# 1993 AHSME Problems/Problem 17

## Problem $[asy] draw((-1,-1)--(1,-1)--(1,1)--(-1,1)--cycle, black+linewidth(.75)); draw((0,-1)--(0,1), black+linewidth(.75)); draw((-1,0)--(1,0), black+linewidth(.75)); draw((-1,-1/sqrt(3))--(1,1/sqrt(3)), black+linewidth(.75)); draw((-1,1/sqrt(3))--(1,-1/sqrt(3)), black+linewidth(.75)); draw((-1/sqrt(3),-1)--(1/sqrt(3),1), black+linewidth(.75)); draw((1/sqrt(3),-1)--(-1/sqrt(3),1), black+linewidth(.75)); [/asy]$

Amy painted a dartboard over a square clock face using the "hour positions" as boundaries.[See figure.] If $t$ is the area of one of the eight triangular regions such as that between 12 o'clock and 1 o'clock, and $q$ is the area of one of the four corner quadrilaterals such as that between 1 o'clock and 2 o'clock, then $\frac{q}{t}=$ $\text{(A) } 2\sqrt{3}-2\quad \text{(B) } \frac{3}{2}\quad \text{(C) } \frac{\sqrt{5}+1}{2}\quad \text{(D) } \sqrt{3}\quad \text{(E) } 2$

## Solution

Assume the length of the side of the square is 2, WLOG. This means the side of one t section is 1. As the lines are at clock face positions, each section has a $\tfrac{360}{12} = 30$ degree angle from the center. So each section t is a $30-60-90$ triangle with a long leg of 1. Therefore, the short leg is $\tfrac{1}{\sqrt3}$.

This makes the area of each $t = \tfrac{1}{2}\cdot b \cdot h = \tfrac{1}{2}\cdot 1 \cdot \tfrac{1}{\sqrt3} = \tfrac{1}{2\sqrt3}$

The total area comprises $4q+8t$, so $$4q+(8\cdot \tfrac{1}{2\sqrt3}) = 2^2=4$$ $$4q + \tfrac{4}{\sqrt3} = 4$$ $$4q = 4 - \tfrac{4}{\sqrt3}$$ $$q = 1 - \tfrac{1}{\sqrt3} = \tfrac{\sqrt3 - 1 }{\sqrt3}$$ $$\frac{q}{t} = \frac{\tfrac{\sqrt3 - 1 }{\sqrt3}}{\tfrac{1}{2\sqrt3}} = 2\cdot (\sqrt3 - 1) = \boxed{2\sqrt3-2}$$ $\fbox{A}$

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