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 == | ||
First, by the [[Law of Cosines]], we have <cmath>\cos BAC = \frac {16^2 + 10^2 - 14^2}{2\cdot 10 \cdot 16} = \frac {256+100-196}{320} = \frac {1}{2},</cmath> so <math>\angle BAC = 60^\circ</math>. | First, by the [[Law of Cosines]], we have <cmath>\cos BAC = \frac {16^2 + 10^2 - 14^2}{2\cdot 10 \cdot 16} = \frac {256+100-196}{320} = \frac {1}{2},</cmath> so <math>\angle BAC = 60^\circ</math>. | ||
− | Let <math>O_1</math> and <math>O_2</math> be the circumcenters of triangles <math>BI_BD</math> and <math>CI_CD</math>, respectively. We first compute <cmath>\angle BO_1D = \angle BO_1I_B + \angle I_BO_1D = 2\angle BDI_B + 2\angle I_BBD.</cmath> Because <math>\angle BDI_B</math> and <math>\angle I_BBD</math> are half of <math>\angle BDA</math> and <math>\angle ABD</math>, respectively, the above expression can be simplified to <cmath>\angle BO_1D = \angle BO_1I_B + \angle I_BO_1D = 2\angle BDI_B + 2\angle I_BBD = \angle ABD + \angle BDA.</cmath> Similarly, <math>\angle CO_2D = \angle ACD + \angle CDA</math>. As a result <cmath>\begin{align*}\angle CPB &= \angle CPD + \angle BPD \\&= \frac {1}{2} \cdot \angle CO_2D + \frac {1}{2} \cdot \angle BO_1D \\&= \frac {1}{2}(\angle ABD + \angle BDA + \angle ACD + \angle CDA) \\&= \frac {1}{2} (2 \cdot 180^\circ - \angle BAC) \\&= \frac {1}{2} \cdot 300^\circ = 150^\circ.\end{align*}</cmath> | + | Let <math>O_1</math> and <math>O_2</math> be the circumcenters of triangles <math>BI_BD</math> and <math>CI_CD</math>, respectively. We first compute <cmath>\angle BO_1D = \angle BO_1I_B + \angle I_BO_1D = 2\angle BDI_B + 2\angle I_BBD.</cmath> Because <math>\angle BDI_B</math> and <math>\angle I_BBD</math> are half of <math>\angle BDA</math> and <math>\angle ABD</math>, respectively, the above expression can be simplified to <cmath>\angle BO_1D = \angle BO_1I_B + \angle I_BO_1D = 2\angle BDI_B + 2\angle I_BBD = \angle ABD + \angle BDA.</cmath> Similarly, <math>\angle CO_2D = \angle ACD + \angle CDA</math>. As a result |
+ | |||
+ | <cmath>\begin{align*}\angle CPB &= \angle CPD + \angle BPD \\&= \frac {1}{2} \cdot \angle CO_2D + \frac {1}{2} \cdot \angle BO_1D \\&= \frac {1}{2}(\angle ABD + \angle BDA + \angle ACD + \angle CDA) \\&= \frac {1}{2} (2 \cdot 180^\circ - \angle BAC) \\&= \frac {1}{2} \cdot 300^\circ = 150^\circ.\end{align*}</cmath> | ||
− | Therefore <math>\angle CPB</math> is constant (<math>150^\circ</math>). Also, <math>P</math> is <math>B</math> or <math>C</math> when <math>D</math> is <math>B</math> or <math>C</math>. Let point <math>L</math> be on the same side of <math>\overline{BC}</math> as <math>A</math> with <math>LC = LB = BC = 14</math>; <math>P</math> is on the circle with <math>L</math> as the center and <math>\overline{LC}</math> as the radius, which is <math>14</math>. The shortest distance from <math>L</math> to <math>\overline{BC}</math> is <math>7\sqrt {3}</math>. | + | Therefore <math>\angle CPB</math> is constant (<math>150^\circ</math>). Also, <math>P</math> is <math>B</math> or <math>C</math> when <math>D</math> is <math>B</math> or <math>C</math>. Let point <math>L</math> be on the same side of <math>\overline{BC}</math> as <math>A</math> with <math>LC = LB = BC = 14</math>; <math>P</math> is on the circle with <math>L</math> as the center and <math>\overline{LC}</math> as the radius, which is <math>14</math>. The shortest (and only) distance from <math>L</math> to <math>\overline{BC}</math> is <math>7\sqrt {3}</math>. |
When the area of <math>\triangle BPC</math> is the maximum, the distance from <math>P</math> to <math>\overline{BC}</math> has to be the greatest. In this case, it's <math>14 - 7\sqrt {3}</math>. The maximum area of <math>\triangle BPC</math> is | When the area of <math>\triangle BPC</math> is the maximum, the distance from <math>P</math> to <math>\overline{BC}</math> has to be the greatest. In this case, it's <math>14 - 7\sqrt {3}</math>. The maximum area of <math>\triangle BPC</math> is |
Latest revision as of 19:35, 2 January 2023
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
![[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]](http://latex.artofproblemsolving.com/8/9/8/8989b91d49ad23f9b7923acdd082fc02c94ab5e6.png)
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 (and only) 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.