2021 JMPSC Accuracy Problems/Problem 12

Problem

A rectangle with base $1$ and height $2$ is inscribed in an equilateral triangle. Another rectangle with height $1$ is also inscribed in the triangle. The base of the second rectangle can be written as a fully simplified fraction $\frac{a+b\sqrt{3}}{c}$ such that $gcd(a,b,c)=1.$ Find $a+b+c$ .

Solution

We are given $DF=1$ , from which in rectangle $EFDA$ we can conclude $AE=1$ . Since $AB=2$ , we have $AB-AE=2-1=1=BE.$

Since $EF$ is parallel to $AC$ and $\angle C =60^\circ$ , we have that $\angle BFE = 60^\circ$ by corresponding angles. Similarly, $\angle BEF = 90^\circ$ and it follows that $\triangle BFE$ is a $30-60-90$ right triangle.

Since the side opposite the $60^\circ$ angle in $\triangle BFE$ is $1$ , we use our $30-60-90$ ratios to find that $EF=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}.$ In rectangle $EFDA$ , we also have $AD=\frac{\sqrt{3}}{3}.$ Analogously, we find that $QP=\frac{\sqrt{3}}{3}.$ Since we are looking for the base $d$ of the horizontal rectangle and we are given $PA=1,$ we have $d=QP+PA+AD=\frac{\sqrt{3}}{3}+1+\frac{\sqrt{3}}{3}=\frac{3+2\sqrt{3}}{3}.$ This gives us an answer of $2+3+3=\boxed{8}.$ ~ samrocksnature

Solution 2

Since the angles of an equilateral triangle are $60^o$ , we have by similar triangles that the length of the segments of the "small" rectangles are $\frac{\sqrt{3}}{3}$ , meaning the answer is $\frac{3+2\sqrt{3}}{3} \implies 3+2+3=\boxed{8}$ $\linebreak$ ~Geometry285

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2021 JMPSC Accuracy Problems/Problem 12

Problem

Solution

Solution 2