Difference between revisions of "2016 AIME II Problems/Problem 5"
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− | Triangle <math>ABC_0</math> has a right angle at <math>C_0</math>. Its side lengths are | + | ==Problem== |
+ | Triangle <math>ABC_0</math> has a right angle at <math>C_0</math>. Its side lengths are pairwise relatively prime positive integers, and its perimeter is <math>p</math>. Let <math>C_1</math> be the foot of the altitude to <math>\overline{AB}</math>, and for <math>n \geq 2</math>, let <math>C_n</math> be the foot of the altitude to <math>\overline{C_{n-2}B}</math> in <math>\triangle C_{n-2}C_{n-1}B</math>. The sum <math>\sum_{n=2}^\infty C_{n-2}C_{n-1} = 6p</math>. Find <math>p</math>. | ||
==Solution 1== | ==Solution 1== | ||
− | Note that by counting the area in 2 ways, the first altitude is <math>\dfrac{ | + | |
+ | Note that by counting the area in 2 ways, the first altitude is <math>x = \frac{ab}{c}</math>. By similar triangles, the common ratio is <math>\rho = \frac{a}{c}</math> for each height, so by the geometric series formula, we have | ||
+ | <cmath>\begin{align} | ||
+ | 6p=\frac{x}{1-\rho} = \frac{ab}{c-a}. | ||
+ | \end{align}</cmath> | ||
+ | Writing <math>p=a+b+c</math> and clearing denominators, we get <cmath>13a=6p .</cmath>Thus <math>p=13q</math>, <math>a=6q</math>, and <math>b+c=7q</math>, i.e. <math>c=7q-b</math>. Plugging these into <math>(1)</math>, we get <math>78q(q-b)=6bq</math>, i.e., <math>14b=13q</math>. Thus <math>q=14r</math> and <math>p=182r</math>, <math>b=13r</math>, <math>a=84r</math>, <math>c=85r</math>. Taking <math>r=1</math> (since <math>a,b,c</math> are relatively prime) we get <math>p=182</math>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | Note that by counting the area in 2 ways, the first altitude is <math>\dfrac{ab}{c}</math>. By similar triangles, the common ratio is <math>\dfrac{a}{c}</math> for reach height, so by the geometric series formula, we have <math>6p=\dfrac{\dfrac{ab}{c}}{1-\dfrac{a}{c}}</math>. Multiplying by the denominator and expanding, the equation becomes <math>\dfrac{ab}{c}=6a+6b+6c-\dfrac{6a^2}{c}-\dfrac{6ab}{c}-6a</math>. Cancelling <math>6a</math> and multiplying by <math>c</math> yields <math>ab=6bc+6c^2-6a^2-6ab</math>, so <math>7ab = 6bc+6b^2</math> and <math>7a=6b+6c</math>. Checking for Pythagorean triples gives <math>13,84,</math> and <math>85</math>, so <math>p=13+84+85=\boxed{182}</math> | ||
+ | |||
+ | Note: a more rigorous solution instead of checking for triples would be to substitute <math>c = \sqrt{a^2 + b^2}</math> and heavily simplifying the equation. Eventually we are left with <math>84a = 13b,</math> and since <math>a, b</math> are relatively prime, we know <math>a = 13</math> and <math>b = 84.</math> From here we can note that <math>(13, 84, 85)</math> is a pythagorean triple or again use the Pythagorean Theorem to find <math>c = 85.</math> Thus, the answer is <math>\boxed{182}.</math> | ||
Solution modified/fixed from Shaddoll's solution. | Solution modified/fixed from Shaddoll's solution. | ||
− | ==Solution | + | ==Solution 3== |
We start by splitting the sum of all <math>C_{n-2}C_{n-1}</math> into two parts: those where <math>n-2</math> is odd and those where <math>n-2</math> is even. | We start by splitting the sum of all <math>C_{n-2}C_{n-1}</math> into two parts: those where <math>n-2</math> is odd and those where <math>n-2</math> is even. | ||
− | First consider the sum of the lengths of the segments for which <math>n-2</math> is odd for each <math>n\geq2</math>. The perimeters of these triangles can be expressed using <math>p</math> and ratios that result because of similar triangles. Considering triangles where <math>n-2</math> is odd, we find that the perimeter for each such <math>n</math> is <math>p(\frac{C_{n-1}C_{n}}{C_{0}B})</math>. Thus, | + | First consider the sum of the lengths of the segments for which <math>n-2</math> is odd for each <math>n\geq2</math>. The perimeters of these triangles can be expressed using <math>p</math> and ratios that result because of similar triangles. Considering triangles where <math>n-2</math> is odd, we find that the perimeter for each such <math>n</math> is <math>p\left(\frac{C_{n-1}C_{n}}{C_{0}B}\right)</math>. Thus, |
<math>p\sum_{n=1}^{\infty}\frac{C_{2n-1}C_{2n}}{C_{0}B}=6p+C_{0}B</math>. | <math>p\sum_{n=1}^{\infty}\frac{C_{2n-1}C_{2n}}{C_{0}B}=6p+C_{0}B</math>. | ||
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Simplifying, | Simplifying, | ||
− | <math>\sum_{n=1}^{\infty}C_{2n-1}C_{2n}=6C_{0}B + \frac{(C_{0}B)^2}{p}=C_{0}B(6+\frac{C_{0}B}{p})</math>. (1) | + | <math>\sum_{n=1}^{\infty}C_{2n-1}C_{2n}=6C_{0}B + \frac{(C_{0}B)^2}{p}=C_{0}B\left(6+\frac{C_{0}B}{p}\right)</math>. (1) |
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Simplifying, | Simplifying, | ||
− | <math>\sum_{n=1}^{\infty}C_{2n-2}C_{2n-1}=C_{0}B(7-\frac{C_{0}B}{p})</math>. (2) | + | <math>\sum_{n=1}^{\infty}C_{2n-2}C_{2n-1}=C_{0}B\left(7-\frac{C_{0}B}{p}\right)</math>. (2) |
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Setting <math>a=C_{0}B</math>, <math>b=C_{0}A</math>, and <math>c=AB</math>, we can now proceed as in Shaddoll's solution, and our answer is <math>p=13+84+85=\boxed{182}</math>. | Setting <math>a=C_{0}B</math>, <math>b=C_{0}A</math>, and <math>c=AB</math>, we can now proceed as in Shaddoll's solution, and our answer is <math>p=13+84+85=\boxed{182}</math>. | ||
− | + | == Solution 4== | |
− | |||
− | == Solution | ||
<asy> | <asy> | ||
size(10cm); | size(10cm); | ||
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label("$C_3$", C3, NE); | label("$C_3$", C3, NE); | ||
label("$C_4$", C4, W); | label("$C_4$", C4, W); | ||
+ | label("$a$", (B+C0)/2, W); | ||
+ | label("$b$", (A+C0)/2, S); | ||
+ | label("$c$", (A+B)/2, NE); | ||
+ | |||
// Drawings | // Drawings | ||
Line 70: | Line 83: | ||
draw(C0--C1--C2, red); | draw(C0--C1--C2, red); | ||
draw(C2--C3--C4--C5--C6--C7--C8); | draw(C2--C3--C4--C5--C6--C7--C8); | ||
− | draw(C0--C2, | + | draw(C0--C2, blue); |
</asy> | </asy> | ||
Let <math>a = BC_0</math>, <math>b = AC_0 </math>, and <math>c = AB</math>. | Let <math>a = BC_0</math>, <math>b = AC_0 </math>, and <math>c = AB</math>. | ||
− | Note that the total length of the red segments in the figure above is equal to the length of the | + | Note that the total length of the red segments in the figure above is equal to the length of the blue segment times <math>\frac{a+c}{b}</math>. |
The desired sum is equal to the total length of the infinite path <math>C_0 C_1 C_2 C_3 \cdots</math>, shown in red in the figure below. | The desired sum is equal to the total length of the infinite path <math>C_0 C_1 C_2 C_3 \cdots</math>, shown in red in the figure below. | ||
− | Since each of the triangles <math>\triangle C_0 C_1 C_2, \triangle C_2 C_3 C_4, \dots</math> on the left are similar, it follows that the total length of the red segments in the figure below is equal to the length of the | + | Since each of the triangles <math>\triangle C_0 C_1 C_2, \triangle C_2 C_3 C_4, \dots</math> on the left are similar, it follows that the total length of the red segments in the figure below is equal to the length of the blue segment times <math>\frac{a+c}{b}</math>. |
In other words, we have that <math>a\left(\frac{a+c}{b}\right) = 6p</math>. | In other words, we have that <math>a\left(\frac{a+c}{b}\right) = 6p</math>. | ||
Line 111: | Line 124: | ||
draw(A--B--C0--cycle); | draw(A--B--C0--cycle); | ||
draw(C0--C1--C2--C3--C4--C5--C6--C7--C8, red); | draw(C0--C1--C2--C3--C4--C5--C6--C7--C8, red); | ||
− | draw(C0--B, | + | draw(C0--B, blue); |
</asy> | </asy> | ||
+ | |||
+ | ==Solution 5== | ||
+ | This solution proceeds from <math>\frac{\frac{ab}{c}}{1-\frac{a}{c}} = \frac{\frac{ab}{c}}{\frac{c-a}{c}} = \frac{ab}{c-a} = 6(a+b+c)</math>. Note the general from for a primitive pythagorean triple, <math>m^2-n^2, 2mn, m^2+n^2</math> and after substitution, letting <math>a = m^2-n^2, b = 2mn, c = m^2+n^2</math> into the previous equation simplifies down very nicely into <math>m = 13n</math>. Thus <math>a = 168n^2, b = 26n^2, c = 170n^2</math>. Since we know all three side lengths are relatively prime, we must divide each by 2 and let n = 1 giving <math>a = 84, b = 13, c = 85</math> yielding <math>p = a + b + c = \boxed{182}</math>. | ||
+ | |||
+ | Alternate way to find a, b, and c (from VarunGotem) : <math>7a=6b+6c</math> implies that <math>6b=7a-6c</math>. Because <math>a^2+b^2=c^2</math>, <math>36a^2+(6b)^2=36c^2</math>. Plugging in gives <math>85a^2-84ac+36c^2=36c^2</math>. Simplifying gives <math>85a-84c=0</math>, and since <math>a</math> and <math>c</math> are relatively prime, <math>a=84</math> and <math>c=85</math>. This means <math>b=13</math> and <math>p=13+84+85=\boxed{182}</math>. | ||
+ | |||
+ | ==Solution 6== | ||
+ | For this problem, first notice that its an infinite geometric series of <math>6(a+b+c)=\frac{ab}{c-b}</math> if <math>c</math> is the hypotenuse. WLOG <math>a<b</math>, we can generalize a pythagorean triple of <math>x^2-y^2, 2xy, x^2+y^2</math>. Let <math>b=2xy</math>, then this generalization gives <math>6(a+b+c)(c-b)=ab</math> | ||
+ | <cmath>(x^2-y^2)2xy=6(2x^2+2xy)(x-y)^2</cmath> | ||
+ | <cmath>(x+y)xy=6(x^2+xy)(x-y)</cmath> | ||
+ | <cmath>xy=6x(x-y)</cmath> | ||
+ | <cmath>7xy=6x^2</cmath> | ||
+ | <cmath>y=\frac{6}{7}x</cmath> | ||
+ | |||
+ | Now this is just clear. Let <math>x=7m</math> and <math>y=6m</math> for <math>m</math> to be a positive integer, the pythagorean triple is <math>13-84-85</math> which yields <math>\boxed{182}</math>. | ||
== See also == | == See also == | ||
{{AIME box|year=2016|n=II|num-b=4|num-a=6}} | {{AIME box|year=2016|n=II|num-b=4|num-a=6}} | ||
+ | {{MAA Notice}} |
Latest revision as of 21:46, 27 January 2022
Contents
Problem
Triangle has a right angle at . Its side lengths are pairwise relatively prime positive integers, and its perimeter is . Let be the foot of the altitude to , and for , let be the foot of the altitude to in . The sum . Find .
Solution 1
Note that by counting the area in 2 ways, the first altitude is . By similar triangles, the common ratio is for each height, so by the geometric series formula, we have Writing and clearing denominators, we get Thus , , and , i.e. . Plugging these into , we get , i.e., . Thus and , , , . Taking (since are relatively prime) we get .
Solution 2
Note that by counting the area in 2 ways, the first altitude is . By similar triangles, the common ratio is for reach height, so by the geometric series formula, we have . Multiplying by the denominator and expanding, the equation becomes . Cancelling and multiplying by yields , so and . Checking for Pythagorean triples gives and , so
Note: a more rigorous solution instead of checking for triples would be to substitute and heavily simplifying the equation. Eventually we are left with and since are relatively prime, we know and From here we can note that is a pythagorean triple or again use the Pythagorean Theorem to find Thus, the answer is
Solution modified/fixed from Shaddoll's solution.
Solution 3
We start by splitting the sum of all into two parts: those where is odd and those where is even.
First consider the sum of the lengths of the segments for which is odd for each . The perimeters of these triangles can be expressed using and ratios that result because of similar triangles. Considering triangles where is odd, we find that the perimeter for each such is . Thus,
.
Simplifying,
. (1)
Continuing with a similar process for the sum of the lengths of the segments for which is even,
.
Simplifying,
. (2)
Adding (1) and (2) together, we find that
.
Setting , , and , we can now proceed as in Shaddoll's solution, and our answer is .
Solution 4
Let , , and . Note that the total length of the red segments in the figure above is equal to the length of the blue segment times .
The desired sum is equal to the total length of the infinite path , shown in red in the figure below. Since each of the triangles on the left are similar, it follows that the total length of the red segments in the figure below is equal to the length of the blue segment times . In other words, we have that .
Guessing and checking Pythagorean triples reveals that , , , and satisfies this equation.
Solution 5
This solution proceeds from . Note the general from for a primitive pythagorean triple, and after substitution, letting into the previous equation simplifies down very nicely into . Thus . Since we know all three side lengths are relatively prime, we must divide each by 2 and let n = 1 giving yielding .
Alternate way to find a, b, and c (from VarunGotem) : implies that . Because , . Plugging in gives . Simplifying gives , and since and are relatively prime, and . This means and .
Solution 6
For this problem, first notice that its an infinite geometric series of if is the hypotenuse. WLOG , we can generalize a pythagorean triple of . Let , then this generalization gives
Now this is just clear. Let and for to be a positive integer, the pythagorean triple is which yields .
See also
2016 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
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.