Difference between revisions of "2014 AIME I Problems/Problem 15"
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== Solution == | == Solution == | ||
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| + | <asy> | ||
| + | pair A = (0,3); | ||
| + | pair B = (0,0); | ||
| + | pair C = (4,0); | ||
| + | draw(A--B--C--cycle); | ||
| + | dotfactor = 3; | ||
| + | dot("$A$",A,dir(135)); | ||
| + | dot("$B$",B,dir(215)); | ||
| + | dot("$C$",C,dir(305)); | ||
| + | pair D = (2.21, 0); | ||
| + | pair E = (0, 1.21); | ||
| + | pair F = (1.71, 1.71); | ||
| + | pair G = (2, 1.5); | ||
| + | dot("$D$",D,dir(270)); | ||
| + | dot("$E$",E,dir(180)); | ||
| + | dot("$F$",F,dir(90)); | ||
| + | dot("$G$",G,dir(0)); | ||
| + | draw(Circle((1.107, 0.607), 1.26)); | ||
| + | draw(D--E); | ||
| + | draw(E--F); | ||
| + | draw(D--F); | ||
| + | draw(E--G); | ||
| + | draw(D--G); | ||
| + | draw(B--F); | ||
| + | draw(B--G); | ||
| + | </asy> | ||
First we note that <math>\triangle DEF</math> is an isosceles right triangle with hypotenuse <math>\overline{DE}</math> the same as the diameter of <math>\omega</math>. We also note that <math>\triangle DGE \sim \triangle ABC</math> since <math>\angle EGD</math> is a right angle and the ratios of the sides are <math>3:4:5</math>. | First we note that <math>\triangle DEF</math> is an isosceles right triangle with hypotenuse <math>\overline{DE}</math> the same as the diameter of <math>\omega</math>. We also note that <math>\triangle DGE \sim \triangle ABC</math> since <math>\angle EGD</math> is a right angle and the ratios of the sides are <math>3:4:5</math>. | ||
| − | From congruent arc intersections, we know that <math>\angle GED \cong \angle GBC</math>, and that from similar triangles <math>\angle GED</math> is also congruent to <math>\angle GCB</math>. Thus, <math>\triangle BGC</math> is an isosceles triangle with <math>BG = GC</math>, so <math>G</math> is the midpoint of <math>\overline{AC}</math> and <math>AG = GC = 5/2</math>. Similarly, we can find from angle chasing that <math>\angle ABF = \angle EDF = \frac{\pi}4</math>. | + | From congruent arc intersections, we know that <math>\angle GED \cong \angle GBC</math>, and that from similar triangles <math>\angle GED</math> is also congruent to <math>\angle GCB</math>. Thus, <math>\triangle BGC</math> is an isosceles triangle with <math>BG = GC</math>, so <math>G</math> is the midpoint of <math>\overline{AC}</math> and <math>AG = GC = 5/2</math>. Similarly, we can find from angle chasing that <math>\angle ABF = \angle EDF = \frac{\pi}4</math>. Therefore, <math>\overline{BF}</math> is the angle bisector of <math>\angle B</math>. From the angle bisector theorem, we have <math>\frac{AF}{AB} = \frac{CF}{CB}</math>, so <math>AF = 15/7</math> and <math>CF = 20/7</math>. |
Lastly, we apply power of a point from points <math>A</math> and <math>C</math> with respect to <math>\omega</math> and have <math>AE \times AB=AF \times AG</math> and <math>CD \times CB=CG \times CF</math>, so we can compute that <math>EB = \frac{17}{14}</math> and <math>DB = \frac{31}{14}</math>. From the Pythagorean Theorem, we result in <math>DE = \frac{25 \sqrt{2}}{14}</math>, so <math>a+b+c=25+2+14= \boxed{041}</math> | Lastly, we apply power of a point from points <math>A</math> and <math>C</math> with respect to <math>\omega</math> and have <math>AE \times AB=AF \times AG</math> and <math>CD \times CB=CG \times CF</math>, so we can compute that <math>EB = \frac{17}{14}</math> and <math>DB = \frac{31}{14}</math>. From the Pythagorean Theorem, we result in <math>DE = \frac{25 \sqrt{2}}{14}</math>, so <math>a+b+c=25+2+14= \boxed{041}</math> | ||
Revision as of 23:43, 18 March 2014
Problem 15
In 
, 
, 
, and 
. Circle 
intersects 
at 
and 
, 
at and 
, and 
at 
and 
. Given that 
and 
, length 
, where 
and 
are relatively prime positive integers, and 
is a positive integer not divisible by the square of any prime. Find 
.
Solution
![[asy] pair A = (0,3); pair B = (0,0); pair C = (4,0); draw(A--B--C--cycle); dotfactor = 3; dot("$A$",A,dir(135)); dot("$B$",B,dir(215)); dot("$C$",C,dir(305)); pair D = (2.21, 0); pair E = (0, 1.21); pair F = (1.71, 1.71); pair G = (2, 1.5); dot("$D$",D,dir(270)); dot("$E$",E,dir(180)); dot("$F$",F,dir(90)); dot("$G$",G,dir(0)); draw(Circle((1.107, 0.607), 1.26)); draw(D--E); draw(E--F); draw(D--F); draw(E--G); draw(D--G); draw(B--F); draw(B--G); [/asy]](http://latex.artofproblemsolving.com/b/7/d/b7d671d77d204fd52915309c2a1c23b3445cdd1d.png)
First we note that 
is an isosceles right triangle with hypotenuse 
the same as the diameter of . We also note that 
since 
is a right angle and the ratios of the sides are 
.
From congruent arc intersections, we know that 
, and that from similar triangles 
is also congruent to 
. Thus, 
is an isosceles triangle with 
, so is the midpoint of and 
. Similarly, we can find from angle chasing that 
. Therefore, 
is the angle bisector of 
. From the angle bisector theorem, we have 
, so 
and 
.
Lastly, we apply power of a point from points 
and 
with respect to and have 
and 
, so we can compute that 
and 
. From the Pythagorean Theorem, we result in 
, so 
See also
| 2014 AIME I (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. 
