Difference between revisions of "1996 AIME Problems/Problem 13"
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By [[Stewart's Theorem]], <math>AE = \frac{\sqrt{2(AB^2 + AC^2) - BC^2}}2 = \frac{\sqrt {57}}{2}</math>, and by the [[Pythagorean Theorem]] on <math>\triangle ABD, \triangle EBD</math>, | By [[Stewart's Theorem]], <math>AE = \frac{\sqrt{2(AB^2 + AC^2) - BC^2}}2 = \frac{\sqrt {57}}{2}</math>, and by the [[Pythagorean Theorem]] on <math>\triangle ABD, \triangle EBD</math>, | ||
− | < | + | <cmath>\begin{align*} |
BD^2 + \left(DE + \frac {\sqrt{57}}2\right)^2 &= 30 \\ | BD^2 + \left(DE + \frac {\sqrt{57}}2\right)^2 &= 30 \\ | ||
BD^2 + DE^2 &= \frac{15}{4} \\ | BD^2 + DE^2 &= \frac{15}{4} \\ | ||
− | \end{align*}</ | + | \end{align*}</cmath> |
Subtracting the two equations yields <math>DE\sqrt{57} + \frac{57}{4} = \frac{105}{4} \Longrightarrow DE = \frac{12}{\sqrt{57}}</math>. Then <math>\frac mn = \frac{1}{2} + \frac{DE}{2AE} = \frac{1}{2} + \frac{\frac{12}{\sqrt{57}}}{2 \cdot \frac{\sqrt{57}}{2}} = \frac{27}{38}</math>, and <math>m+n = \boxed{065}</math>. | Subtracting the two equations yields <math>DE\sqrt{57} + \frac{57}{4} = \frac{105}{4} \Longrightarrow DE = \frac{12}{\sqrt{57}}</math>. Then <math>\frac mn = \frac{1}{2} + \frac{DE}{2AE} = \frac{1}{2} + \frac{\frac{12}{\sqrt{57}}}{2 \cdot \frac{\sqrt{57}}{2}} = \frac{27}{38}</math>, and <math>m+n = \boxed{065}</math>. |
Revision as of 11:47, 13 March 2015
Problem
In triangle , , , and . There is a point for which bisects , and is a right angle. The ratio can be written in the form , where and are relatively prime positive integers. Find .
Solution
Let be the midpoint of . Since , then and share the same height and have equal bases, and thus have the same area. Similarly, and share the same height, and have bases in the ratio , so (see area ratios). Now,
By Stewart's Theorem, , and by the Pythagorean Theorem on ,
Subtracting the two equations yields . Then , and .
See also
1996 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.