Difference between revisions of "1985 AHSME Problems/Problem 3"
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==Problem== | ==Problem== | ||
− | 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 | + | 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== | ||
− | + | 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}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 17:40, 19 March 2024
Problem
In right with legs and , arcs of circles are drawn, one with center and radius , the other with center and radius . They intersect the hypotenuse in and . Then has length
Solution
Firstly, the Pythagorean theorem gives Also, and since they are both radii of the respective circles. Thus , and so .
See Also
1985 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
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All AHSME Problems and Solutions |
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