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Difference between revisions of "1997 AHSME Problems/Problem 19"
(Area solution)
 
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<math> \textbf{(A)}\ 2.18\qquad\textbf{(B)}\ 2.24\qquad\textbf{(C)}\ 2.31\qquad\textbf{(D)}\ 2.37\qquad\textbf{(E)}\ 2.41 </math>
 
<math> \textbf{(A)}\ 2.18\qquad\textbf{(B)}\ 2.24\qquad\textbf{(C)}\ 2.31\qquad\textbf{(D)}\ 2.37\qquad\textbf{(E)}\ 2.41 </math>
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==Solution==
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<asy>
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defaultpen(linewidth(.8pt));
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dotfactor=3;
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pair A = origin;
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pair B = (1,0);
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pair C = (0,sqrt(3));
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pair O = (2.33,2.33);
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pair D = (2.33,0);
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pair E = (0, 2.33);
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pair F = (0.35, 1.1);
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dot(A);dot(B);dot(C);dot(O);dot(D);dot(E);dot(F);
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label("$A$",A,SW);label("$B$",B,SE);label("$C$",C,W);label("$O$",O,NW);label("$D$",D,SW);label("$E$",E,SW);label("$F$", F, W);
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label("$1$",midpoint(A--B),S);label("$60^\circ$",B,2W + N);
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draw((3,0)--A--(0,3));
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draw(B--C);draw(O--E);draw(O--D);
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draw(Arc(O,2.33,163,288.5));</asy>
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Draw radii <math>OE</math> and <math>OD</math> to the axes, and label the point of tangency to triangle <math>ABC</math> point <math>F</math>.  Let the radius of the circle <math>O</math> be <math>r</math>.  Square <math>OEAD</math> has side length <math>r</math>.
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Because <math>BD</math> and <math>BF</math> are tangents from a common point <math>B</math>, <math>BD = BF</math>.
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<math>AD = AB + BD</math>
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<math>r = 1 + BD</math>
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<math>r = 1 + BF</math>
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Similarly, <math>CF = CE</math>, and we can write:
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<math>AE = AC + CE</math>
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<math>r = \sqrt{3} + CF</math>
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Equating the radii lengths, we have <math>1 + BF = \sqrt{3} + CF</math>
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This means <math>BF - CF = \sqrt{3} - 1</math>
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<math>BF + CF = 2</math> by the 30-60-90 triangle.
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Therefore, <math>2BF = 2 + \sqrt{3} - 1</math>, and we get <math>BF = \frac{1}{2} + \frac{\sqrt{3}}{2}</math>
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The radius of the circle is <math>AD</math>, which is <math>BF + 1 = \frac{3}{2} + \frac{\sqrt{3}}{2}</math>
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Using decimal approximations, <math>r \approx 1.5 + \frac{1.73^+}{2} \approx 2.37</math>, and the answer is <math>\boxed{D}</math>.
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== Solution 2 ==
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----
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From the diagram above, it is more direct to note that BC = CF + BF = r - <math>\sqrt{3}</math> + r - 1 = 2
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== Solution 3 ==
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The total area of both kites is <math>BC \cdot OF= 2r</math>. Thus, <math>r^2 = \frac{\sqrt{3}}{2} + 2r</math>
  
 
== See also ==
 
== See also ==
 
{{AHSME box|year=1997|num-b=18|num-a=20}}
 
{{AHSME box|year=1997|num-b=18|num-a=20}}
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{{MAA Notice}}

Latest revision as of 19:28, 10 October 2024

Problem

A circle with center $O$ is tangent to the coordinate axes and to the hypotenuse of the $30^\circ$-$60^\circ$-$90^\circ$ triangle $ABC$ as shown, where $AB=1$. To the nearest hundredth, what is the radius of the circle?


[asy] defaultpen(linewidth(.8pt)); dotfactor=3; pair A = origin; pair B = (1,0); pair C = (0,sqrt(3)); pair O = (2.33,2.33); dot(A);dot(B);dot(C);dot(O); label("$A$",A,SW);label("$B$",B,SE);label("$C$",C,W);label("$O$",O,NW); label("$1$",midpoint(A--B),S);label("$60^\circ$",B,2W + N); draw((3,0)--A--(0,3)); draw(B--C); draw(Arc(O,2.33,163,288.5));[/asy]

$\textbf{(A)}\ 2.18\qquad\textbf{(B)}\ 2.24\qquad\textbf{(C)}\ 2.31\qquad\textbf{(D)}\ 2.37\qquad\textbf{(E)}\ 2.41$

Solution

[asy] defaultpen(linewidth(.8pt)); dotfactor=3; pair A = origin; pair B = (1,0); pair C = (0,sqrt(3)); pair O = (2.33,2.33); pair D = (2.33,0); pair E = (0, 2.33); pair F = (0.35, 1.1); dot(A);dot(B);dot(C);dot(O);dot(D);dot(E);dot(F); label("$A$",A,SW);label("$B$",B,SE);label("$C$",C,W);label("$O$",O,NW);label("$D$",D,SW);label("$E$",E,SW);label("$F$", F, W); label("$1$",midpoint(A--B),S);label("$60^\circ$",B,2W + N); draw((3,0)--A--(0,3)); draw(B--C);draw(O--E);draw(O--D); draw(Arc(O,2.33,163,288.5));[/asy]

Draw radii $OE$ and $OD$ to the axes, and label the point of tangency to triangle $ABC$ point $F$. Let the radius of the circle $O$ be $r$. Square $OEAD$ has side length $r$.

Because $BD$ and $BF$ are tangents from a common point $B$, $BD = BF$.

$AD = AB + BD$

$r = 1 + BD$

$r = 1 + BF$

Similarly, $CF = CE$, and we can write:

$AE = AC + CE$

$r = \sqrt{3} + CF$

Equating the radii lengths, we have $1 + BF = \sqrt{3} + CF$

This means $BF - CF = \sqrt{3} - 1$

$BF + CF = 2$ by the 30-60-90 triangle.

Therefore, $2BF = 2 + \sqrt{3} - 1$, and we get $BF = \frac{1}{2} + \frac{\sqrt{3}}{2}$

The radius of the circle is $AD$, which is $BF + 1 = \frac{3}{2} + \frac{\sqrt{3}}{2}$

Using decimal approximations, $r \approx 1.5 + \frac{1.73^+}{2} \approx 2.37$, and the answer is $\boxed{D}$.

Solution 2


From the diagram above, it is more direct to note that BC = CF + BF = r - $\sqrt{3}$ + r - 1 = 2

Solution 3

The total area of both kites is $BC \cdot OF= 2r$. Thus, $r^2 = \frac{\sqrt{3}}{2} + 2r$

See also

1997 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
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All AHSME Problems and Solutions

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