Difference between revisions of "2021 AMC 12B Problems/Problem 17"
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<cmath>14 = \frac12\cdot XY\cdot (AB+CD) = \frac12\left(\frac 8r + \frac 4s\right)(r+s) = 6 + 4\cdot\frac rs + 2\cdot\frac sr.</cmath> | <cmath>14 = \frac12\cdot XY\cdot (AB+CD) = \frac12\left(\frac 8r + \frac 4s\right)(r+s) = 6 + 4\cdot\frac rs + 2\cdot\frac sr.</cmath> | ||
Thus, the ratio <math>\rho := \tfrac rs</math> satisfies <math>2\rho + \rho^{-1} = 4</math>; solving yields <math>\rho = \boxed{2+\sqrt 2\textbf{ (B)}}</math>. | Thus, the ratio <math>\rho := \tfrac rs</math> satisfies <math>2\rho + \rho^{-1} = 4</math>; solving yields <math>\rho = \boxed{2+\sqrt 2\textbf{ (B)}}</math>. | ||
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+ | == Video Solution by OmegaLearn (Triangle Ratio and Trapezoid Area) == | ||
+ | https://youtu.be/MpMdRI9wC54 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
==See Also== | ==See Also== |
Revision as of 22:37, 11 February 2021
Contents
[hide]Problem
Let be an isoceles trapezoid having parallel bases and with Line segments from a point inside to the vertices divide the trapezoid into four triangles whose areas are and starting with the triangle with base and moving clockwise as shown in the diagram below. What is the ratio
Solution
Without loss let have vertices , , , and , with and . Also denote by the point in the interior of .
Let and be the feet of the perpendiculars from to and , respectively. Observe that and . Now using the formula for the area of a trapezoid yields Thus, the ratio satisfies ; solving yields .
Video Solution by OmegaLearn (Triangle Ratio and Trapezoid Area)
~ pi_is_3.14
See Also
2021 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 16 |
Followed by Problem 18 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.