Difference between revisions of "1996 AIME Problems/Problem 13"
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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>. | ||
+ | |||
+ | |||
+ | == Solution 2== | ||
+ | Because the problem asks for a ratio, we can divide each side length by <math>\sqrt{3}</math> to make things simpler. We now have a triangle with sides <math>\sqrt{10}</math>, <math>\sqrt{5}</math>, and <math>\sqrt{2}</math>. | ||
+ | |||
+ | We use the same graph as above. | ||
+ | |||
+ | Draw perpendicular from <math>C</math> to <math>AE</math>. Denote this point as <math>F</math>. We know that <math>DE = EF = x</math> and <math>BD = CF = z</math> and also let <math>AE = y</math>. | ||
+ | |||
+ | Using Pythagorean theorem, we get three equations, | ||
+ | |||
+ | <center><math>(y+x)^2 + z^2 = 10</math> | ||
+ | |||
+ | <math>(y-x)^2 + z^2 = 2</math> | ||
+ | |||
+ | <math>x^2 + z^2 = \frac{5}{4}</math></center> | ||
+ | |||
+ | Adding the first and second, we obtain <math>x^2 + y^2 + z^2 = 6</math>, and then subtracting the third from this we find that <math>y = \frac{\sqrt{19}}{2}</math>. (Note, we could have used [[Stewart's Theorem]] to achieve this result). | ||
+ | |||
+ | Subtracting the first and second, we see that <math>xy = 2</math>, and then we find that <math>x = \frac{4}{\sqrt{19}}</math> | ||
+ | |||
+ | Using base ratios, we then quickly find that the desired ratio is <math>\frac{27}{38}</math> so our answer is <math>\boxed{065}</math> | ||
== See also == | == See also == |
Latest revision as of 01:18, 25 February 2016
Contents
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
(see 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
,
Subtracting the two equations yields . Then
, and
.
Solution 2
Because the problem asks for a ratio, we can divide each side length by to make things simpler. We now have a triangle with sides
,
, and
.
We use the same graph as above.
Draw perpendicular from to
. Denote this point as
. We know that
and
and also let
.
Using Pythagorean theorem, we get three equations,
![$(y+x)^2 + z^2 = 10$](http://latex.artofproblemsolving.com/3/b/5/3b57d7bc20921b5d8a4123c069b99335ceb564a7.png)
![$x^2 + z^2 = \frac{5}{4}$](http://latex.artofproblemsolving.com/7/3/3/733acc19c8fe72c8a52840ca98c88a96601ef256.png)
Adding the first and second, we obtain , and then subtracting the third from this we find that
. (Note, we could have used Stewart's Theorem to achieve this result).
Subtracting the first and second, we see that , and then we find that
Using base ratios, we then quickly find that the desired ratio is so our answer is
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.