Difference between revisions of "2015 AIME I Problems/Problem 11"
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==Problem== | ==Problem== | ||
Triangle <math>ABC</math> has positive integer side lengths with <math>AB=AC</math>. Let <math>I</math> be the intersection of the bisectors of <math>\angle B</math> and <math>\angle C</math>. Suppose <math>BI=8</math>. Find the smallest possible perimeter of <math>\triangle ABC</math>. | Triangle <math>ABC</math> has positive integer side lengths with <math>AB=AC</math>. Let <math>I</math> be the intersection of the bisectors of <math>\angle B</math> and <math>\angle C</math>. Suppose <math>BI=8</math>. Find the smallest possible perimeter of <math>\triangle ABC</math>. | ||
+ | |||
+ | ==Solution 1== | ||
+ | |||
+ | Let <math>D</math> be the midpoint of <math>\overline{BC}</math>. Then by SAS Congruence, <math>\triangle ABD \cong \triangle ACD</math>, so <math>\angle ADB = \angle ADC = 90^o</math>. | ||
+ | |||
+ | Now let <math>BD=y</math>, <math>AB=x</math>, and <math>\angle IBD = \dfrac{\angle ABD}{2} = \theta</math>. | ||
+ | |||
+ | Then <math>\mathrm{cos}{(\theta)} = \dfrac{y}{8}</math> | ||
+ | |||
+ | and <math>\mathrm{cos}{(2\theta)} = \dfrac{y}{x} = 2\mathrm{cos^2}{(\theta)} - 1 = \dfrac{y^2-32}{32}</math>. | ||
+ | |||
+ | Cross-multiplying yields <math>32y = x(y^2-32)</math>. | ||
+ | |||
+ | Since <math>x,y>0</math>, <math>y^2-32</math> must be positive, so <math>y > 5.5</math>. | ||
+ | |||
+ | Additionally, since <math>\triangle IBD</math> has hypotenuse <math>\overline{IB}</math> of length <math>8</math>, <math>BD=y < 8</math>. | ||
+ | |||
+ | Therefore, given that <math>BC=2y</math> is an integer, the only possible values for <math>y</math> are <math>6</math>, <math>6.5</math>, <math>7</math>, and <math>7.5</math>. | ||
+ | |||
+ | However, only one of these values, <math>y=6</math>, yields an integral value for <math>AB=x</math>, so we conclude that <math>y=6</math> and <math>x=\dfrac{32(6)}{(6)^2-32}=48</math>. | ||
+ | |||
+ | Thus the perimeter of <math>\triangle ABC</math> must be <math>2(x+y) = \boxed{108}</math>. | ||
+ | |||
+ | ==Solution 2 (No Trig)== | ||
+ | Let <math>AB=x</math> and the foot of the altitude from <math>A</math> to <math>BC</math> be point <math>E</math> and <math>BE=y</math>. Since ABC is isosceles, <math>I</math> is on <math>AE</math>. By Pythagorean Theorem, <math>AE=\sqrt{x^2-y^2}</math>. Let <math>IE=a</math> and <math>IA=b</math>. By Angle Bisector theorem, <math>\frac{y}{a}=\frac{x}{b}</math>. Also, <math>a+b=\sqrt{x^2-y^2}</math>. Solving for <math>a</math>, we get <math>a=\frac{\sqrt{x^2-y^2}}{1+\frac{x}{y}}</math>. Then, using Pythagorean Theorem on <math>\triangle BEI</math> we have <math>y^2+\left(\frac{\sqrt{x^2-y^2}}{1+\frac{x}{y}}\right)^2=8^2=64</math>. Simplifying, we have <math>y^2+y^2\frac{x^2-y^2}{(x+y)^2}=64</math>. Factoring out the <math>y^2</math>, we have <math>y^2\left(1+\frac{x^2-y^2}{(x+y)^2}\right)=64</math>. Adding 1 to the fraction and simplifying, we have <math>\frac{y^2x(x+y)}{(x+y)^2}=32</math>. Crossing out the <math>x+y</math>, and solving for <math>x</math> yields <math>32y = x(y^2-32)</math>. Then, we continue as Solution 1 does. | ||
+ | |||
+ | ==Solution 3== | ||
+ | Let <math>AB=x</math>, call the midpoint of <math>BC</math> point <math>E</math>, call the point where the incircle meets <math>AB</math> point <math>D</math>, | ||
+ | |||
+ | and let <math>BE=y</math>. We are looking for the minimum value of <math>2(x+y)</math>. <math>AE</math> is an altitude because the triangle | ||
+ | |||
+ | is isosceles. By Pythagoras on <math>BEI</math>, the inradius is <math>\sqrt{64-y^2}</math> and by Pythagoras on <math>ABE</math>, <math>AE</math> is | ||
+ | |||
+ | <math>\sqrt{x^2-y^2}</math>. By equal tangents, <math>BE=BD=y</math>, so <math>AD=x-y</math>. Since <math>ID</math> is an inradius, <math>ID=IE</math> and | ||
+ | using pythagoras on <math>ADI</math> yields <math>AI=</math><math>\sqrt{x^2-2xy+64}</math>. <math>ADI</math> is similar to <math>AEB</math> by <math>AA</math>, so we | ||
+ | |||
+ | can write <math>\frac{x-y}{\sqrt{x^2-2xy+64}}=\frac{\sqrt{x^2-y^2}}{x}</math>. Simplifying, <math>\frac{x}{\sqrt{x^2-2xy+64}}=\sqrt{\frac{x+y}{x-y}}</math>. | ||
+ | |||
+ | Squaring, subtracting 1 from both sides, and multiplying everything out, we get <math>yx^2-2xy^2+64y=yx^2 -32x+32y-xy^2</math>, which turns into <math>32y=x(y^2-32)</math>. Finish as in Solution 1. | ||
+ | ==Solution 4== | ||
+ | Angle bisectors motivate trig bash. | ||
+ | Define angle <math>IBC = x</math>. Foot of perpendicular from <math>I</math> to <math>BC</math> is point <math>P</math>. | ||
+ | <math>\overline{BC} = 2\overline{BP} = 2(8\cos(x)) = N</math>, where <math>N</math> is an integer. Thus, <math>\cos(x) = \frac{N}{16}</math>. Via double angle, we calculate <math>\overline{AB}</math> to be <math>\frac{8\cos(x)}{2\cos(x)^2 - 1} = \frac{64N}{N^2 - 128}</math>. This is to be an integer. We can bound <math>N</math> now, as <math>N > 11</math> to avoid negative values and <math>N < 16</math> due to triangle inequality. Testing, <math>N = 12</math> works, giving <math>\overline{AB} = 48, \overline{BC} = 12</math>. | ||
+ | Our answer is <math>2 * 48 + 12 = \boxed{108}</math>. | ||
+ | - whatRthose | ||
+ | |||
+ | ==Solution 5== | ||
+ | |||
+ | [[File:2015 AIME I 11.png|270px|right]] | ||
+ | Let <math>M</math> be midpoint <math>BC, BM = x, AB = y, \angle IBM = \alpha.</math> | ||
+ | |||
+ | <math>BI</math> is the bisector of <math>\angle ABM</math> in <math>\triangle ABM.</math> | ||
+ | <math>BI = \frac {2 xy \cos \alpha}{x+y} = 8, \cos \alpha = \frac {x}{8} \implies \frac {x^2 y}{x+y} = 32.</math> | ||
+ | <cmath>y = \frac {32 x} {x^2 - 32}.</cmath> | ||
+ | <math>BC = 2x</math> is integer, <math>5.5^2 < 32 \implies x \ge 6.</math> | ||
+ | <math>BM < BI \implies x =\{ 6, 6.5, 7, 7.5 \}.</math> | ||
+ | |||
+ | If <math>x > 6</math> then <math>y</math> is not integer. | ||
+ | <cmath>x = 6 \implies y = 48 \implies 2(x+y) = \boxed{\textbf{108}}.</cmath> | ||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | |||
+ | |||
+ | |||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://youtu.be/R8kvayz7Rtw?si=hFg4yGZO4dxyxAuG | ||
+ | |||
+ | ~MathProblemSolvingSkills.com | ||
+ | |||
==See Also== | ==See Also== | ||
{{AIME box|year=2015|n=I|num-b=10|num-a=12}} | {{AIME box|year=2015|n=I|num-b=10|num-a=12}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
+ | |||
+ | [[Category: Intermediate Geometry Problems]] |
Latest revision as of 16:43, 27 September 2023
Contents
Problem
Triangle has positive integer side lengths with . Let be the intersection of the bisectors of and . Suppose . Find the smallest possible perimeter of .
Solution 1
Let be the midpoint of . Then by SAS Congruence, , so .
Now let , , and .
Then
and .
Cross-multiplying yields .
Since , must be positive, so .
Additionally, since has hypotenuse of length , .
Therefore, given that is an integer, the only possible values for are , , , and .
However, only one of these values, , yields an integral value for , so we conclude that and .
Thus the perimeter of must be .
Solution 2 (No Trig)
Let and the foot of the altitude from to be point and . Since ABC is isosceles, is on . By Pythagorean Theorem, . Let and . By Angle Bisector theorem, . Also, . Solving for , we get . Then, using Pythagorean Theorem on we have . Simplifying, we have . Factoring out the , we have . Adding 1 to the fraction and simplifying, we have . Crossing out the , and solving for yields . Then, we continue as Solution 1 does.
Solution 3
Let , call the midpoint of point , call the point where the incircle meets point ,
and let . We are looking for the minimum value of . is an altitude because the triangle
is isosceles. By Pythagoras on , the inradius is and by Pythagoras on , is
. By equal tangents, , so . Since is an inradius, and using pythagoras on yields . is similar to by , so we
can write . Simplifying, .
Squaring, subtracting 1 from both sides, and multiplying everything out, we get , which turns into . Finish as in Solution 1.
Solution 4
Angle bisectors motivate trig bash. Define angle . Foot of perpendicular from to is point . , where is an integer. Thus, . Via double angle, we calculate to be . This is to be an integer. We can bound now, as to avoid negative values and due to triangle inequality. Testing, works, giving . Our answer is . - whatRthose
Solution 5
Let be midpoint
is the bisector of in is integer,
If then is not integer. vladimir.shelomovskii@gmail.com, vvsss
Video Solution
https://youtu.be/R8kvayz7Rtw?si=hFg4yGZO4dxyxAuG
~MathProblemSolvingSkills.com
See Also
2015 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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