1970 AHSME Problems/Problem 7

Problem

Inside square $ABCD$ with side $s$ , quarter-circle arcs with radii $s$ and centers at $A$ and $B$ are drawn. These arcs intersect at a point $X$ inside the square. How far is $X$ from the side of $CD$ ?

$\text{(A) } \tfrac{1}{2} s(\sqrt{3}+4)\quad \text{(B) } \tfrac{1}{2} s\sqrt{3}\quad \text{(C) } \tfrac{1}{2} s(1+\sqrt{3})\quad \text{(D) } \tfrac{1}{2} s(\sqrt{3}-1)\quad \text{(E) } \tfrac{1}{2} s(2-\sqrt{3})$

Solution

All answers are proportional to $s$ , so for ease, let $s=1$ .

Let $ABCD$ be oriented so that $A(0, 0), B(1, 0), C(1, 1), D(0, 1)$ .

The circle centered at $A$ with radius $1$ is $x^2 + y^2 = 1$ . The circle centered at $B$ with radius $1$ is $(x - 1)^2 + y^2 = 1$ . Solving each equation for $1 - y^2$ to find the intersection leads to $(x - 1)^2 = x^2$ , which leads to $x = \frac{1}{2}$ .

Plugging that back in to $x^2 + y^2 = 1$ leads to $y^2 = 1 - \frac{1}{4}$ , or $y = \pm \frac{\sqrt{3}}{2}$ . Since we want the intersection within the square where $0 < x, y < 1$ , we take the positive solution, and the intersection is at $X(\frac{1}{2}, \frac{\sqrt{3}}{2})$ .

The distance from $X$ to side $CD$ , which lies along the line $y=1$ , is $1 - \frac{\sqrt{3}}{2}$ , or $\frac{2 -\sqrt{3}}{2}$ . This is answer $\fbox{E}$ when $s=1$ .

1970 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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1970 AHSME Problems/Problem 7

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