Notice that this means the value we are looking for is the square of , which is just .
The area conditions give us two equations: and .
This means that and that .
Taking the second equation, we get , so since , .
Since , we get .
The value we are looking for is just so the answer is .
Solution 2
Like in Solution 1, we have and .
Squaring both equations yields and .
Let and . Then , and , so .
We are looking for the value of , so the answer is .
Solution 3
Firstly, let the right triangles be and , with being the smaller triangle. As in Solution 1, let and . Additionally, let and .
We are given that and , so using , we have and . Dividing the two equations, we get = , so .
Thus is a right triangle, meaning that . Now by the Pythagorean Theorem in , .
The problem requires the square of the product of the third side lengths of each triangle, which is . By substitution, we see that = . We also know .
Since we want , multiplying both sides by gets us . Now squaring gives .
Video Solution
~IceMatrix
See Also
2019 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
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