Difference between revisions of "2005 AIME I Problems/Problem 15"
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Use Power of a point similar to the first solution to find that <math>AB = 30</math> and that the side <math>AC = 2 \cdot x \sqrt{2}</math>, where <math>x</math> is one third of the median's length. Then use systems of law of cosines, creating two triangles, with <math>10-10-3x</math> with angle <math>\theta</math>, and <math>10-20-2 \cdot x \sqrt{2}</math> with the same angle. Solving the system yields <math>x = 4 \sqrt{2}</math>. Solving using Heron's Formula gets the answer <math>24 \sqrt{14}</math>, or <math>\boxed{038}</math>. | Use Power of a point similar to the first solution to find that <math>AB = 30</math> and that the side <math>AC = 2 \cdot x \sqrt{2}</math>, where <math>x</math> is one third of the median's length. Then use systems of law of cosines, creating two triangles, with <math>10-10-3x</math> with angle <math>\theta</math>, and <math>10-20-2 \cdot x \sqrt{2}</math> with the same angle. Solving the system yields <math>x = 4 \sqrt{2}</math>. Solving using Heron's Formula gets the answer <math>24 \sqrt{14}</math>, or <math>\boxed{038}</math>. | ||
+ | == Solution 3 == | ||
+ | Assume <math>AD = 3m</math>. And WLOG, let E be be between B & D (Draw your own diagram, in the diagram on top, E is between C & D) | ||
+ | |||
+ | We use power of a point, and get that <math>AB = 10</math>, <math>BC = 20</math>, <math>AC = 10+2\sqrt{2} m</math>. Now, we can apply [[Stewart's Theorem]]. | ||
+ | <cmath>2000 + 180 m^2 = 10(10+2\sqrt{2}m)^{2} + 1000</cmath> | ||
+ | <cmath>1000 + 180 m^2 = 1000 + 400\sqrt{2}m + 80 m^{2}</cmath> | ||
+ | <cmath>100 m^2 = 400\sqrt{2}m</cmath> | ||
+ | |||
+ | <math>m = 4\sqrt{2}</math> or <math>m = 0</math> if <math>m = 0</math>, we get a degenerate triangle, so <math>m = 4\sqrt{2}</math>, and thus AC = 26. You can now use [[Heron's Formula]] to finish. | ||
+ | |||
+ | -Alexlikemath | ||
== See also == | == See also == | ||
{{AIME box|year=2005|n=I|num-b=14|after=Last Question}} | {{AIME box|year=2005|n=I|num-b=14|after=Last Question}} |
Revision as of 11:38, 11 December 2019
Problem
Triangle has The incircle of the triangle evenly trisects the median If the area of the triangle is where and are integers and is not divisible by the square of a prime, find
Solution 1
Let , and be the points of tangency of the incircle with , and , respectively. Without loss of generality, let , so that is between and . Let the length of the median be . Then by two applications of the Power of a Point Theorem, , so . Now, and are two tangents to a circle from the same point, so and thus . Then so and thus .
Now, by Stewart's Theorem in triangle with cevian , we have
Our earlier result from Power of a Point was that , so we combine these two results to solve for and we get
Thus or . We discard the value as extraneous (it gives us an equilateral triangle) and are left with , so our triangle has area and so the answer is .
Solution 2
Use Power of a point similar to the first solution to find that and that the side , where is one third of the median's length. Then use systems of law of cosines, creating two triangles, with with angle , and with the same angle. Solving the system yields . Solving using Heron's Formula gets the answer , or .
Solution 3
Assume . And WLOG, let E be be between B & D (Draw your own diagram, in the diagram on top, E is between C & D)
We use power of a point, and get that , , . Now, we can apply Stewart's Theorem.
or if , we get a degenerate triangle, so , and thus AC = 26. You can now use Heron's Formula to finish.
-Alexlikemath
See also
2005 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 |
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