Difference between revisions of "2009 AIME I Problems/Problem 12"
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In right <math>\triangle ABC</math> with hypotenuse <math>\overline{AB}</math>, <math>AC = 12</math>, <math>BC = 35</math>, and <math>\overline{CD}</math> is the altitude to <math>\overline{AB}</math>. Let <math>\omega</math> be the circle having <math>\overline{CD}</math> as a diameter. Let <math>I</math> be a point outside <math>\triangle ABC</math> such that <math>\overline{AI}</math> and <math>\overline{BI}</math> are both tangent to circle <math>\omega</math>. The ratio of the perimeter of <math>\triangle ABI</math> to the length <math>AB</math> can be expressed in the form <math>\frac {m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>. | In right <math>\triangle ABC</math> with hypotenuse <math>\overline{AB}</math>, <math>AC = 12</math>, <math>BC = 35</math>, and <math>\overline{CD}</math> is the altitude to <math>\overline{AB}</math>. Let <math>\omega</math> be the circle having <math>\overline{CD}</math> as a diameter. Let <math>I</math> be a point outside <math>\triangle ABC</math> such that <math>\overline{AI}</math> and <math>\overline{BI}</math> are both tangent to circle <math>\omega</math>. The ratio of the perimeter of <math>\triangle ABI</math> to the length <math>AB</math> can be expressed in the form <math>\frac {m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>. | ||
+ | ==Solution 1== | ||
+ | First, note that <math>AB=37</math>; let the tangents from <math>I</math> to <math>\omega</math> have length <math>x</math>. Then the perimeter of <math>\triangle ABI</math> is equal to <cmath>2(x+AD+DB)=2(x+37).</cmath> It remains to compute <math>\dfrac{2(x+37)}{37}=2+\dfrac{2}{37}x</math>. | ||
− | == Solution | + | Observe <math>CD=\dfrac{12\cdot 35}{37}=\dfrac{420}{37}</math>, so the radius of <math>\omega</math> is <math>\dfrac{210}{37}</math>. We may also compute <math>AD=\dfrac{12^{2}}{37}</math> and <math>DB=\dfrac{35^{2}}{37}</math> by similar triangles. Let <math>O</math> be the center of <math>\omega</math>; notice that <cmath>\tan(\angle DAO)=\dfrac{DO}{AD}=\dfrac{210/37}{144/37}=\dfrac{35}{24}</cmath> so it follows <cmath>\sin(\angle DAO)=\dfrac{35}{\sqrt{35^{2}+24^{2}}}=\dfrac{35}{\sqrt{1801}}</cmath> while <math>\cos(\angle DAO)=\dfrac{24}{\sqrt{1801}}</math>. By the double-angle formula <math>\sin(2\theta)=2\sin\theta\cos\theta</math>, it turns out that <cmath>\sin(\angle BAI)=\sin(2\angle DAO)=\dfrac{2\cdot 35\cdot 24}{1801}=\dfrac{1680}{1801}</cmath> |
+ | |||
+ | Using the area formula <math>\dfrac{1}{2}ab\sin(C)</math> in <math>\triangle ABI</math>, <cmath>[ABI]=\left(\dfrac{1}{2}\right)\left(\dfrac{144}{37}+x\right)(37)\left(\dfrac{1680}{1801}\right)=\left(\dfrac{840}{1801}\right)(144+37x).</cmath> But also, using <math>rs</math>, <cmath>[ABI]=\left(\dfrac{210}{37}\right)(37+x).</cmath> Now we can get <cmath>\dfrac{[ABI]}{210}=\dfrac{4(144+37x)}{1801}=\dfrac{37+x}{37}</cmath> so multiplying everything by <math>37\cdot 1801=66637</math> lets us solve for <math>x</math>: <cmath>21312+5476x=66637+1801x.</cmath> We have <math>x=\dfrac{66637-21312}{5476-1801}=\dfrac{45325}{3675}=\dfrac{37}{3}</math>, and now <cmath>2+\dfrac{2}{37}x=2+\dfrac{2}{3}=\dfrac{8}{3}</cmath> giving the answer, <math>\boxed{011}</math>. | ||
+ | |||
+ | [[File:AIME 2009-I12 Geogebra Diagram.png]] | ||
+ | |||
+ | ==Solution 2== | ||
Let <math>O</math> be center of the circle and <math>P</math>,<math>Q</math> be the two points of tangent such that <math>P</math> is on <math>BI</math> and <math>Q</math> is on <math>AI</math>. We know that <math>AD:CD = CD:BD = 12:35</math>. | Let <math>O</math> be center of the circle and <math>P</math>,<math>Q</math> be the two points of tangent such that <math>P</math> is on <math>BI</math> and <math>Q</math> is on <math>AI</math>. We know that <math>AD:CD = CD:BD = 12:35</math>. | ||
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Hence the perimeter is <math>1225*2 + 144*2 + \frac {1369}{3} *2 = 1369* \frac {8}{3}</math>, and <math>BC</math> is <math>1369</math>. Hence <math>\frac {m}{n} = \frac {8}{3}</math>, <math>m + n = 11</math>. | Hence the perimeter is <math>1225*2 + 144*2 + \frac {1369}{3} *2 = 1369* \frac {8}{3}</math>, and <math>BC</math> is <math>1369</math>. Hence <math>\frac {m}{n} = \frac {8}{3}</math>, <math>m + n = 11</math>. | ||
− | ==Solution | + | ==Solution 3== |
− | |||
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. | 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. | ||
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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> | 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> | ||
− | == | + | Note: If you don't want to solve the quadratic, you can continue with <math>37+x = 2 \sqrt{x(x+37)}</math> and divide both sides by <math>\sqrt{x+37}</math> to get <math>\sqrt{37+x} = 2 \sqrt{x}</math>. Square both sides to get <math>37+x = 4x</math> and solve to get <math>x=\frac{37}{3}</math>. |
+ | ==Solution 4== | ||
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. | 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. | ||
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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> | 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> | ||
− | == Solution | + | ==Solution 5== |
− | |||
<asy> | <asy> | ||
size(300); | size(300); | ||
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and <math>8+3=\boxed{011}</math>. | and <math>8+3=\boxed{011}</math>. | ||
− | == Solution | + | ==Solution 6== |
− | |||
This solution is not a real solution and is solving the problem with a ruler and compass. | This solution is not a real solution and is solving the problem with a ruler and compass. | ||
Draw <math>AC = 4.8, BC = 14, AB = 14.8</math>. Then, drawing the tangents and intersecting them, we get that <math>IA</math> is around <math>6.55</math> and <math>IB</math> is around <math>18.1</math>. We then find the ratio to be around <math>\frac{39.45}{14.8}</math>. Using long division, we find that this ratio is approximately 2.666, which you should recognize as <math>\frac{8}{3}</math>. Since this seems reasonable, we find that the answer is <math>\boxed{11}</math> ~ilp | Draw <math>AC = 4.8, BC = 14, AB = 14.8</math>. Then, drawing the tangents and intersecting them, we get that <math>IA</math> is around <math>6.55</math> and <math>IB</math> is around <math>18.1</math>. We then find the ratio to be around <math>\frac{39.45}{14.8}</math>. Using long division, we find that this ratio is approximately 2.666, which you should recognize as <math>\frac{8}{3}</math>. Since this seems reasonable, we find that the answer is <math>\boxed{11}</math> ~ilp | ||
+ | |||
+ | ==Solution 7== | ||
+ | Denoting three tangents has length <math>h_1,h_2,h_3</math> while <math>h_1,h_3</math> lies on <math>AB</math> with <math>h_1>h_3</math>.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>.As we know that the diameter of the circle is the height of <math>\triangle ACB</math> from <math>C</math> to <math>AB</math>. Assume that <math>\tan\alpha=\frac{h_1}{r}</math> and <math>\tan\beta=\frac{h_3}{r}</math> and <math>\tan\omega=\frac{h_2}{r}</math>. But we know that <math>\tan(\alpha+\beta)=-\tan(180-\alpha-\beta)=-\tan\omega</math> According to the basic computation, we can get that <math>\tan(\alpha)=\frac{35}{6}</math>; <math>\tan(\beta)=\frac{24}{35}</math> | ||
+ | So we know that <math>\tan(\omega)=\frac{1369}{630}</math> according to the tangent addition formula. Hence, it is not hard to find that the length of <math>h_2</math> is <math>\frac{37}{3}</math>. According to basic addition and division, we get the answer is <math>\frac{8}{3}</math> which leads to <math>8+3=\boxed{11}</math> ~bluesoul | ||
+ | |||
+ | ==Solution 8 (no trig and intuitive)== | ||
+ | Notice <math>CD</math> is the altitude of <math>\triangle ABC</math>, and so <math>CD = \frac{AC \cdot BC}{AB} = \frac{210}{37}.</math> This means in the inradius of the circle is <math>r=\frac{CD}{2}=\frac{210}{37}.</math> | ||
+ | |||
+ | Next, by <math>\triangle ABC \sim \triangle ACD \sim \triangle CBD</math>, we have <math>AD=CD \cdot \frac{12}{35} = \frac{144}{37}</math> and <math>BD = CD \cdot \frac{35}{12} = \frac{1225}{37}</math>. | ||
+ | |||
+ | Lastly, denote the tangent point of <math>AI, BI</math> with the circle as <math>P, Q</math> respectively. Then by the properties of inscribed circles, we have that <math>AP=AD=\frac{144}{37}</math> and <math>BQ=BD=\frac{1225}{37}</math>, and <math>PI=QI=x</math>. We want to make an algebraic expression utilizing what we know about ABI. | ||
+ | |||
+ | We use the inradius formula: <math>r = \frac{[\triangle ABI]}{s}</math>. We have <math>s=37+x</math>. With Heron's (very basic calculations) we get <math>[\triangle ABI] = \sqrt{(x+37)(x)(\frac{1225}{37})(\frac{144}{37})} = \frac{420}{37} \sqrt{x(37+x)}.</math> Using the inradius formula, we get <cmath>\frac{210}{37} = \frac{\frac{420}{37} \sqrt{x(37+x)}}{37+x}</cmath> which simplifies to <math>x=\frac{37}{3}</math>. The answer is then <math>\frac{8}{3}. \boxed{11}</math>~ brocolimanx | ||
+ | |||
== 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}} |
Latest revision as of 22:35, 3 September 2024
Contents
Problem
In right with hypotenuse , , , and is the altitude to . Let be the circle having as a diameter. Let be a point outside such that and are both tangent to circle . The ratio of the perimeter of to the length can be expressed in the form , where and are relatively prime positive integers. Find .
Solution 1
First, note that ; let the tangents from to have length . Then the perimeter of is equal to It remains to compute .
Observe , so the radius of is . We may also compute and by similar triangles. Let be the center of ; notice that so it follows while . By the double-angle formula , it turns out that
Using the area formula in , But also, using , Now we can get so multiplying everything by lets us solve for : We have , and now giving the answer, .
Solution 2
Let be center of the circle and , be the two points of tangent such that is on and is on . We know that .
Since the ratios between corresponding lengths of two similar diagrams are equal, we can let and . Hence and the radius .
Since we have and , we have .
Hence . let , then we have Area = = . Then we get .
Now the equation looks very complex but we can take a guess here. Assume that is a rational number (If it's not then the answer to the problem would be irrational which can't be in the form of ) that can be expressed as such that . Look at both sides; we can know that has to be a multiple of and not of and it's reasonable to think that is divisible by so that we can cancel out the on the right side of the equation.
Let's see if fits. Since , and . Amazingly it fits!
Since we know that , the other solution of this equation is negative which can be ignored. Hence .
Hence the perimeter is , and is . Hence , .
Solution 3
As in Solution , let and be the intersections of with and respectively.
First, by pythagorean theorem, . Now the area of is , so and the inradius of is .
Now from we find that so and similarly, .
Note , , and . So we have , . Now we can compute the area of in two ways: by heron's formula and by inradius times semiperimeter, which yields
The quadratic formula now yields . Plugging this back in, the perimeter of is so the ratio of the perimeter to is and our answer is
Note: If you don't want to solve the quadratic, you can continue with and divide both sides by to get . Square both sides to get and solve to get .
Solution 4
As in Solution , let and be the intersections of with and 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 . Let . Let . The semi-perimeter of is . Since the lengths of the sides of are , and , the square of its area by Heron's formula is .
The radius of is . Therefore . As is the in-circle of , the area of is also , and so the square area is .
Therefore Dividing both sides by we get: and so . The semi-perimeter of is therefore and the whole perimeter is . Now , so the ratio of the perimeter of to the hypotenuse is and our answer is
Solution 5
Let , let , and let . Let . We find . Let , , and be the angles , , and respectively. Then , so The perimeter of is . The desired ratio is then We need to find . In , . We get Similarly, . Then Computing in two ways we get , so . Using this value of we get . Thus and .
Solution 6
This solution is not a real solution and is solving the problem with a ruler and compass.
Draw . Then, drawing the tangents and intersecting them, we get that is around and is around . We then find the ratio to be around . Using long division, we find that this ratio is approximately 2.666, which you should recognize as . Since this seems reasonable, we find that the answer is ~ilp
Solution 7
Denoting three tangents has length while lies on with .The area of is , so and the inradius of is .As we know that the diameter of the circle is the height of from to . Assume that and and . But we know that According to the basic computation, we can get that ; So we know that according to the tangent addition formula. Hence, it is not hard to find that the length of is . According to basic addition and division, we get the answer is which leads to ~bluesoul
Solution 8 (no trig and intuitive)
Notice is the altitude of , and so This means in the inradius of the circle is
Next, by , we have and .
Lastly, denote the tangent point of with the circle as respectively. Then by the properties of inscribed circles, we have that and , and . We want to make an algebraic expression utilizing what we know about ABI.
We use the inradius formula: . We have . With Heron's (very basic calculations) we get Using the inradius formula, we get which simplifies to . The answer is then ~ brocolimanx
See also
2009 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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.