1996 AIME Problems/Problem 13
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
![[asy] pointpen = black; pathpen = black + linewidth(0.7); pair B=(0,0), C=(15^.5, 0), A=IP(CR(B,30^.5),CR(C,6^.5)), E=(B+C)/2, D=foot(B,A,E); D(MP("A",A)--MP("B",B,SW)--MP("C",C)--A--MP("D",D)--B); D(MP("E",E)); MP("\sqrt{30}",(A+B)/2,NW); MP("\sqrt{6}",(A+C)/2,SE); MP("\frac{\sqrt{15}}2",(E+C)/2); D(rightanglemark(B,D,A)); [/asy]](http://latex.artofproblemsolving.com/7/e/9/7e995d2f972dad7f2aefe99478218d9c0a690451.png)
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
(where
denotes area; the concept of comparing bases and heights is known as area ratios). Now,
![$\dfrac{[ADB]}{[ABC]} = \frac{[ABE] + [BDE]}{2[ABE]} = \frac{1}{2} + \frac{DE}{2AE}.$](http://latex.artofproblemsolving.com/1/2/a/12a88550c43fd7bd333f379af275aa2895d4102e.png)
By Stewart's Theorem, , and by the Pythagorean Theorem on
,
BD^2 + \left(DE + \frac {\sqrt{57}}2\right)^2 &= 30 \\ BD^2 + DE^2 &= \frac{15}{4} \\
\end{align*}$ (Error compiling LaTeX. Unknown error_msg)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 |