Difference between revisions of "2009 AIME I Problems/Problem 12"
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− | == Solution == | + | == Solution 1== |
− | + | 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>. | |
+ | |||
+ | Since the ratios between corresponding lengths of two similar diagrams are equal, we can let <math>AD = 144, CD = 420</math> and <math>BD = 1225</math>. Hence <math>AQ = 144, BP = 1225, AB = 1369</math> and the radius <math>r = OD = 210</math>. | ||
+ | |||
+ | Since we have <math>\tan OAB = \frac {35}{24}</math> and <math>\tan OBA = \frac{6}{35}</math> , we have <math>\sin {(OAB + OBA)} = \frac {1369}{\sqrt {(1801*1261)}},</math><math>\cos {(OAB + OBA)} = \frac {630}{\sqrt {(1801*1261)}}</math>. | ||
+ | |||
+ | Hence <math>\sin I = \sin {(2OAB + 2OBA)} = \frac {2*1369*630}{1801*1261}</math>. let <math>IP = IQ = x</math> , then we have Area<math>(IBC)</math> = <math>(2x + 1225*2 + 144*2)*\frac {210}{2}</math> = <math>(x + 144)(x + 1225)* \sin {\frac {I}{2}}</math>. Then we get <math>x + 1369 = \frac {3*1369*(x + 144)(x + 1225)}{1801*1261}</math>. | ||
+ | |||
+ | Now the equation looks very complex but we can take a guess here. Assume that <math>x</math> 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 <math>\frac {m}{n}</math>) | ||
+ | that can be expressed as <math>\frac {a}{b}</math> such that <math>(a,b) = 1</math>. Look at both sides; we can know that <math>a</math> has to be a multiple of <math>1369</math> and not of <math>3</math> and it's reasonable to think that <math>b</math> is divisible by <math>3</math> so that we can cancel out the <math>3</math> on the right side of the equation. | ||
+ | |||
+ | Let's see if <math>x = \frac {1369}{3}</math> fits. Since <math>\frac {1369}{3} + 1369 = \frac {4*1369}{3}</math>, and <math>\frac {3*1369*(x + 144)(x + 1225)}{1801*1261} = \frac {3*1369* \frac {1801}{3} * \frac {1261*4}{3}} {1801*1261} = \frac {4*1369}{3}</math>. Amazingly it fits! | ||
+ | |||
+ | Since we know that <math>3*1369*144*1225 - 1369*1801*1261 < 0</math>, the other solution of this equation is negative which can be ignored. Hence <math>x = 1369/3</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 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== | ||
+ | |||
+ | 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> | ||
+ | |||
+ | == Solution 4 == | ||
+ | |||
+ | We shall yet again let <math>P</math> and <math>Q</math> be the intersections of <math>AI</math> and <math>BI</math> to <math>\omega</math>, respectively. We want to find the perimeter of <math>ABI</math>, which is <math>AD+BD+BQ+QI+IP+PA</math>. We can easily find <math>AD</math> and <math>BD</math> using the fact that <math>ABC</math>, <math>ACD</math>, and <math>BCD</math> are all similar triangles. We get <math>AD=\frac{144}{37}</math> and <math>\frac{1225}{37}</math>. Since <math>AP</math> and <math>AD</math> are tangents to <math>\omega</math>, <math>AP=AD=\frac{144}{37}</math>, and similarly <math>BQ=BD=\frac{1225}{37}</math>. We now wish to find <math>IP</math> and <math>IQ</math>. | ||
+ | |||
+ | Let the center of the given circle be <math>O</math>. We know that <math>\angle AOP=\angle AOD</math>, <math>\angle BOQ=\angle BOD</math>, and <math>\angle IOQ=\angle IOP</math>. Since all six angles sum to <math>360^{\circ}</math>, <math>\angle AOP+\angle BOQ+\angle IOP=180^{\circ}</math>. If we knew the radius of circle <math>\omega</math> now, then we could find <math>\tan{\angle AOP}</math> and <math>\tan{\angle BOQ}</math>, and then we can use the sum (or difference) of tangents formula to find <math>\tan{\angle IOP}</math>, which reveals <math>IP</math>. This means we should find the radius of <math>\omega</math>. We can easily see that the height of triangle <math>ABC</math> from <math>C</math> has length <math>\frac{12*35}{37}</math>, so the radius of <math>\omega</math> is <math>\frac{210}{37}</math>. Now we shall proceed with the above plan. | ||
+ | |||
+ | <math>\tan{\angle AOP}=\frac{144}{210}</math>. <math>\tan{\angle BOQ}=\frac{1225}{210}</math>. | ||
+ | |||
+ | <math>\tan{\angle IOP}=\tan{(180^{\circ}-\angle AOP-\angle BOQ)}=-\tan{(\angle AOP+\angle BOQ)}</math> | ||
+ | |||
+ | <math>=-\frac{\frac{144}{210}+\frac{1225}{210}}{1-\frac{144}{210}*\frac{1225}{210}}=-\frac{1369}{210-\frac{144*1225}{210}}=\frac{1369}{\frac{144*1225}{210}-210}=\frac{37*37}{35*18}</math>. | ||
+ | |||
+ | Therefore <math>OP=\frac{210}{37},IP=\frac{37}{3}</math>, and the perimeter of <math>AIB</math> is <math>2*\frac{37}{3}+2*\frac{144}{37}+2*\frac{1225}{37}=37*\frac{8}{3}</math>. Since <math>AB=37</math>, the desired ratio is <math>\frac{8}{3}</math>, and <math>8+3=\boxed{011}</math>. | ||
+ | |||
+ | == Solution 5 == | ||
+ | |||
+ | 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 | ||
== 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]] | ||
+ | {{MAA Notice}} |
Latest revision as of 11:04, 30 December 2020
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
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 2
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
Solution 3
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 4
We shall yet again let and be the intersections of and to , respectively. We want to find the perimeter of , which is . We can easily find and using the fact that , , and are all similar triangles. We get and . Since and are tangents to , , and similarly . We now wish to find and .
Let the center of the given circle be . We know that , , and . Since all six angles sum to , . If we knew the radius of circle now, then we could find and , and then we can use the sum (or difference) of tangents formula to find , which reveals . This means we should find the radius of . We can easily see that the height of triangle from has length , so the radius of is . Now we shall proceed with the above plan.
. .
.
Therefore , and the perimeter of is . Since , the desired ratio is , and .
Solution 5
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
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.