Prove that if the sides $a,b,c$ of a triangle satisfy $a^2+b^2=kc^2$, then $k>\frac {1}{2}$.

Question

Prove that if the sides $a,b,c$ of a triangle satisfy $a^2+b^2=kc^2$, then $k>\frac {1}{2}$.

Source : CRUX (Page Number $1$;Question Number $74$)

Obviously, for $k=1$, a right angled triangle exists. I tried assuming $k<\frac {1}{2}$ and finding some contradiction, but all I got was $a \leq b, $ which seems pretty alright to me. I strongly suspect that there has to be some elegant proof by contradiction for this problem.

Can anyone provide a guideline as to what should be done ?

Any help would me gratefully acknowledged :) .

Answer 1

By triangle inequality we can write:

$$c<a+b \to c^2<a^2+b^2+2ab$$

Multiply both sides by $k$ and use $kc^2=a^2+b^2$:

$$kc^2<k(a^2+b^2)+(2ab)k \to (1-k)(a^2+b^2)<(2ab)k$$

If $k \le 1/2$ then $1-k \ge 1/2$, so we have:

$$\frac{a^2+b^2}{2}\le (1-k)(a^2+b^2)<(2ab)k \le ab$$

what give us

$$\frac{a^2+b^2}{2}< ab \to (a-b)^2<0$$

what is a contradiction.

Answer 2

HINT:

$$c^2=a^2+b^2-2ab\cos C$$

$$(a^2+b^2)(1-k)=2abk\cos C\le2abk\le(a^2+b^2)k$$