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 [[hypotenuse]] | + | 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 |
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==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}} |
Latest revision as of 00:35, 3 April 2018
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
First of all, from the Pythagorean Theorem, . Also, since and are radii of the same circle, . Therefore, . Also, since and are radii of the same circle, . We therefore have .
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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