Difference between revisions of "2021 JMPSC Accuracy Problems/Problem 12"
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We are given <math>DF=1</math>, from which in rectangle <math>EFDA</math> we can conclude <math>AE=1</math>. Since <math>AB=2</math>, we have <cmath>AB-AE=2-1=1=BE.</cmath> | We are given <math>DF=1</math>, from which in rectangle <math>EFDA</math> we can conclude <math>AE=1</math>. Since <math>AB=2</math>, we have <cmath>AB-AE=2-1=1=BE.</cmath> | ||
Revision as of 22:17, 10 July 2021
Problem
A rectangle with base and height is inscribed in an equilateral triangle. Another rectangle with height is also inscribed in the triangle. The base of the second rectangle can be written as a fully simplified fraction such that Find .
Solution
We are given , from which in rectangle we can conclude . Since , we have
Since is parallel to and , we have that by corresponding angles. Similarly, and it follows that is a right triangle.
Since the side opposite the angle in is , we use our ratios to find that In rectangle , we also have Analogously, we find that Since we are looking for the base of the horizontal rectangle and we are given we have This gives us an answer of