Difference between revisions of "2014 AIME I Problems/Problem 15"
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== Problem 15 == | == Problem 15 == | ||
− | In <math>\triangle ABC</math>, <math>AB = 3</math>, <math>BC = 4</math>, and <math>CA = 5</math>. Circle <math>\omega</math> intersects <math>\overline{AB}</math> at <math>E</math> and <math>B</math>, <math>\overline{BC}</math> at <math>B</math> and <math>D</math>, and <math>\overline{AC}</math> at <math>F</math> and <math>G</math>. Given that <math>EF=DF</math> and <math>\frac{DG}{EG} = \frac{3}{4}</math>, <math>DE=\frac{a\sqrt{b}}{c}</math>, where <math>a</math> and <math>c</math> are relatively prime positive integers, and <math>b</math> is a positive integer not divisible by the square of any prime. Find <math>a+b+c</math>. | + | In <math>\triangle ABC</math>, <math>AB = 3</math>, <math>BC = 4</math>, and <math>CA = 5</math>. Circle <math>\omega</math> intersects <math>\overline{AB}</math> at <math>E</math> and <math>B</math>, <math>\overline{BC}</math> at <math>B</math> and <math>D</math>, and <math>\overline{AC}</math> at <math>F</math> and <math>G</math>. Given that <math>EF=DF</math> and <math>\frac{DG}{EG} = \frac{3}{4}</math>, length <math>DE=\frac{a\sqrt{b}}{c}</math>, where <math>a</math> and <math>c</math> are relatively prime positive integers, and <math>b</math> is a positive integer not divisible by the square of any prime. Find <math>a+b+c</math>. |
− | == Solution == | + | == Solution 1 == |
+ | |||
+ | Since <math>\angle DBE = 90^\circ</math>, <math>DE</math> is the diameter of <math>\omega</math>. Then <math>\angle DFE=\angle DGE=90^\circ</math>. But <math>DF=FE</math>, so <math>\triangle DEF</math> is a 45-45-90 triangle. Letting <math>DG=3x</math>, we have that <math>EG=4x</math>, <math>DE=5x</math>, and <math>DF=EF=\frac{5x}{\sqrt{2}}</math>. | ||
+ | |||
+ | Note that <math>\triangle DGE \sim \triangle ABC</math> by SAS similarity, so <math>\angle BAC = \angle GDE</math> and <math>\angle ACB = \angle DEG</math>. Since <math>DEFG</math> is a cyclic quadrilateral, <math>\angle BAC = \angle GDE=180^\circ-\angle EFG = \angle AFE</math> and <math>\angle ACB = \angle DEG = \angle GFD</math>, implying that <math>\triangle AFE</math> and <math>\triangle CDF</math> are isosceles. As a result, <math>AE=CD=\frac{5x}{\sqrt{2}}</math>, so <math>BE=3-\frac{5x}{\sqrt{2}}</math> and <math>BD =4-\frac{5x}{\sqrt{2}}</math>. | ||
+ | |||
+ | Finally, using the Pythagorean Theorem on <math>\triangle BDE</math>, | ||
+ | <cmath> \left(3-\frac{5x}{\sqrt{2}}\right)^2 + \left(4-\frac{5x}{\sqrt{2}}\right)^2 = (5x)^2</cmath> | ||
+ | Solving for <math>x</math>, we get that <math>x=\frac{5\sqrt{2}}{14}</math>, so <math>DE=5x=\frac{25\sqrt{2}}{14}</math>. Thus, the answer is <math>25+2+14=\boxed{041}</math>. | ||
+ | |||
+ | == Solution 2 == | ||
+ | |||
+ | <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((1.109, 0.609), 1.28)); | ||
+ | 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>. | ||
+ | |||
+ | 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> | ||
+ | |||
+ | |||
+ | Also: <math>FG=\frac{20}{7}-\frac{5}{2}=\frac{5}{2}-\frac{15}{7}=\frac{5}{14}</math>. We can also use Ptolemy's Theorem on quadrilateral <math>DEFG</math> to figure what <math>FG</math> is in terms of <math>d</math>: | ||
+ | <cmath>DE\cdot FG+DG\cdot EF=DF\cdot EG</cmath> | ||
+ | <cmath>d\cdot FG+\frac{3d}{5}\cdot \frac{d}{\sqrt{2}}=\frac{4d}{5}\cdot \frac{d}{\sqrt{2}}</cmath> | ||
+ | <cmath>d\cdot FG+\frac{3d^2}{5\sqrt{2}}=\frac{4d^2}{5\sqrt{2}}\implies FG=\frac{d}{5\sqrt{2}}</cmath> | ||
+ | Thus <math>\frac{d}{5\sqrt{2}}=\frac{5}{14}\rightarrow d=5\sqrt{2}\cdot\frac{5}{14}=\frac{25\sqrt{2}}{14}</math>. <math>a+b+c=25+2+14= \boxed{041}</math> | ||
+ | |||
+ | ===Solution 3=== | ||
+ | Call <math>DE=x</math> and as a result <math>DF=EF=\frac{x\sqrt{2}}{2}, EG=\frac{4x}{5}, GD=\frac{3x}{5}</math>. Since <math>EFGD</math> is cyclic we just need to get <math>DG</math> and using LoS(for more detail see the <math>2</math>nd paragraph of Solution <math>2</math>) we get <math>AG=\frac{5}{2}</math> and using a similar argument(use LoS again) and subtracting you get <math>FG=\frac{5}{14}</math> so you can use Ptolemy to get <math>x=\frac{25\sqrt{2}}{14} \implies \boxed{041}</math>. | ||
+ | ~First | ||
+ | |||
+ | == See also == | ||
+ | {{AIME box|year=2014|n=I|num-b=14|after=Last Question}} | ||
+ | {{MAA Notice}} |
Revision as of 16:57, 9 August 2018
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 1
Since , is the diameter of . Then . But , so is a 45-45-90 triangle. Letting , we have that , , and .
Note that by SAS similarity, so and . Since is a cyclic quadrilateral, and , implying that and are isosceles. As a result, , so and .
Finally, using the Pythagorean Theorem on , Solving for , we get that , so . Thus, the answer is .
Solution 2
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
Also: . We can also use Ptolemy's Theorem on quadrilateral to figure what is in terms of :
Thus .
Solution 3
Call and as a result . Since is cyclic we just need to get and using LoS(for more detail see the nd paragraph of Solution ) we get and using a similar argument(use LoS again) and subtracting you get so you can use Ptolemy to get . ~First
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 | ||
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