Difference between revisions of "2005 AIME I Problems/Problem 15"
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== Problem == | == Problem == | ||
− | + | Triangle <math> ABC </math> has <math> BC=20. </math> The [[incircle]] of the triangle evenly [[trisect]]s the [[median of a triangle | median]] <math> AD. </math> If the area of the triangle is <math> m \sqrt{n} </math> where <math> m </math> and <math> n </math> are integers and <math> n </math> is not [[divisor | divisible]] by the [[perfect square | square]] of a prime, find <math> m+n. </math> | |
− | == Solution == | + | == Solution 1== |
− | + | <center><asy> | |
+ | size(300); pointpen=black;pathpen=black+linewidth(0.65); pen s = fontsize(10); | ||
+ | pair A=(0,0),B=(26,0),C=IP(circle(A,10),circle(B,20)),D=(B+C)/2,I=incenter(A,B,C); | ||
+ | path cir = incircle(A,B,C); pair E1=IP(cir,B--C),F=IP(cir,A--C),G=IP(cir,A--B),P=IP(A--D,cir),Q=OP(A--D,cir); | ||
+ | D(MP("A",A,s)--MP("B",B,s)--MP("C",C,N,s)--cycle); D(cir); | ||
+ | D(A--MP("D",D,NE,s)); D(MP("E",E1,NE,s)); D(MP("F",F,NW,s)); D(MP("G",G,s)); D(MP("P",P,SW,s)); D(MP("Q",Q,SE,s)); | ||
+ | MP("10",(B+D)/2,NE); MP("10",(C+D)/2,NE); | ||
+ | </asy></center><!-- Asymptote replacement for Image:2005_I_AIME-15.png by azjps --> | ||
− | + | Let <math>E</math>, <math>F</math> and <math>G</math> be the points of tangency of the incircle with <math>BC</math>, <math>AC</math> and <math>AB</math>, respectively. Without loss of generality, let <math>AC < AB</math>, so that <math>E</math> is between <math>D</math> and <math>C</math>. Let the length of the median be <math>3m</math>. Then by two applications of the [[Power of a Point Theorem]], <math>DE^2 = 2m \cdot m = AF^2</math>, so <math>DE = AF</math>. Now, <math>CE</math> and <math>CF</math> are two tangents to a circle from the same point, so by the Two Tangent Theorem <math>CE = CF = c</math> and thus <math>AC = AF + CF = DE + CE = CD = 10</math>. Then <math>DE = AF = AG = 10 - c</math> so <math>BG = BE = BD + DE = 20 - c</math> and thus <math>AB = AG + BG = 30 - 2c</math>. | |
− | <math> | + | Now, by [[Stewart's Theorem]] in triangle <math>\triangle ABC</math> with [[cevian]] <math>\overline{AD}</math>, we have |
− | < | + | <cmath>(3m)^2\cdot 20 + 20\cdot10\cdot10 = 10^2\cdot10 + (30 - 2c)^2\cdot 10.</cmath> |
− | + | Our earlier result from Power of a Point was that <math>2m^2 = (10 - c)^2</math>, so we combine these two results to solve for <math>c</math> and we get | |
− | < | + | <cmath>9(10 - c)^2 + 200 = 100 + (30 - 2c)^2 \quad \Longrightarrow \quad c^2 - 12c + 20 = 0.</cmath> |
− | Thus <math>c = 2</math> or <math> | + | Thus <math>c = 2</math> or <math> = 10</math>. We discard the value <math>c = 10</math> as extraneous (it gives us a line) and are left with <math>c = 2</math>, so our triangle has area <math>\sqrt{28 \cdot 18 \cdot 8 \cdot 2} = 24\sqrt{14}</math> and so the answer is <math>24 + 14 = \boxed{038}</math>. |
+ | == Solution 2 == | ||
+ | |||
+ | 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 == | ||
+ | WLOG let E be be between C & D (as in solution 1). Assume <math>AD = 3m</math>. We use power of a point to get that | ||
+ | <math>AG = DE = \sqrt{2}m </math> and <math>AB = AG + GB = AG + BE = 10+2\sqrt{2} m</math> | ||
+ | |||
+ | Since now we have <math>AC = 10</math>, <math>BC = 20, AB = 10+2\sqrt{2} m </math> in triangle <math>\triangle ABC</math> and [[cevian]] <math>AD = 3m</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 AB = 26. You can now use [[Heron's Formula]] to finish. The answer is <math>24 \sqrt{14}</math>, or <math>\boxed{038}</math>. | ||
+ | |||
+ | -Alexlikemath | ||
== See also == | == See also == | ||
− | + | {{AIME box|year=2005|n=I|num-b=14|after=Last Question}} | |
− | |||
[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 20:20, 25 June 2020
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 by the Two Tangent Theorem 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 a line) 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
WLOG let E be be between C & D (as in solution 1). Assume . We use power of a point to get that and
Since now we have , in triangle and cevian . Now, we can apply Stewart's Theorem.
or if , we get a degenerate triangle, so , and thus AB = 26. You can now use Heron's Formula to finish. The answer is , or .
-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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