2021 JMPSC Accuracy Problems/Problem 12

Revision as of 17:24, 11 July 2021 by Samrocksnature (talk | contribs)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)


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 $a+b+c$.




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

See also

  1. Other 2021 JMPSC Accuracy Problems
  2. 2021 JMPSC Accuracy Answer Key
  3. All JMPSC Problems and Solutions

The problems on this page are copyrighted by the Junior Mathematicians' Problem Solving Competition. JMPSC.png