# 1954 AHSME Problems/Problem 46

## Problem 46

In the diagram, if points $A, B$ and $C$ are points of tangency, then $x$ equals:

$[asy] unitsize(5cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); dotfactor=3; pair A=(-3*sqrt(3)/32,9/32), B=(3*sqrt(3)/32, 9/32), C=(0,9/16); pair O=(0,3/8); draw((-2/3,9/16)--(2/3,9/16)); draw((-2/3,1/2)--(-sqrt(3)/6,1/2)--(0,0)--(sqrt(3)/6,1/2)--(2/3,1/2)); draw(Circle(O,3/16)); draw((-2/3,0)--(2/3,0)); label("A",A,SW); label("B",B,SE); label("C",C,N); label("\frac{3}{8}",O); draw(O+.07*dir(60)--O+3/16*dir(60),EndArrow(3)); draw(O+.07*dir(240)--O+3/16*dir(240),EndArrow(3)); label("\frac{1}{2}",(.5,.25)); draw((.5,.33)--(.5,.5),EndArrow(3)); draw((.5,.17)--(.5,0),EndArrow(3)); label("x",midpoint((.5,.5)--(.5,9/16))); draw((.5,5/8)--(.5,9/16),EndArrow(3)); label("60^{\circ}",(0.01,0.12)); dot(A); dot(B); dot(C);[/asy]$

$\textbf{(A)}\ \frac{3}{16}"\qquad\textbf{(B)}\ \frac{1}{8}"\qquad\textbf{(C)}\ \frac{1}{32}"\qquad\textbf{(D)}\ \frac{3}{32}"\qquad\textbf{(E)}\ \frac{1}{16}"$

## Solution 1

First we extend the line with $A$ and the line $B$ so that they both meet the line with $C$, forming an equilateral triangle. Let the vertices of this triangle be $D$, $E$, and $F$. We know it is equilateral because of the angle of $60^\circ$ shown, and because the tangent lines $\overline{EF}$ and $\overline{DE}$ are congruent. $[asy] unitsize(5cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); dotfactor=3; pair A=(-3*sqrt(3)/32,9/32), B=(3*sqrt(3)/32, 9/32), C=(0,9/16); pair O=(0,3/8); draw((-2/3,9/16)--(2/3,9/16)); draw((-2/3,1/2)--(-sqrt(3)/6,1/2)--(0,0)--(sqrt(3)/6,1/2)--(2/3,1/2)); draw((9/(16*3^(1/2)),9/16)--(0,0)--(-9/(16*3^(1/2)),9/16)); draw(Circle(O,3/16)); draw((-2/3,0)--(2/3,0)); label("A",A,SW); label("B",B,SE); label("C",C,N); label("\frac{3}{8}",O); draw(O+.07*dir(60)--O+3/16*dir(60),EndArrow(3)); draw(O+.07*dir(240)--O+3/16*dir(240),EndArrow(3)); label("\frac{1}{2}",(.5,.25)); draw((.5,.33)--(.5,.5),EndArrow(3)); draw((.5,.17)--(.5,0),EndArrow(3)); label("x",midpoint((.5,.5)--(.5,9/16))); draw((.5,5/8)--(.5,9/16),EndArrow(3)); label("60^{\circ}",(0.01,0.12)); dot(A); dot(B); dot(C); label("D",(9/(16*3^(1/2)),9/16),NE); label("E",(0,0),S); label("F",(-9/(16*3^(1/2)),9/16),NW);[/asy]$ We can see, because $A$, $B$, and $C$ are points of tangency, that circle $ABC$ is inscribed in $\triangle DEF$. The height of an equilateral triangle is exactly $3$ times the radius of a circle inscribed in it. Let the height of $\triangle DEF$ be $h$. We can see that the radius of the circle equals $\frac{3}{16}$. Thus $$h = \frac{3\cdot3}{16} = \frac 9{16}.$$ Subtracting $\frac 12$ from $h$ gives us $$x = h-\frac 12 = \frac 1{16},$$ so our answer is $\boxed{\text{E}}$.