Difference between revisions of "2019 AMC 10B Problems/Problem 15"
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==Problem== | ==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) | + | 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) sides 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> | <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> | ||
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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>. | 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>. | ||
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==Solution 2== | ==Solution 2== | ||
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==Solution 3== | ==Solution 3== | ||
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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>. | 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>. | ||
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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>. | 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>. | ||
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+ | [Note: there is a mismatch of variables near the beginning that someone can fix: xw/2 is the area of the small triangle, which is actually the 30-60-90.] | ||
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+ | ==Video Solution== | ||
+ | https://youtu.be/mXvetCMMzpU | ||
+ | |||
+ | ~IceMatrix | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2019|ab=B|num-b=14|num-a=16}} | {{AMC10 box|year=2019|ab=B|num-b=14|num-a=16}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 22:59, 22 March 2023
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) sides 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 .
[Note: there is a mismatch of variables near the beginning that someone can fix: xw/2 is the area of the small triangle, which is actually the 30-60-90.]
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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