Difference between revisions of "2005 AIME II Problems/Problem 14"
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Drop the altitude from A and call the base of the altitude Q. Also, drop the altitudes from E and D to AB and AC respectively. Call the feet of the altitudes R and S respectively. | Drop the altitude from A and call the base of the altitude Q. Also, drop the altitudes from E and D to AB and AC respectively. Call the feet of the altitudes R and S respectively. | ||
− | From here, we can use Heron's Formula to find the altitude. The area of the triangle is <math>\sqrt{21*6*7*8}</math> = 84. We can then use similar triangles with triangle AQC and triangle DSC to find DS=<math>\frac{24}{5}</math>. Consequently, from Pythagorean theorem, SC = <math>\frac{18}{5}</math> and AS = 14-SC = <math>\frac{52}{5}</math>. We can also use pythagorean triangle on triangle AQB | + | From here, we can use Heron's Formula to find the altitude. The area of the triangle is <math>\sqrt{21*6*7*8}</math> = 84. We can then use similar triangles with triangle AQC and triangle DSC to find DS=<math>\frac{24}{5}</math>. Consequently, from Pythagorean theorem, SC = <math>\frac{18}{5}</math> and AS = 14-SC = <math>\frac{52}{5}</math>. We can also use pythagorean triangle on triangle AQB to determine that BQ = <math>\frac{33}{5}</math>. |
Label AR as y and RE as x. RB then equals 13-y. Then, we have two similar triangles. | Label AR as y and RE as x. RB then equals 13-y. Then, we have two similar triangles. |
Revision as of 20:04, 11 December 2016
Problem
In triangle and
Point
is on
with
Point
is on
such that
Given that
where
and
are relatively prime positive integers, find
Solution 1
![[asy] import olympiad; import cse5; import geometry; size(150); defaultpen(fontsize(10pt)); defaultpen(0.8); dotfactor = 4; pair A = origin; pair C = rotate(15,A)*(A+dir(-50)); pair B = rotate(15,A)*(A+dir(-130)); pair D = extension(A,A+dir(-68),B,C); pair E = extension(A,A+dir(-82),B,C); label("$A$",A,N); label("$B$",B,SW); label("$D$",D,SE); label("$E$",E,S); label("$C$",C,SE); draw(A--B--C--cycle); draw(A--E); draw(A--D); draw(anglemark(B,A,E,5)); draw(anglemark(D,A,C,5)); [/asy]](http://latex.artofproblemsolving.com/f/5/3/f53f3016596f4d0c3c6dc23b6e0b5b41fabe85f2.png)
By the Law of Sines and since , we have
Substituting our knowns, we have . The answer is
.
Solution 2 (Similar Triangles)
Drop the altitude from A and call the base of the altitude Q. Also, drop the altitudes from E and D to AB and AC respectively. Call the feet of the altitudes R and S respectively.
From here, we can use Heron's Formula to find the altitude. The area of the triangle is = 84. We can then use similar triangles with triangle AQC and triangle DSC to find DS=
. Consequently, from Pythagorean theorem, SC =
and AS = 14-SC =
. We can also use pythagorean triangle on triangle AQB to determine that BQ =
.
Label AR as y and RE as x. RB then equals 13-y. Then, we have two similar triangles.
Firstly: . From there, we have
.
Next: . From there, we have
.
Solve the system to get and
. Notice that 463 is prime, so even though we use pythagorean theorem on x and 13-y, the denominator won't change. The answer we desire is
.
See also
2005 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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.