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

Revision as of 05:26, 30 September 2022

Solution 3

As in Solution $2$ , let $P$ and $Q$ be the intersections of $\omega$ with $BI$ and $AI$ 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 $x = \overline{AD} = \overline{AQ}$ . Let $y = \overline{BD} = \overline{BP}$ . Let $z = \overline{PI} = \overline{QI}$ . The semi-perimeter of $ABI$ is $x + y + z$ . Since the lengths of the sides of $ABI$ are $x + y$ , $y + z$ and $x + z$ , the square of its area by Heron's formula is $(x+y+z)xyz$ .

The radius $r$ of $\omega$ is $\overline{CD}/2$ . Therefore $r^2 = xy/4$ . As $\omega$ is the in-circle of $ABI$ , the area of $ABI$ is also $r(x+y+z)$ , and so the square area is $r^2(x+y+z)^2$ .

Therefore $(x+y+z)xyz = r^2(x+y+z)^2 = \frac{xy(x+y+z)^2}{4}$ Dividing both sides by $xy(x+y+z)/4$ we get: $4z = (x+y+z),$ and so $z = (x+y)/3$ . The semi-perimeter of $ABI$ is therefore $\frac{4}{3}(x+y)$ and the whole perimeter is $\frac{8}{3}(x+y)$ . Now $x + y = \overline{AB}$ , so the ratio of the perimeter of $ABI$ to the hypotenuse $\overline{AB}$ is $8/3$ and our answer is $8+3=\boxed{011}$

and $8+3=\boxed{011}$ .

@@ Line 1: / Line 1: @@
-==Solution 2==
-As in Solution <math>1</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.
-First, by pythagorean theorem, <math>AB = \sqrt{12^2+35^2} = 37</math>.  Now the area of <math>ABC</math> is <math>1/2*12*35 = 1/2*37*CD</math>, so <math>CD=\frac{420}{37}</math> and the inradius of <math>\triangle ABI</math> is <math>r=\frac{210}{37}</math>.
-Now from <math>\triangle CDB \sim \triangle ACB</math> we find that <math>\frac{BC}{BD} = \frac{AB}{BC}</math> so <math>BD = BC^2/AB = 35^2/37</math> and similarly, <math>AD = 12^2/37</math>.
-Note <math>IP=IQ=x</math>, <math>BP=BD</math>, and <math>AQ=AD</math>. So we have <math>AI = 144/37+x</math>, <math>BI = 1225/37+x</math>.  Now we can compute the area of <math>\triangle ABI</math> in two ways: by heron's formula and by inradius times semiperimeter, which yields
-<math>rs=210/37(37+x) = \sqrt{(37+x)(37-144/37)(37-1225/37)x}</math>
-<math>210/37(37+x) = 12*35/37 \sqrt{x(37+x)}</math>
-<math>37+x = 2 \sqrt{x(x+37)}</math>
-<math>x^2+74x+1369 = 4x^2 + 148x</math>
-<math>3x^2 + 74x - 1369 = 0</math>
-The quadratic formula now yields <math>x=37/3</math>.  Plugging this back in, the perimeter of <math>ABI</math> is <math>2s=2(37+x)=2(37+37/3) = 37(8/3)</math> so the ratio of the perimeter to <math>AB</math> is <math>8/3</math> and our answer is <math>8+3=\boxed{011}</math>
 ==Solution 3==

2009 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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Difference between revisions of "2009 AIME I Problems/Problem 12"

Revision as of 05:26, 30 September 2022

Solution 3

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