Difference between revisions of "2021 JMPSC Accuracy Problems/Problem 12"
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+ | #[[2021 JMPSC Sprint Problems|Other 2021 JMPSC Sprint Problems]] | ||
+ | #[[2021 JMPSC Sprint Answer Key|2021 JMPSC Sprint Answer Key]] | ||
+ | #[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]] | ||
+ | {{JMPSC Notice}} |
Revision as of 16:19, 11 July 2021
Contents
[hide]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 ~ samrocksnature
Solution 2
Since the angles of an equilateral triangle are , we have by similar triangles that the length of the segments of the "small" rectangles are , meaning the answer is ~Geometry285
See also
The problems on this page are copyrighted by the Junior Mathematicians' Problem Solving Competition.