Difference between revisions of "2009 AIME I Problems/Problem 15"
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== Problem == | == Problem == | ||
In triangle <math>ABC</math>, <math>AB = 10</math>, <math>BC = 14</math>, and <math>CA = 16</math>. Let <math>D</math> be a point in the interior of <math>\overline{BC}</math>. Let points <math>I_B</math> and <math>I_C</math> denote the incenters of triangles <math>ABD</math> and <math>ACD</math>, respectively. The circumcircles of triangles <math>BI_BD</math> and <math>CI_CD</math> meet at distinct points <math>P</math> and <math>D</math>. The maximum possible area of <math>\triangle BPC</math> can be expressed in the form <math>a - b\sqrt {c}</math>, where <math>a</math>, <math>b</math>, and <math>c</math> are positive integers and <math>c</math> is not divisible by the square of any prime. Find <math>a + b + c</math>. | In triangle <math>ABC</math>, <math>AB = 10</math>, <math>BC = 14</math>, and <math>CA = 16</math>. Let <math>D</math> be a point in the interior of <math>\overline{BC}</math>. Let points <math>I_B</math> and <math>I_C</math> denote the incenters of triangles <math>ABD</math> and <math>ACD</math>, respectively. The circumcircles of triangles <math>BI_BD</math> and <math>CI_CD</math> meet at distinct points <math>P</math> and <math>D</math>. The maximum possible area of <math>\triangle BPC</math> can be expressed in the form <math>a - b\sqrt {c}</math>, where <math>a</math>, <math>b</math>, and <math>c</math> are positive integers and <math>c</math> is not divisible by the square of any prime. Find <math>a + b + c</math>. | ||
+ | |||
+ | ==Diagram== | ||
+ | <center><asy> | ||
+ | defaultpen(fontsize(11)+0.8); size(300); | ||
+ | pair A,B,C,D,Ic,Ib,P; | ||
+ | A=MP("A",origin,down+left); B=MP("B",8*right,down+right); C=MP("C",IP(CR(A,5), CR(B,7)),2*up); real t=0.505; D=MP("",B+t*(C-B),SW); draw(A--B--C--A--D); path c1=incircle(A,D,C); path c2=incircle(A,D,B); draw(c1, gray+0.25); draw(c2, gray+0.25); Ic=MP("I_C",incenter(A,D,C),down+left); Ib=MP("I_B",incenter(A,D,B),left); path c3=circumcircle(Ic,D,C); path c4=circumcircle(Ib,D,B); draw(c3, fuchsia+0.2); draw(c4, fuchsia+0.2); P=MP("P",OP(c3,c4),up); draw(arc(circumcenter(B,C,P),B,C), royalblue+0.5+dashed); draw(C--Ic--D--P--C^^P--Ic, black+0.3); draw(B--Ib--D--P--B^^P--Ib, black+0.3); label("10",A--B,down); label("16",A--C,left); | ||
+ | </asy></center> | ||
+ | |||
== Solution 1 == | == Solution 1 == |
Revision as of 21:54, 5 June 2022
Contents
[hide]Problem
In triangle , , , and . Let be a point in the interior of . Let points and denote the incenters of triangles and , respectively. The circumcircles of triangles and meet at distinct points and . The maximum possible area of can be expressed in the form , where , , and are positive integers and is not divisible by the square of any prime. Find .
Diagram
Solution 1
First, by the Law of Cosines, we have so .
Let and be the circumcenters of triangles and , respectively. We first compute Because and are half of and , respectively, the above expression can be simplified to Similarly, . As a result
Therefore is constant (). Also, is or when is or . Let point be on the same side of as with ; is on the circle with as the center and as the radius, which is . The shortest distance from to is .
When the area of is the maximum, the distance from to has to be the greatest. In this case, it's . The maximum area of is and the requested answer is .
Solution 2
From Law of Cosines on , Now, Since and are cyclic quadrilaterals, it follows that Next, applying Law of Cosines on , By AM-GM, , so Finally, and the maximum area would be so the answer is .
Solution 3
Proceed as in Solution 2 until you find . The locus of points that give is a fixed arc from to ( will move along this arc as moves along ) and we want to maximise the area of []. This means we want to be farthest distance away from as possible, so we put in the middle of the arc (making isosceles). We know that and , so . Let be the foot of the perpendicular from to line . Then the area of [] is the same as because base has length . We can split into two triangles and , with and . Then, the area of [] is equal to , and so the answer is .
See also
2009 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.