Difference between revisions of "2021 JMPSC Accuracy Problems/Problem 12"
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Since <math>EF</math> is parallel to <math>AC</math> and <math>\angle C =60^\circ</math>, we have that <math>\angle BFE = 60^\circ</math> by corresponding angles. Similarly, <math>\angle BEF = 90^\circ</math> and it follows that <math>\triangle BFE</math> is a <math>30-60-90</math> right triangle. | Since <math>EF</math> is parallel to <math>AC</math> and <math>\angle C =60^\circ</math>, we have that <math>\angle BFE = 60^\circ</math> by corresponding angles. Similarly, <math>\angle BEF = 90^\circ</math> and it follows that <math>\triangle BFE</math> is a <math>30-60-90</math> right triangle. | ||
− | Since the side opposite the <math>60^\circ</math> angle in <math>\triangle BFE</math> is <math>1</math>, we use our <math>30-60-90</math> ratios to find that <math>EF=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}.</math> In rectangle <math>EFDA</math>, we also have <cmath>AD=\frac{\sqrt{3}}{3}.</cmath> Analogously, we find that <cmath>QP=\frac{\sqrt{3}}{3}.</cmath> Since we are looking for the base <math>d</math> of the horizontal rectangle and we are given <cmath>PA=1,</cmath> we have <cmath>d=QP+PA+AD=\frac{\sqrt{3}}{3}+1+\frac{\sqrt{3}}{3}=\frac{3+2\sqrt{3}}{3}.</cmath> This gives us an answer of <math>2+3=\boxed{8}.</math> | + | Since the side opposite the <math>60^\circ</math> angle in <math>\triangle BFE</math> is <math>1</math>, we use our <math>30-60-90</math> ratios to find that <math>EF=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}.</math> In rectangle <math>EFDA</math>, we also have <cmath>AD=\frac{\sqrt{3}}{3}.</cmath> Analogously, we find that <cmath>QP=\frac{\sqrt{3}}{3}.</cmath> Since we are looking for the base <math>d</math> of the horizontal rectangle and we are given <cmath>PA=1,</cmath> we have <cmath>d=QP+PA+AD=\frac{\sqrt{3}}{3}+1+\frac{\sqrt{3}}{3}=\frac{3+2\sqrt{3}}{3}.</cmath> This gives us an answer of <math>2+3=\boxed{8}.</math> ~ samrocksnature |
Revision as of 22:18, 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
~ samrocksnature