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Difference between revisions of "1954 AHSME Problems/Problem 46"
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== Problem 46==
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In the diagram, if points <math>A, B</math> and <math>C</math> are points of tangency, then <math>x</math> equals:
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<asy>
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unitsize(5cm);
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defaultpen(linewidth(.8pt)+fontsize(8pt));
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dotfactor=3;
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pair A=(-3*sqrt(3)/32,9/32), B=(3*sqrt(3)/32, 9/32), C=(0,9/16);
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pair O=(0,3/8);
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draw((-2/3,9/16)--(2/3,9/16));
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draw((-2/3,1/2)--(-sqrt(3)/6,1/2)--(0,0)--(sqrt(3)/6,1/2)--(2/3,1/2));
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draw(Circle(O,3/16));
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draw((-2/3,0)--(2/3,0));
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label("$A$",A,SW);
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label("$B$",B,SE);
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label("$C$",C,N);
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label("$\frac{3}{8}$",O);
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draw(O+.07*dir(60)--O+3/16*dir(60),EndArrow(3));
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draw(O+.07*dir(240)--O+3/16*dir(240),EndArrow(3));
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label("$\frac{1}{2}$",(.5,.25));
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draw((.5,.33)--(.5,.5),EndArrow(3));
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draw((.5,.17)--(.5,0),EndArrow(3));
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label("$x$",midpoint((.5,.5)--(.5,9/16)));
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draw((.5,5/8)--(.5,9/16),EndArrow(3));
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label("$60^{\circ}$",(0.01,0.12));
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dot(A);
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dot(B);
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dot(C);</asy>
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<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==
 
==Solution 1==
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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);
 +
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);
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draw((-2/3,9/16)--(2/3,9/16));
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draw((-2/3,1/2)--(-sqrt(3)/6,1/2)--(0,0)--(sqrt(3)/6,1/2)--(2/3,1/2));
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draw((9/(16*3^(1/2)),9/16)--(0,0)--(-9/(16*3^(1/2)),9/16));
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draw(Circle(O,3/16));
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draw((-2/3,0)--(2/3,0));
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label("$A$",A,SW);
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label("$B$",B,SE);
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label("$C$",C,N);
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label("$\frac{3}{8}$",O);
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draw(O+.07*dir(60)--O+3/16*dir(60),EndArrow(3));
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draw(O+.07*dir(240)--O+3/16*dir(240),EndArrow(3));
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label("$\frac{1}{2}$",(.5,.25));
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draw((.5,.33)--(.5,.5),EndArrow(3));
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draw((.5,.17)--(.5,0),EndArrow(3));
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label("$x$",midpoint((.5,.5)--(.5,9/16)));
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draw((.5,5/8)--(.5,9/16),EndArrow(3));
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label("$60^{\circ}$",(0.01,0.12));
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dot(A);
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dot(B);
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dot(C);
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label("$D$",(9/(16*3^(1/2)),9/16),NE);
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label("$E$",(0,0),S);
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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 17:01, 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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