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Difference between revisions of "1985 AHSME Problems/Problem 3"

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==Problem==
 
==Problem==
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In right <math> \triangle ABC </math> with legs <math> 5 </math> and <math> 12 </math>, arcs of circles are drawn, one with center <math> A </math> and radius <math> 12 </math>, the other with center <math> B </math> and radius <math> 5 </math>. They intersect the [[hypotenuse]] at <math> M </math> and <math> N </math>. Then, <math> MN </math> has length:
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In right <math>\triangle ABC</math> with legs <math>5</math> and <math>12</math>, arcs of circles are drawn, one with center <math>A</math> and radius <math>12</math>, the other with center <math>B</math> and radius <math>5</math>. They intersect the hypotenuse in <math>M</math> and <math>N</math>. Then <math>MN</math> has length
  
 
<asy>
 
<asy>
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==Solution==
 
==Solution==
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First of all, from the [[Pythagorean Theorem]], <math> AB=\sqrt{AC^2+BC^2}=\sqrt{12^2+5^2}=\sqrt{144+25}=\sqrt{169}=13 </math>. Also, since <math> AM </math> and <math> AC </math> are radii of the same circle, <math> AM=AC=12 </math>. Therefore, <math> MB=AB-AM=13-12=1 </math>. Also, since <math> BN </math> and <math> BC </math> are radii of the same circle, <math> BN=BC=5 </math>. We therefore have <math> MN=BN-BM=5-1=4, \boxed{\text{D}} </math>.
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Firstly, the Pythagorean theorem gives <cmath>\begin{align*}AB &=\sqrt{AC^2+BC^2} \\ &= \sqrt{12^2+5^2} \\ & =\sqrt{144+25} \\ &=\sqrt{169} \\ &= 13.\end{align*}</cmath> Also, <math>AM = AC = 12</math> and <math>BN = BC = 5</math> since they are both radii of the respective circles. Thus <math>MB = AB-AM = 13-12 = 1</math>, and so <math>MN = BN-BM = 5-1 = \boxed{\text{(D)} \ 4}</math>.
  
 
==See Also==
 
==See Also==
 
{{AHSME box|year=1985|num-b=2|num-a=4}}
 
{{AHSME box|year=1985|num-b=2|num-a=4}}
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{{MAA Notice}}

Latest revision as of 17:40, 19 March 2024

Problem

In right $\triangle ABC$ with legs $5$ and $12$, arcs of circles are drawn, one with center $A$ and radius $12$, the other with center $B$ and radius $5$. They intersect the hypotenuse in $M$ and $N$. Then $MN$ has length

[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair A=origin, B=(12,7), C=(12,0), M=12*dir(A--B), N=B+B.y*dir(B--A); real r=degrees(B); draw(A--B--C--cycle^^Arc(A,12,0,r)^^Arc(B,B.y,180+r,270)); pair point=incenter(A,B,C); label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$M$", M, dir(point--M)); label("$N$", N, dir(point--N)); label("$12$", (6,0), S); label("$5$", (12,3.5), E);[/asy]

$\mathrm{(A)\ } 2 \qquad \mathrm{(B) \ }\frac{13}{5} \qquad \mathrm{(C) \ } 3 \qquad \mathrm{(D) \ } 4 \qquad \mathrm{(E) \ }\frac{24}{5}$

Solution

Firstly, the Pythagorean theorem gives \begin{align*}AB &=\sqrt{AC^2+BC^2} \\ &= \sqrt{12^2+5^2} \\ & =\sqrt{144+25} \\ &=\sqrt{169} \\ &= 13.\end{align*} Also, $AM = AC = 12$ and $BN = BC = 5$ since they are both radii of the respective circles. Thus $MB = AB-AM = 13-12 = 1$, and so $MN = BN-BM = 5-1 = \boxed{\text{(D)} \ 4}$.

See Also

1985 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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