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

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==Solution 1==
 
==Solution 1==
  
First we extend the line with <math>A</math> and the line <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>
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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);
 
unitsize(5cm);
 
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defaultpen(linewidth(.8pt)+fontsize(8pt));

Latest revision as of 18:01, 24 May 2020

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 with $B$ 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 $\boxed{\text{E}}$.

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