Difference between revisions of "2019 AMC 10B Problems/Problem 16"
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draw((0,5)--(2.5,3),black); | draw((0,5)--(2.5,3),black); | ||
dot((0,5)); | dot((0,5)); | ||
+ | label ("$B$", (0,8),N); | ||
+ | label ("$C$", (0,0),SW); | ||
+ | label ("$A$", (4,0),SE); | ||
+ | label ("$E$", (0,5),W); | ||
+ | label ("$D$", (2.5,3),NE); | ||
</asy> | </asy> | ||
~ By Little Mouse | ~ By Little Mouse | ||
− | + | <!-- EDITED BY LVLUO --> | |
==Solution 1== | ==Solution 1== | ||
Without loss of generality, let <math>AC = CD = 4</math> and <math>DE = EB = 3</math>. Let <math>\angle A = \alpha</math> and <math>\angle B = \beta = 90^{\circ} - \alpha</math>. As <math>\triangle ACD</math> and <math>\triangle DEB</math> are isosceles, <math>\angle ADC = \alpha</math> and <math>\angle BDE = \beta</math>. Then <math>\angle CDE = 180^{\circ} - \alpha - \beta = 90^{\circ}</math>, so <math>\triangle CDE</math> is a <math>3-4-5</math> triangle with <math>CE = 5</math>. | Without loss of generality, let <math>AC = CD = 4</math> and <math>DE = EB = 3</math>. Let <math>\angle A = \alpha</math> and <math>\angle B = \beta = 90^{\circ} - \alpha</math>. As <math>\triangle ACD</math> and <math>\triangle DEB</math> are isosceles, <math>\angle ADC = \alpha</math> and <math>\angle BDE = \beta</math>. Then <math>\angle CDE = 180^{\circ} - \alpha - \beta = 90^{\circ}</math>, so <math>\triangle CDE</math> is a <math>3-4-5</math> triangle with <math>CE = 5</math>. |
Revision as of 14:46, 21 June 2022
Contents
[hide]Problem
In with a right angle at , point lies in the interior of and point lies in the interior of so that and the ratio . What is the ratio
Diagram
~ By Little Mouse
Solution 1
Without loss of generality, let and . Let and . As and are isosceles, and . Then , so is a triangle with .
Then , and is a triangle.
In isosceles triangles and , drop altitudes from and onto ; denote the feet of these altitudes by and respectively. Then by AAA similarity, so we get that , and . Similarly we get , and .
Alternatively, once finding the length of one could use the Pythagorean Theorem to find and consequently , and then compute the ratio.
Solution 2
Let , and . (For this solution, is above , and is to the right of ). Also let , so , which implies . Similarly, , which implies . This further implies that .
Now we see that . Thus is a right triangle, with side lengths of , , and (by the Pythagorean Theorem, or simply the Pythagorean triple ). Therefore (by definition), , and . Hence (by the double angle formula), giving .
By the Law of Cosines in , if , we have Now . Thus the answer is .
Solution 3
WLOG, let , and . . Because of this, is a 3-4-5 right triangle. Draw the altitude of . is by the base-height triangle area formula. is similar to (AA). So . is of . Therefore, is .
~Thegreatboy90
Solution 4 (a bit long)
WLOG, and . Notice that in , we have . Since and , we find that and , so and is right. Therefore, by 3-4-5 triangle, and . Define point F such that is an altitude; we know the area of the whole triangle is and we know the hypotenuse is , so . By the geometric mean theorem, . Solving the quadratic we get , so . For now, assume . Then . splits into two parts (quick congruence by Leg-Angle) so and . . Now we know and , we can find or .
Video Solution 1
~IceMatrix
Video Solution 2
https://youtu.be/4_x1sgcQCp4?t=4245
~ pi_is_3.14
See Also
2019 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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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