Let $n\ge 1$ be an integer and consider the sum $$x=\sum_{k\ge 0} \dbinom{n}{2k} 2^{n-2k}3^k=\dbinom{n}{0}2^n+\dbinom{n}{2}2^{n-2}\cdot{}3+\dbinom{n}{4}2^{n-k}\cdot{}3^2 + \cdots{}.$$Show that $2x-1,2x,2x+1$ form the sides of a triangle whose area and inradius are also integers.
Problem
Source: 2017 India National Olympiad
Tags: binomial coefficients, binomial theorem, geometry, algebra, Integers, Heron's formula
Ghoshadi
16.01.2017 10:17
Check out at post inmo 2017 forum
tarzanjunior
16.01.2017 10:28
Here's the link
PSY-Math
16.01.2017 10:38
Ghoshadi wrote: Check out at post inmo 2017 forum Didn't understand What do you mean by check out?
Ghoshadi
16.01.2017 10:42
I mean, if you or anyone(watching the problem here,) want to discuss, then he/she is invited to that forum...tarzanjunior already gave a link...so...
toto1234567890
19.01.2017 18:34
Just $ x=\frac{(2+\sqrt{3})^n+(2-\sqrt{3})^n}{2} $ kills it
Tintarn
20.01.2017 01:47
This particular problem was also discussed here.
roben
12.04.2017 10:20
................
DynamoBlaze
03.10.2018 12:45
Can someone post a proper solution?
Tintarn
03.10.2018 14:32
DynamoBlaze wrote: Can someone post a proper solution? Huh? I have given a complete solution in the thread I linked above, a long time ago?!?
DynamoBlaze
03.10.2018 16:45
Tintarn wrote: DynamoBlaze wrote: Can someone post a proper solution? Huh? I have given a complete solution in the thread I linked above, a long time ago?!? Yes, and it is very inconvenient to find. Besides, this is in the Olympiad Collection, so it should have a solution here.
Krish230905
01.09.2020 16:56
Using Binomial Expansion, We find that $x = \frac{(2+ \sqrt3)^{n}+ (2- \sqrt3)^{n}}{2}$ For the results to be proved, we need $x^2=3k^2+1$,Where $k$ is a integer. $x^2=\frac{(2+ \sqrt3)^{2n}+ (2- \sqrt3)^{2n}+2}{4}$ For $3k^2+1 = \frac{(2+ \sqrt3)^{2n}+ (2- \sqrt3)^{2n}+2}{4}$,We need, $3k^2 = \left(\frac{(2+ \sqrt3)^{n} - (2- \sqrt3)^{n}}{2}\right)^2$,or, $\sqrt 3k = \left(\frac{(2+ \sqrt3)^{n} - (2- \sqrt3)^{n}}{2}\right)$, Which is trivial.
MatBoy-123
05.04.2021 06:53
Its a Pell's Equation..
Mogmog8
09.06.2022 20:46
Notice $$x=\frac{1}{2}\left(\sum_{i=1}^n{\binom{n}{i}2^i\left(\sqrt{3}\right)^{n-i}}+\sum_{i=1}^n{\binom{n}{i}2^i\left(-\sqrt{3}\right)^{n-i}}\right)=\frac{1}{2}\left((2+\sqrt{3})^n+(2-\sqrt{3})^n\right).$$Hence, $x$ is a solution to the pell equation $x^2-3y^2=1$ since $(2,1)$ is the fundamental solution. Therefore, $\tfrac{1}{3}(x^2-1)=y^2.$ By Heron, we see $$A=\sqrt{3x(x-1)(x+1)x}=3x\sqrt{\tfrac{1}{3}(x^2-1)}=3xy$$and $r=\frac{A}{s}=y,$ where $A,s,r$ are the area, semiperimeter, and inradius, respectively. $\square$