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

Latest revision as of 00:42, 6 August 2023

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

Do note that by counting the area in 2 ways, the first altitude is $x = \frac{ab}{c}$ . By similar triangles, the common ratio is $\rho = \frac{a}{c}$ for each height, so by the geometric series formula, we have $\begin{align} 6p=\frac{x}{1-\rho} = \frac{ab}{c-a}. \end{align}$ Writing $p=a+b+c$ and clearing denominators, we get $13a=6p .$ Thus $p=13q$ , $a=6q$ , and $b+c=7q$ , i.e. $c=7q-b$ . Plugging these into $(1)$ , we get $78q(q-b)=6bq$ , i.e., $14b=13q$ . Thus $q=14r$ and $p=182r$ , $b=13r$ , $a=84r$ , $c=85r$ . Taking $r=1$ (since $a,b,c$ are relatively prime) we get $p=\boxed{182}$ .

Solution 2

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 $13a = 84b,$ and since $a, b$ are relatively prime, we know $a = 84$ and $b = 13.$ 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 3

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 4

$[asy] size(10cm); // Setup pair A, B; pair C0, C1, C2, C3, C4, C5, C6, C7, C8; A = (5, 0); B = (0, 3); C0 = (0, 0); C1 = foot(C0, A, B); C2 = foot(C1, C0, B); C3 = foot(C2, C1, B); C4 = foot(C3, C2, B); C5 = foot(C4, C3, B); C6 = foot(C5, C4, B); C7 = foot(C6, C5, B); C8 = foot(C7, C6, B); // Labels label("$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 5

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 $p=13+84+85=\boxed{182}$ .

Solution 6

For this problem, first notice that its an infinite geometric series of $6(a+b+c)=\frac{ab}{c-b}$ if $c$ is the hypotenuse. WLOG $a<b$ , we can generalize a pythagorean triple of $x^2-y^2, 2xy, x^2+y^2$ . Let $b=2xy$ , then this generalization gives $6(a+b+c)(c-b)=ab$ $(x^2-y^2)2xy=6(2x^2+2xy)(x-y)^2$ $(x+y)xy=6(x^2+xy)(x-y)$ $xy=6x(x-y)$ $7xy=6x^2$ $y=\frac{6}{7}x$

Now this is just clear. Let $x=7m$ and $y=6m$ for $m$ to be a positive integer, the pythagorean triple is $13-84-85$ which yields $\boxed{182}$ .

@@ Line 1: / Line 1: @@
-Triangle <math>ABC_0</math> has a right angle at <math>C_0</math>. Its side lengths are pariwise 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_{i=1}^\infty C_{n-2}C_{n-1} = 6p</math>. Find <math>p</math>.
+==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==
-Note that by counting the area in 2 ways, the first altitude is <math>\dfrac{ac}{b}</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>
+Do 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}
+p=\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=\boxed{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>13a = 84b,</math> and since <math>a, b</math> are relatively prime, we know <math>a = 84</math> and <math>b = 13.</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 2==
+==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.
-Considering the sum of the lengths of the segments for which <math>n-2</math> is odd, for each <math>n\geq2</math>, consider the perimeters of the triangles <math>C_{n-2}C_{n-1}C_{n}</math> and <math>C_{n-1}C_{n}C_{n+1}</math>. The perimeters of these triangles can be expressed using <math>p</math> and ratios that result because of similar triangles. Considering triangles of the form <math>C_{n-2}C_{n-1}C_{n}</math>, we find that the perimeter 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>.
@@ Line 17: / Line 28: @@
 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)
-Continuing with a similar process for the sum of the lengths of the segments for which <math>n-2</math> is even, the following results:
+Continuing with a similar process for the sum of the lengths of the segments for which <math>n-2</math> is even,
 <math>p\sum_{n=1}^{\infty}\frac{C_{2n-2}C_{2n-1}}{C_{0}B}=6p+C_{0}A+AB=7p-C_{0}B</math>.
@@ Line 27: / Line 38: @@
 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)
@@ Line 36: / Line 47: @@
 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 4==
+<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 <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 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.
+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>.
+Guessing and checking Pythagorean triples reveals that <math>a = 84</math>, <math>b=13</math>, <math>c = 85</math>, and <math>p = a + b + c = \boxed{182}</math> 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 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 ==
 {{AIME box|year=2016|n=II|num-b=4|num-a=6}}
+{{MAA Notice}}

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