# Difference between revisions of "1954 AHSME Problems/Problem 46"

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− | E | + | == Problem 46== |

+ | |||

+ | In the diagram, if points <math>A, B</math> and <math>C</math> are points of tangency, then <math>x</math> 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> | ||

+ | |||

+ | <math> \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}" </math> | ||

+ | |||

+ | ==Solution 1== | ||

+ | |||

+ | First we extend the line with <math>A</math> and the line with <math>B</math> so that they both meet the line with <math>C</math>, forming an equilateral triangle. Let the vertices of this triangle be <math>D</math>, <math>E</math>, and <math>F</math>. We know it is equilateral because of the angle of <math>60^\circ</math> shown, and because the tangent lines <math>\overline{EF}</math> and <math>\overline{DE}</math> 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 <math>A</math>, <math>B</math>, and <math>C</math> are points of tangency, that circle <math>ABC</math> is inscribed in <math>\triangle DEF</math>. The height of an equilateral triangle is exactly <math>3</math> times the radius of a circle inscribed in it. Let the height of <math>\triangle DEF</math> be <math>h</math>. We can see that the radius of the circle equals <math>\frac{3}{16}</math>. Thus <cmath>h = \frac{3\cdot3}{16} = \frac 9{16}.</cmath> Subtracting <math>\frac 12</math> from <math>h</math> gives us <cmath>x = h-\frac 12 = \frac 1{16},</cmath> so our answer is <math>\boxed{\text{E}}</math>. |

## Latest revision as of 18:01, 24 May 2020

## Problem 46

In the diagram, if points and are points of tangency, then equals:

## Solution 1

First we extend the line with and the line with so that they both meet the line with , forming an equilateral triangle. Let the vertices of this triangle be , , and . We know it is equilateral because of the angle of shown, and because the tangent lines and are congruent. We can see, because , , and are points of tangency, that circle is inscribed in . The height of an equilateral triangle is exactly times the radius of a circle inscribed in it. Let the height of be . We can see that the radius of the circle equals . Thus Subtracting from gives us so our answer is .