Difference between revisions of "2005 AIME I Problems/Problem 15"
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== Solution == | == Solution == | ||
− | Let <math>E</math> and <math> | + | 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 <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>. |
− | {{ | + | |
+ | Now, by [[Stewart's Theorem]] in triangle <math>\triangle ABC</math> with [[cevian]] <math>\overline AD</math>, we have | ||
+ | |||
+ | <math>(3m)^2\cdot 20 + 20\cdot10\cdot10 = 10^2\cdot10 + (30 - 2c)^2\cdot 10</math>. 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 | ||
+ | |||
+ | <math>9(10 - c)^2 + 200 = 100 + (30 - 2c)^2</math> | ||
+ | |||
+ | or equivalently | ||
+ | |||
+ | <math>c^2 - 12c + 20 = 0</math>. | ||
+ | |||
+ | Thus <math>c = 2</math> or <math>c = 10</math>. We discard the value <math>c = 10</math> as extraneous (it gives us an equilateral triangle) and are left with <math>c = 2</math>, so our triangle has sides of length 10, 20 and 26. Applying [[Heron's formula]] or the equivalent gives that the area is <math>A = \sqrt{28 \cdot 18 \cdot 8 \cdot 2} = 24\sqrt{14}</math> and so the answer is <math>24 + 14 = 038</math>. | ||
+ | |||
== See also == | == See also == |
Revision as of 17:49, 17 January 2007
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
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
or equivalently
.
Thus or
. We discard the value
as extraneous (it gives us an equilateral triangle) and are left with
, so our triangle has sides of length 10, 20 and 26. Applying Heron's formula or the equivalent gives that the area is
and so the answer is
.