Question

Number theory / Geometry question.

Answer 1

The perimeter of the triangle is \(240\).Find the number of possible distinct triangles.the sides of triangles are in integers.

Answer 2

\[
c^2=a^2+b^2-2ab\cos(C)
\]
So \(\cos(C)\) is an integer?

Answer 3

No actually \(\cos(C)\in \mathbb{Q}\land \cos(C)\not\in \mathbb{R} \setminus \mathbb{Q}\).

Answer 4

\(a+b+c=240\)

from triangle inequality
\(c\lt a+b \implies 2c\lt a+b+c\implies c \lt 120\)
so each side must be less than \(120\) (not considering deranged triangles)

Answer 5

\[
c=240-a-b\\
c^2=57600-480(a+b)+(a+b)^2\\
c^2=57600-480a-480b+a^2+2ab+b^2\\
57600-480a-480b+a^2+2ab+b^2=a^2+b^2-2ab\cos(C)\\
57600-480a-480b+2ab=-2ab\cos(C)\\
\]
\[
0<C<\pi\\
\cos\left(\frac{2\pi}{3,4,6}\right)=-\frac{1}{2},0,\frac{1}{2}
\]
\[
\cos\left(\frac{2\pi}{3}\right)=-\frac{1}{2}\\
57600-480a-480b+2ab=ab\\
57600-480a-480b+ab=0
\]
\[
\cos\left(\frac{\pi}{2}\right)=0\\
c^2=a^2+b^2\\
57600-480a-480b+2ab=0\\
\]
\[
\cos\left(\frac{2\pi}{3}\right)=-\frac{1}{2}\\
57600-480a-480b+2ab=-ab\\
57600-480a-480b+3ab=0
\]
Don't know how to solve the equation. The only rational values of \(\cos(\theta)=0,\dfrac{1}{2},-\dfrac{1}{2},1,-1\) as mentioned in here.

Answer 6

Actually no. The argument to cosine need not be multiples of pi.

Answer 7

yeah there are infinitely many rational values on cosine graph

Answer 8

cos(x) = a/b
can be solved for all rational numbers a/b such that -1<= a/b <=1

Answer 9

The first obvious triangle is 80-80-80.

Answer 10

Can a,b,c be any integers such that \(a+b+c=240\) and \(c\leq a+b\)? So\[
\cos(C)=\frac{c^2-a^2-b^2}{2ab}\]

Answer 11

*\(c<a+b\)

Answer 12

So as long as \(-1\leq\dfrac{c^2-a^2-b^2}{2ab}\leq 1\), the equation has solutions in integers?

Answer 13

\[
\begin{align*}
\frac{c^2-a^2-b^2}{2ab}&=\frac{(240-a-b)^2-a^2-b^2}{2ab}\\
&=\frac{57600-480a-480b+2ab}{2ab}\\
&=\frac{28800}{ab}-\frac{240}{a}-\frac{240}{b}+1
\end{align*}
\]
\[
\begin{align*}
\frac{28800}{ab}-\frac{240}{a}-\frac{240}{b}+1&\leq1\\
\frac{28800}{ab}-\frac{240}{a}-\frac{240}{b}&\leq0\\
28800-240b-240a&\leq0\\
120-a-b&\leq0\\
120&\leq a+b\leq 239
\end{align*}
\]
\[
\begin{align*}
\frac{28800}{ab}-\frac{240}{a}-\frac{240}{b}+1&\geq -1\\
\frac{28800}{ab}-\frac{240}{a}-\frac{240}{b}&\geq -2\\
\frac{120}{a}+\frac{120}{b}-\frac{14400}{ab}&\leq1\\
120b+120a-14400&\leq ab\\
120b+120b-14400-ab&\leq 0\\
ab-120a-120b+14400&\geq 0\\
(a-120)(b-120)&\geq 0\\
a\geq 120 \land b\geq 120
\end{align*}
\]

Answer 14

I am very tired now so please check my work.

Answer 15

last line could be \(a\le 120 \land b\le 120\) also right

Answer 16

It cannot be the case that \(a\geq 120 \land b\geq 120\) as \(c>0\)/

Answer 17

Yes looks good to me, I think we end up solving below equation in positive integers :
\[a+b+c=240\\~\\a\lt 120,~b\lt 120,~c\lt 120\]

Answer 18

Can \(a=120\) or \(b=120\)?

Answer 19

not possible because, that gives a deranged triangle
a = 120 = b+c

Answer 20

What is a deranged triangle?

Answer 21

when the vertices of triangle are collinear..

Answer 22

Couldn't it be the case that a=120, b=119 and c=1?