Difference between revisions of "2016 AIME II Problems/Problem 5"

(Solution 1)
(One intermediate revision by one other user not shown)
Line 4: Line 4:
 
==Solution 1==
 
==Solution 1==
 
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 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.
Line 36: Line 38:
  
 
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 by brightaz
 
  
 
== Solution 3 ==
 
== Solution 3 ==

Revision as of 18:29, 11 August 2021

Problem

Triangle $ABC_0$ has a right angle at $C_0$. Its side lengths are pairwise relatively prime positive integers, and its perimeter is $p$. Let $C_1$ be the foot of the altitude to $\overline{AB}$, and for $n \geq 2$, let $C_n$ be the foot of the altitude to $\overline{C_{n-2}B}$ in $\triangle C_{n-2}C_{n-1}B$. The sum $\sum_{n=2}^\infty C_{n-2}C_{n-1} = 6p$. Find $p$.

Solution 1

Note that by counting the area in 2 ways, the first altitude is $\dfrac{ab}{c}$. By similar triangles, the common ratio is $\dfrac{a}{c}$ for reach height, so by the geometric series formula, we have $6p=\dfrac{\dfrac{ab}{c}}{1-\dfrac{a}{c}}$. Multiplying by the denominator and expanding, the equation becomes $\dfrac{ab}{c}=6a+6b+6c-\dfrac{6a^2}{c}-\dfrac{6ab}{c}-6a$. Cancelling $6a$ and multiplying by $c$ yields $ab=6bc+6c^2-6a^2-6ab$, so $7ab = 6bc+6b^2$ and $7a=6b+6c$. Checking for Pythagorean triples gives $13,84,$ and $85$, so $p=13+84+85=\boxed{182}$

Note: a more rigorous solution instead of checking for triples would be to substitute $c = \sqrt{a^2 + b^2}$ and heavily simplifying the equation. Eventually we are left with $84a = 13b,$ and since $a, b$ are relatively prime, we know $a = 13$ and $b = 84.$ From here we can note that $(13, 84, 85)$ is a pythagorean triple or again use the Pythagorean Theorem to find $c = 85.$ Thus, the answer is $\boxed{182}.$

Solution modified/fixed from Shaddoll's solution.

Solution 2

We start by splitting the sum of all $C_{n-2}C_{n-1}$ into two parts: those where $n-2$ is odd and those where $n-2$ is even.


First consider the sum of the lengths of the segments for which $n-2$ is odd for each $n\geq2$. The perimeters of these triangles can be expressed using $p$ and ratios that result because of similar triangles. Considering triangles where $n-2$ is odd, we find that the perimeter for each such $n$ is $p\left(\frac{C_{n-1}C_{n}}{C_{0}B}\right)$. Thus,

$p\sum_{n=1}^{\infty}\frac{C_{2n-1}C_{2n}}{C_{0}B}=6p+C_{0}B$.


Simplifying,

$\sum_{n=1}^{\infty}C_{2n-1}C_{2n}=6C_{0}B + \frac{(C_{0}B)^2}{p}=C_{0}B\left(6+\frac{C_{0}B}{p}\right)$. (1)


Continuing with a similar process for the sum of the lengths of the segments for which $n-2$ is even,

$p\sum_{n=1}^{\infty}\frac{C_{2n-2}C_{2n-1}}{C_{0}B}=6p+C_{0}A+AB=7p-C_{0}B$.


Simplifying,

$\sum_{n=1}^{\infty}C_{2n-2}C_{2n-1}=C_{0}B\left(7-\frac{C_{0}B}{p}\right)$. (2)


Adding (1) and (2) together, we find that

$6p=13C_{0}B \Rightarrow p=\frac{13C_{0}B}{6}=C_{0}B+C_{0}A+AB \Rightarrow \frac{7C_{0}B}{6}=C_{0}A+AB \Rightarrow 7C_{0}B=6C_{0}A + 6AB$.

Setting $a=C_{0}B$, $b=C_{0}A$, and $c=AB$, we can now proceed as in Shaddoll's solution, and our answer is $p=13+84+85=\boxed{182}$.

Solution 3

[asy] size(10cm);  // Setup pair A, B; pair C0, C1, C2, C3, C4, C5, C6, C7, C8; A = (5, 0); B = (0, 3); C0 = (0, 0); C1 = foot(C0, A, B); C2 = foot(C1, C0, B); C3 = foot(C2, C1, B); C4 = foot(C3, C2, B); C5 = foot(C4, C3, B); C6 = foot(C5, C4, B); C7 = foot(C6, C5, B); C8 = foot(C7, C6, B);  // Labels label("$A$", A, SE); label("$B$", B, NW); label("$C_0$", C0, SW); label("$C_1$", C1, NE); label("$C_2$", C2, W); label("$C_3$", C3, NE); label("$C_4$", C4, W); label("$a$", (B+C0)/2, W); label("$b$", (A+C0)/2, S); label("$c$", (A+B)/2, NE);   // Drawings draw(A--B--C0--cycle); draw(C0--C1--C2, red); draw(C2--C3--C4--C5--C6--C7--C8); draw(C0--C2, blue); [/asy]

Let $a = BC_0$, $b = AC_0$, and $c = AB$. Note that the total length of the red segments in the figure above is equal to the length of the blue segment times $\frac{a+c}{b}$.

The desired sum is equal to the total length of the infinite path $C_0 C_1 C_2 C_3 \cdots$, shown in red in the figure below. Since each of the triangles $\triangle C_0 C_1 C_2, \triangle C_2 C_3 C_4, \dots$ 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 $\frac{a+c}{b}$. In other words, we have that $a\left(\frac{a+c}{b}\right) = 6p$.

Guessing and checking Pythagorean triples reveals that $a = 84$, $b=13$, $c = 85$, and $p = a + b + c = \boxed{182}$ satisfies this equation.

[asy] size(10cm);  // Setup pair A, B; pair C0, C1, C2, C3, C4, C5, C6, C7, C8; A = (5, 0); B = (0, 3); C0 = (0, 0); C1 = foot(C0, A, B); C2 = foot(C1, C0, B); C3 = foot(C2, C1, B); C4 = foot(C3, C2, B); C5 = foot(C4, C3, B); C6 = foot(C5, C4, B); C7 = foot(C6, C5, B); C8 = foot(C7, C6, B);  // Labels label("$A$", A, SE); label("$B$", B, NW); label("$C_0$", C0, SW); label("$a$", (B+C0)/2, W); label("$b$", (A+C0)/2, S); label("$c$", (A+B)/2, NE);  // Drawings draw(A--B--C0--cycle); draw(C0--C1--C2--C3--C4--C5--C6--C7--C8, red); draw(C0--B, blue); [/asy]

Solution 4

This solution proceeds from $\frac{\frac{ab}{c}}{1-\frac{a}{c}} = \frac{\frac{ab}{c}}{\frac{c-a}{c}} = \frac{ab}{c-a} = 6(a+b+c)$. Note the general from for a primitive pythagorean triple, $m^2-n^2, 2mn, m^2+n^2$ and after substitution, letting $a = m^2-n^2, b = 2mn, c = m^2+n^2$ into the previous equation simplifies down very nicely into $m = 13n$. Thus $a = 168n^2, b = 26n^2, c = 170n^2$. Since we know all three side lengths are relatively prime, we must divide each by 2 and let n = 1 giving $a = 84, b = 13, c = 85$ yielding $p = a + b + c = \boxed{182}$.

Alternate way to find a, b, and c (from VarunGotem) : $7a=6b+6c$ implies that $6b=7a-6c$. Because $a^2+b^2=c^2$, $36a^2+(6b)^2=36c^2$. Plugging in gives $85a^2-84ac+36c^2=36c^2$. Simplifying gives $85a-84c=0$, and since $a$ and $c$ are relatively prime, $a=84$ and $c=85$. This means $b=13$ and $p=13+84+85=\boxed{182}$.

See also

2016 AIME II (ProblemsAnswer KeyResources)
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. AMC logo.png

Invalid username
Login to AoPS