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Difference between revisions of "1970 AHSME Problems/Problem 7"

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== Solution ==
 
== Solution ==
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<math>\fbox{E}</math>
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All answers are proportional to <math>s</math>, so for ease, let <math>s=1</math>.
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Let <math>ABCD</math> be oriented so that <math>A(0, 0), B(1, 0), C(1, 1), D(0, 1)</math>.
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The circle centered at <math>A</math> with radius <math>1</math> is <math>x^2 + y^2 = 1</math>.  The circle centered at <math>B</math> with radius <math>1</math> is <math>(x - 1)^2 + y^2 = 1</math>.  Solving each equation for <math>1 - y^2</math> to find the intersection leads to <math>(x - 1)^2 = x^2</math>, which leads to <math>x = \frac{1}{2}</math>.
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Plugging that back in to <math>x^2 + y^2 = 1</math> leads to <math>y^2 = 1 - \frac{1}{4}</math>, or <math>y = \pm \frac{\sqrt{3}}{2}</math>.  Since we want the intersection within the square where <math>0 < x, y < 1</math>, we take the positive solution, and the intersection is at <math>X(\frac{1}{2}, \frac{\sqrt{3}}{2})</math>.
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The distance from <math>X</math> to side <math>CD</math>, which lies along the line <math>y=1</math>, is <math>1 - \frac{\sqrt{3}}{2}</math>, or <math>\frac{2  -\sqrt{3}}{2}</math>.  This is answer <math>\fbox{E}</math> when <math>s=1</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 19:54, 13 July 2019

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 $s=1$.

See also

1970 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35
All AHSME Problems and Solutions

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