Difference between revisions of "2021 JMPSC Accuracy Problems/Problem 12"

Revision as of 23:17, 10 July 2021

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 $2+3=\boxed{8}.$

@@ Line 6: / Line 6: @@
 ==Solution==
+<center>
-[[File:Accuracy12sol.jpeg|200px]]
+[[File:Accuracy12sol.jpeg|500px]]
+</center>
 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>

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Difference between revisions of "2021 JMPSC Accuracy Problems/Problem 12"

Revision as of 23:17, 10 July 2021

Problem

Solution