Difference between revisions of "Mock AIME I 2015 Problems/Problem 3"
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==Solution== | ==Solution== | ||
+ | First note that <math>\triangle APB</math> is a 3-4-5 scaled right triangle with <math>\overline{B P}=15</math> and <math>\triangle APC</math> is a 20-21-29 right triangle with <math>\overline{P C}=21</math> since <math>\angle APB</math> and <math>\angle APC</math> are inscribed in semicircles, and thus right angles. Also, since <math>\angle APB</math> and <math>\angle APC</math> are right, points B, P and C are collinear and <math>BC=BP+PC=15+21=\boxed{036}.</math> | ||
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+ | Solution by D. Adrian Tanner | ||
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+ | ==Original Solution== | ||
First note that <math>\triangle APB</math> is a 3-4-5 right triangle and <math>\triangle APC</math> is a 20-21-29 right triangle. Thus we have <math>\cos \angle BAP = \dfrac{20}{25}=\dfrac{4}{5}</math> and <math>\cos \angle PAC = \dfrac{20}{29}</math>. Similarly, we have <math>\sin \angle BAP = \dfrac{15}{25}=\dfrac{3}{5}</math> and <math>\sin \angle PAC = \dfrac{21}{29}</math>. Applying the Law of Cosines on <math>\angle BAC</math> gives us <cmath>BC=\sqrt{25^2+29^2-2\cdot25\cdot29\cdot\cos (\angle BAP+\angle PAC)}.</cmath> We also know that <cmath>\cos (\angle BAP+\angle PAC)=\cos\angle BAP\cos\angle PAC - \sin \angle BAP \sin \angle PAC=\dfrac{20}{25}\cdot\dfrac{20}{29}-\dfrac{15}{25}\cdot\dfrac{21}{29}=\dfrac{17}{145}.</cmath> Hence, we have <cmath>BC=\sqrt{25^2+29^2-2\cdot25\cdot29\cdot\dfrac{17}{145}}=\sqrt{1296}=\boxed{036}.</cmath> | First note that <math>\triangle APB</math> is a 3-4-5 right triangle and <math>\triangle APC</math> is a 20-21-29 right triangle. Thus we have <math>\cos \angle BAP = \dfrac{20}{25}=\dfrac{4}{5}</math> and <math>\cos \angle PAC = \dfrac{20}{29}</math>. Similarly, we have <math>\sin \angle BAP = \dfrac{15}{25}=\dfrac{3}{5}</math> and <math>\sin \angle PAC = \dfrac{21}{29}</math>. Applying the Law of Cosines on <math>\angle BAC</math> gives us <cmath>BC=\sqrt{25^2+29^2-2\cdot25\cdot29\cdot\cos (\angle BAP+\angle PAC)}.</cmath> We also know that <cmath>\cos (\angle BAP+\angle PAC)=\cos\angle BAP\cos\angle PAC - \sin \angle BAP \sin \angle PAC=\dfrac{20}{25}\cdot\dfrac{20}{29}-\dfrac{15}{25}\cdot\dfrac{21}{29}=\dfrac{17}{145}.</cmath> Hence, we have <cmath>BC=\sqrt{25^2+29^2-2\cdot25\cdot29\cdot\dfrac{17}{145}}=\sqrt{1296}=\boxed{036}.</cmath> |
Latest revision as of 22:31, 10 June 2015
Problem 3
Let be points in the plane such that , , and . Semicircles with diameters and intersect at a point with . Find the length of line segment .
Solution
First note that is a 3-4-5 scaled right triangle with and is a 20-21-29 right triangle with since and are inscribed in semicircles, and thus right angles. Also, since and are right, points B, P and C are collinear and
Solution by D. Adrian Tanner
Original Solution
First note that is a 3-4-5 right triangle and is a 20-21-29 right triangle. Thus we have and . Similarly, we have and . Applying the Law of Cosines on gives us We also know that Hence, we have