Suppose to have a triangle ABC such that the angle $\angle B = 2\angle C$ and the length $BC=\gamma$. I want to determine the angle $\angle C$ such that the sum of the squares of the heights from vertices $B$ and $C$ is maximal and evaluate the value.

My attempt is to use standard trig to get (call $BH$ and $CK$ the heights and $x:=\angle C$):

$BH = \gamma \sin(x)$

$CK = \gamma \sin(2x)$

Then write $$f(x)=\gamma^2\sin^2(x) + \gamma^2\sin^2(2x)$$ and I wanna maximize this. I evaluate the derivative but I get stuck. How can I proceed?


1 Answer 1


You can use $$\sin^2(2x)=(2\sin x\cos x)^2=4\sin^2 x(1-\sin^2 x)$$ Then you get a quadratic in $\sin^2 x$, for which it would be easy to find the maximum.

  • $\begingroup$ I would be interested also if the rest is correct $\endgroup$ Commented Jul 5, 2022 at 16:21
  • $\begingroup$ It looks fine to me. $\endgroup$
    – Andrei
    Commented Jul 5, 2022 at 17:04
  • $\begingroup$ Thanks thanks!! $\endgroup$ Commented Jul 6, 2022 at 12:24

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