Difference between revisions of "2009 AIME I Problems/Problem 12"

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(Solution 3)
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==Solution 3==
 
 
As in Solution <math>2</math>, let <math>P</math> and <math>Q</math> be the intersections of <math>\omega</math> with <math>BI</math> and <math>AI</math> respectively.
 
 
Recall that the distance from a point outside a circle to that circle is the same along both tangent lines to the circle drawn from the point.
 
 
Recall also that the length of the altitude to the hypotenuse of a right-angle triangle is the geometric mean of the two segments into which it cuts the hypotenuse.
 
 
Let <math>x = \overline{AD} = \overline{AQ}</math>. Let <math>y = \overline{BD} = \overline{BP}</math>. Let <math>z = \overline{PI} = \overline{QI}</math>. The semi-perimeter of <math>ABI</math> is <math>x + y + z</math>.
 
Since the lengths of the sides of <math>ABI</math> are <math>x + y</math>, <math>y + z</math> and <math>x + z</math>, the square of its area by Heron's formula is <math>(x+y+z)xyz</math>.
 
 
The radius <math>r</math> of <math>\omega</math> is <math>\overline{CD}/2</math>. Therefore <math>r^2 = xy/4</math>. As <math>\omega</math> is the in-circle of <math>ABI</math>, the area of <math>ABI</math> is also <math>r(x+y+z)</math>, and so the square area is <math>r^2(x+y+z)^2</math>.
 
 
Therefore <cmath>(x+y+z)xyz = r^2(x+y+z)^2 = \frac{xy(x+y+z)^2}{4}</cmath> Dividing both sides by <math>xy(x+y+z)/4</math> we get: <cmath>4z = (x+y+z),</cmath> and so <math>z = (x+y)/3</math>. The semi-perimeter of <math>ABI</math> is therefore <math>\frac{4}{3}(x+y)</math> and the whole perimeter is <math>\frac{8}{3}(x+y)</math>. Now <math>x + y = \overline{AB}</math>, so the ratio of the perimeter of <math>ABI</math> to the hypotenuse <math>\overline{AB}</math> is <math>8/3</math> and our answer is <math>8+3=\boxed{011}</math>
 
 
 
and <math>8+3=\boxed{011}</math>.
 
 
 
== See also ==
 
== See also ==
 
{{AIME box|year=2009|n=I|num-b=11|num-a=13}}
 
{{AIME box|year=2009|n=I|num-b=11|num-a=13}}
 
[[Category: Intermediate Geometry Problems]]
 
[[Category: Intermediate Geometry Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 05:27, 30 September 2022

See also

2009 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
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All AIME Problems and Solutions

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