Prove that there exist infinitely many positive integers $ n$ such that $ p = nr,$ where $ p$ and $ r$ are respectively the semiperimeter and the inradius of a triangle with integer side lengths.
Problem
Source: IMO Shortlist 2000, N5
Tags: Diophantine equation, number theory, inradius, perimeter, Triangle, IMO Shortlist
Mewto55555
25.06.2010 00:18
First, note that if we prove there are infinitely many $n$ for triangles with rational side lengths, we can scale up the triangle so that the sides are all integers, and the condition is true.
Let the sides of the triangle be $a$, $b$, and $c$.
As $A=rp$, we have $n=\frac{p}{r}=\frac{p}{\frac{A}{p}}=\frac{p^2}{A}$
By Heron's, $A=\sqrt{p(p-a)(p-b)(p-c)}$ so $n=\frac{p^2}{\sqrt{p(p-a)(p-b)(p-c)}}=\sqrt{\frac{p^3}{(p-a)(p-b)(p-c)}}$
Remembering that $p=\frac{a+b+c}{2}$, we have:
$n=\sqrt{{\frac{\frac{(a+b+c)^3}{8}}{\frac{-a+b+c}{2}\cdot\frac{a-b+c}{2}\cdot\frac{a+b-c}{2}}}}=\sqrt{\frac{a+b+c}{-a+b+c} \cdot \frac{a+b+c}{a-b+c} \cdot \frac{a+b+c}{a+b-c}}$
Let $a$,$b$,$c$ be the (necessarily) rational solutions to the simultaneous equations:
$a+b+c=(x^2)(a+b-c)$
$a+b+c=(y^2)(a-b+c)$
$a+b+c=(z^2)(-a+b+c)$
with $x,y,z>1$ and integers.
It remains to be proven that $a,b,c$ can then be rational sides of a triangle.
Note that $x^2=\frac{a+b+c}{a+b-c}=1+\frac{2c}{a+b-c}$
Similarly $y^2=1+\frac{2b}{a-b+c}$ and $z^2=1+\frac{2a}{-a+b+c}$
Letting $f,g,h$ equal $\frac{x^2-1}{2},\frac{y^2-1}{2},\frac{z^2-1}{2}$ respectively, we have
$f(a+b-c)=c$
$g(a-b+c)=b$
$h(-a+b+c)=a$
If $a<0$, then because $h>0$, we have $b+c-a<0 \implies a > b+c$. Thus, as least one of $b$ or $c$ is negative. WLOG, let $b<0$.
Then, $g(a-b+c)<0 \implies a+c< b<a-c \implies c<0$. Thus, if one of them is negative, they all are. If this is the case, we can simply use the triangle with sides $-a$,$-b$, and $-c$.
If $a=0$, then $h(b+c)=0 \implies b=-c$. From our first equation, we have $f(b+b)=-b \implies 2bf+b=0$. If $b\neq 0$, then $f=\frac{-1}{2} < 0$. However, as $f>0$, we must have $b=0$. Similarly, $c=0$. However, this would give $n$ as $\frac{0}{0}$ which is undefined. Therefore, we discard all solutions with $a=b=c=0$. However, as each chosen $x,y,z$ yields infinitely many (albeit similar) triangles, we still have a non-zero triangle for each choice of $x,y,z$.
If $a>0$, then we similarly can find $b>0$, $c>0$. Thus, $a+b-c>0$, $a-b+c>0$, and $-a+b+c>0$, so a triangle can be formed.
Thus, there are infinitely many $n$ such that $p=nr$.
$\blacksquare$
math154
11.07.2010 20:56
In fact if we really do have
\begin{align*}
a+b+c=x^2(a+b-c)=y^2(a-b+c)=z^2(-a+b+c),
\end{align*}then
\[\frac{a+b+c}{x^2}+\frac{a+b+c}{y^2}+\frac{a+b+c}{z^2}=(a+b-c)+(a-b+c)+(-a+b+c)=a+b+c,\]so
\[\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=1,\]which has finitely many solutions... it actually has no solutions since at least one of $x^2,y^2,z^2$ has to be less than or equal to $3$ (i.e. equal to $1$, a clear contradiction).
Here's my solution... If we let $x,y,z$ be the distances from the vertices to their respective tangency points with the incircle, then we want to find infinitely many $n$ such that solutions to \[(x+y+z)^3=n^2xyz\]exist. After playing around with this a bit we think to try $x=k(y+z)$ and then $k=yz-1$ so \[n^2=\frac{(x+y+z)^3}{xyz}=\frac{(yz)^3(y+z)^3}{(yz-1)(y+z)yz}=(yz)^2\frac{(y+z)^2}{yz-1}.\]It suffices to find some fixed value of $t$ so that \[t^2=\frac{(y+z)^2}{yz-1}\]has infinitely many integer solutions since then we would have $n=yzt$ with infinitely many values. Indeed, $t=3$ works since for \[9=t^2=\frac{(y+z)^2}{yz-1}\Longleftrightarrow y^2-7yz+(z^2+9)=0\]we can take \[y=\frac{7z+\sqrt{(7z)^2-4(z^2+9)}}{2}=\frac{7z+3\sqrt{5z^2-4}}{2},\]which has infinitely many solutions from the Pell Equation $w^2=5z^2-4$. Note that as $z$ increases, so does $y$, so $n=yzt=3yz$ does take on infinitely many values, and we are done.
fermat285
10.08.2012 07:21
Let F_n be the n-th Fibonacci Sequence. When we set \[x=3F_{2n+1}^{3},\: y=F_{2n-1},\: z=F_{2n+3},\] \[\frac{(x+y+z)^{3}}{xyz}=\frac{(3F_{2n+1}(F_{2n+1}^{2}+1))^{3}}{3F_{2n+1}^{3}(F_{2n+1}^{2}+1)}=(3(F_{2n+1}^{2}+1))^{2}.\] \[(\: \because F_{2n-1}+F_{2n+3}=3F_{2n+1}\: \: and \: \: F_{2n-1}F_{2n+3}=F_{2n+1}^{2}+1)\]
va2010
31.10.2015 18:42
The above solution is motivated by Vieta Jumping, if anyone was curious. Its quite clean: Observe that the condition rearranges to $(a+b-c)(a-b+c)(-a+b+c) | (a+b+c)^3$. Now, WLOG $a+b-c = 1$, so $c = a+b-1$. Now we need $(2a-1)(2b-1) | (2(a+b)-1)^2$, and let $2a-1 = x$ and $2b-1 = y$. Now we've got some condition like $xy | (x+y-1)^2$. If we let $x$ and $y$ be $a^2$ and $b^2$, we want to have $\frac{a^2+b^2+1}{ab}$ be an integer, which is easy through vieta jumping.
dbz
03.01.2016 15:15
Please review my solution: $\frac{p}{r}=n$ We also know that $s*r=\triangle$ $\Longrightarrow$ $\frac{p^2}{\triangle}=n$ and ${r^2}*{\triangle}=n$ $\Longrightarrow$ $n$ depends on $p,r,\triangle$ for which there are infinite choices.Hence we can find infintely many $n$.
toto1234567890
04.01.2016 16:27
Well, as I know this is IMO SL But I don't remember the year
trumpeter
05.01.2016 02:22
toto1234567890 wrote: Well, as I know this is IMO SL But I don't remember the year If you just scroll up to the top, you will see that this says "IMO Shortlist 2000, N5"...
toto1234567890
05.01.2016 12:06
trumpeter wrote: toto1234567890 wrote: Well, as I know this is IMO SL But I don't remember the year If you just scroll up to the top, you will see that this says "IMO Shortlist 2000, N5"... Oh, sorry.
the_executioner
05.01.2016 12:32
va2010 wrote: The above solution is motivated by Vieta Jumping, if anyone was curious. Its quite clean: Observe that the condition rearranges to $(a+b-c)(a-b+c)(-a+b+c) | (a+b+c)^3$. Now, WLOG $a+b-c = 1$, so $c = a+b-1$. Now we need $(2a-1)(2b-1) | (2(a+b)-1)^2$, and let $2a-1 = x$ and $2b-1 = y$. Now we've got some condition like $xy | (x+y-1)^2$. If we let $x$ and $y$ be $a^2$ and $b^2$, we want to have $\frac{a^2+b^2+1}{ab}$ be an integer, which is easy through vieta jumping. Is the condition $(a+b-c)(a-b+c)(-a+b+c) | (a+b+c)^3$ just sufficient? I am getting $\frac{(a+b+c)^3}{(a+b-c)(a-b+c)(-a+b+c)}=n^2$ which I believe is not the same thing. Also i think that by your statement we can only conclude that there are infinite values of $a,b$ but we can't say anything about $n$. If i am wrong , can you please explain the solution to me?
the_executioner
05.01.2016 14:31
dbz wrote: Please review my solution: $\frac{p}{r}=n$ We also know that $s*r=\triangle$ $\Longrightarrow$ $\frac{p^2}{\triangle}=n$ and ${r^2}*{\triangle}=n$ $\Longrightarrow$ $n$ depends on $p,r,\triangle$ for which there are infinite choices.Hence we can find infintely many $n$. You must notice that $p,r,\triangle$ are not completely independent of each other. They all depend upon the sides of the triangles and whether $\frac{p^2}{\triangle }$ is an integer or not. So your solution is wrong.
va2010
07.01.2016 03:10
the_executioner wrote: va2010 wrote: The above solution is motivated by Vieta Jumping, if anyone was curious. Its quite clean: Observe that the condition rearranges to $(a+b-c)(a-b+c)(-a+b+c) | (a+b+c)^3$. Now, WLOG $a+b-c = 1$, so $c = a+b-1$. Now we need $(2a-1)(2b-1) | (2(a+b)-1)^2$, and let $2a-1 = x$ and $2b-1 = y$. Now we've got some condition like $xy | (x+y-1)^2$. If we let $x$ and $y$ be $a^2$ and $b^2$, we want to have $\frac{a^2+b^2+1}{ab}$ be an integer, which is easy through vieta jumping. Is the condition $(a+b-c)(a-b+c)(-a+b+c) | (a+b+c)^3$ just sufficient? I am getting $\frac{(a+b+c)^3}{(a+b-c)(a-b+c)(-a+b+c)}=n^2$ which I believe is not the same thing. Also i think that by your statement we can only conclude that there are infinite values of $a,b$ but we can't say anything about $n$. If i am wrong , can you please explain the solution to me? You are right. I wasn't thinking clearly But the ending is still quite similar.
HKIS200543
06.08.2020 19:24
Claim. The Pell equation \[ x^2 - 5y^2 = -4 \]has infinitely many solutions in integers $(x_k, y_k)$, one set of which can be described by the recursive relation $(x_1, y_1) = (4,2)$ and \[ \begin{cases} x_{k+1} = 9x_k + 20y_k \\ y_{k+1} = 4x_k + 9y_k . \end{cases} \] Proof. We proceed by induction on $n$ that the recursive relation produces $(x_k, y_k)$ which satisfy the Pell equation. The base case $k=1$ is a simple computation. For the inductive step, we just expand \begin{align*} x_{k+1}^2 - 5y_{k+1}^2 &= (9x_k + 20y_k)^2 -5(4x_k + 9y_k)^2 \\ &= (81x_k^2 + 360 x_k y_k + 400y_k^2) - (80x_k^2 + 360x_ky_k + 405y_k^2) \\ &= x_k^2 - 5y_k^2 = -4. \end{align*}The recursive relation clearly produces increasing sequences $(x_k)$ and $(y_k)$, so we are done. Let $A,B,C$ be the side-lengths of the triangle and conduct a Ravi substitution such that $A = b+c, B = c+a,$ and $C = a+b$. Now set \begin{align*} a &= \frac{3y_k + \sqrt{9y_k^2 - 4(y_k^2 + 1)}}{2} = \frac{3y_k + x_k}{2} \\ b &= 3y_k - a \\ c &= (ab-1) (a+b) \end{align*}Trivially, $x_k$ and $y_k$ are the same parity by induction, so these three quantities are all integers. Obviously $s = a + b + c$. We also know that $[ABC] = sr$. Using Heron's formula, this implies $r = \sqrt{\frac{abc}{a+b+c}}$. Thus \[ n^2 = \frac{(a+b+c)^3}{abc}. \]Now, we just directly substitute to solve for $n$: \begin{align*} n^2 &= \frac{(ab)^3(a+b)^3}{ab(ab-1)(a+b)} \\ &= \frac{(ab)^2(a+b)^2}{ab-1} \\ &= \frac{(ab)^2(3y_k)^2}{y_k^2}, \end{align*}with the final equality coming from \[ ab - 1 = \frac{3y_k + x_k}{2} \cdot \frac{3y_k - x_k}{2} -1 = \frac{9y_k^2 - x_k^2 -4}{4} = y_k^2. \]Thus \[ n = 3ab = \frac{3(9y_k^2 - x_k^2)}{4} = 3y_k^2 + 3 \]As we vary $k$, this produces infinite and increasing (so distinct) $n$, so we are done.
DanjelZone
08.10.2020 20:33
Can someone check? Seems too easy Search for right angle triangles with legs $a,b$ and hypotenuse $c$. Then we know $\frac{a+b+c}{2}=\frac{2S}{a+b+c}$ which implies $\frac{(a+b+c)^2}{4}=\frac{abn}{2}$. After simplifying we get $2a^2+2b^2+2ab+\sqrt{a^2+b^2}(2a+2b)=2abn$. For some pythagorean triplet $a=3x,b=4x,c=5x$ we get $72x^2=12xn$ which implies $6x=n$ for any $x$ which implies infinite possibilities for n.
djmathman
08.10.2020 21:04
DanjelZone wrote: For some pythagorean triplet $a=3x,b=4x,c=5x$ we get $72x^2=12xn$ which implies $6x=n$ for any $x$ which implies infinite possibilities for n. The right hand side is actually $12x^2$, so you only get $n=6$ from this. (Intuitively, your solution can't work because the value of $n$ is invariant under scaling.)
math_comb01
24.03.2023 21:51
Let $S^2= (p)(p-x)(p-y)(p-z)$ by Heron's formula, we let $p-x,p-y,p-z = a,b,c$ so we get $n= \frac {p^2}{S} = \frac {(a+b+c)^{\frac{3}{2}}}{abc}$ so problem condition is equivalent to $n^2abc=(a+b+c)^3$, for simplicity let $c=k(a+b)$ then $n^2kab=(k+1)^3(a+b)^2$ now set $n=3k+3$ to get a quadratic $(k+1)(a+b)^2=9kab$and set $u=x/y$ then take its discriminant $D=k(5k-4)$ set $k=v^2$ to get a classic pell's eqn with fibbonacci as solns