Difference between revisions of "2019 AIME II Problems/Problem 15"
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By SpecialBeing2017 | By SpecialBeing2017 | ||
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+ | ==Solution 2== | ||
+ | |||
+ | Let <math>\overline{AP}=a, \overline{PB} = b, \overline{AQ} = c</math> and <math>\overline{QC} = d</math> | ||
+ | |||
+ | By power of point, we have | ||
+ | <math>\overline{AP}\cdot \overline{PB}=\overline{XP}\cdot \overline{YP}</math> and <math>\overline{AQ}\cdot \overline{QC}=\overline{YQ}\cdot \overline{XQ}</math> | ||
+ | |||
+ | Therefore, substituting in the values: | ||
+ | |||
+ | <math>ab = 400</math> | ||
+ | |||
+ | <math>cd = 525</math> | ||
+ | |||
+ | Notice than quadrilateral <math>BPQC</math> is cyclic. | ||
+ | |||
+ | From this fact, we can deduce that <math>\angle PQA= \angle B</math> and <math>\angle QPA = \angle C</math> | ||
+ | |||
+ | Therefore <math>\triangle ABC</math> is similar to <math>\triangle AQP</math>. | ||
+ | |||
+ | Therefore: | ||
+ | <math>\frac{a}{c+d}=\frac{c}{a+b} \implies a^2 + ab = c^2 +cd \implies a^2 + 400 = c^2 + 525 \implies \bf{a^2 = c^2 + 125}</math> | ||
+ | |||
+ | Now using Law of Cosines on <math>\triangle AQP</math> we get: | ||
+ | |||
+ | <math>625 = a^2 + c^2 - 2ac\cos{A}</math> | ||
+ | |||
+ | Notice <math>\cos{A} = \frac{c}{a+b}</math> | ||
+ | |||
+ | Substituting and Simplifying: | ||
+ | |||
+ | <math>625 = a^2 + c^2 - 2ac\frac{c}{a+b}</math> | ||
+ | |||
+ | <math>625 = a^2 + c^2 - 2ac\frac{c}{a+\frac{400}{a}}</math> | ||
+ | |||
+ | <math>625 = c^2 + 125 + c^2 - 2\frac{(ac)^2}{a^2+400}</math> | ||
+ | |||
+ | <math>625 = c^2 + 125 + c^2 - 2\frac{c^2(c^2+125)}{c^2+125+400}</math> | ||
+ | |||
+ | Now we solve for <math>c</math> using regular algebra which actually turns out to be very easy. | ||
+ | |||
+ | We get <math>c = 5\sqrt{35}</math> and from the above relations between the variables we quickly determine <math>d = 3\sqrt{35}</math>, <math>a = 10\sqrt{10}</math> and <math>b = 4\sqrt{10}</math> | ||
+ | |||
+ | Therefore <math>AB\cdot AC = (a+b)\cdot(c+d) = 560\sqrt{14}</math> | ||
+ | |||
+ | So the answer is <math>560 + 14 = \boxed{574} </math> | ||
+ | |||
+ | |||
==See Also== | ==See Also== | ||
{{AIME box|year=2019|n=II|num-b=14|after=Last Question}} | {{AIME box|year=2019|n=II|num-b=14|after=Last Question}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 16:23, 27 June 2019
Contents
Problem
In acute triangle points and are the feet of the perpendiculars from to and from to , respectively. Line intersects the circumcircle of in two distinct points, and . Suppose , , and . The value of can be written in the form where and are positive integers, and is not divisible by the square of any prime. Find .
Solution
Let
Therefore
By power of point, we have Which are simplified to
Or
(1)
Or
Let Then,
In triangle , by law of cosine
Pluging (1)
Or
Substitute everything by
The quadratic term is cancelled out after simplified
Which gives
Plug back in,
Then
So the final answer is
By SpecialBeing2017
Solution 2
Let and
By power of point, we have and
Therefore, substituting in the values:
Notice than quadrilateral is cyclic.
From this fact, we can deduce that and
Therefore is similar to .
Therefore:
Now using Law of Cosines on we get:
Notice
Substituting and Simplifying:
Now we solve for using regular algebra which actually turns out to be very easy.
We get and from the above relations between the variables we quickly determine , and
Therefore
So the answer is
See Also
2019 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Question | |
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.