In square $ABCD$, the length of its sides is $5$.

$E$, $F$ are two points on $AB$ and $AD$ in such a way so that $\angle ECF = 45^{\circ}$.

Find the maximum value of the perimeter of $\Delta AEF$.


Let say the $\angle ECB=\theta$ then $\angle DCF=45^{\circ}-\theta$. Now note that $AE=5(1-\tan \theta )$, $AF=5(1-\tan(45^{\circ}-\theta))$ and $EF=\sqrt{AE^2+AF^2}=5(\tan\theta + \tan(45^{\circ}-\theta))$ by elementary trig, the simplification for EF goes like this
\begin{align*} EF^2&=AE^2+AF^2=5^2((1-\tan \theta)^2)+(1-\tan (45^{\circ}-\theta))^2)\\ &=5^2(2+\tan^2 \theta + \tan^2 (45^{\circ}-\theta) -2(\tan\theta+\tan(45^{\circ}-\theta)))\\ &=5^2(2+\tan^2 \theta + \tan^2 (45^{\circ}-\theta) -2\tan45^{\circ}(1-\tan \theta \tan(45^{\circ}-\theta) )) ,\text{using}\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B}\\ &=5^2(\tan \theta + \tan(45^{\circ} - \theta))^2 \end{align*}

So perimeter of trianlge $AEF$ is always of $10$ now matter where $E$ or $F$ are on $AB$ and $AD$.

  • $\begingroup$ how EF = (AE^2 + AF^2)^1/2 ????? $\endgroup$ – Rangan Aryan Jan 7 '17 at 21:59
  • $\begingroup$ Pythagoras theorem, angle A is 90 $\endgroup$ – user1131274 Jan 7 '17 at 22:01
  • $\begingroup$ $AE+AF+EF≠10$. EF is wrong. $\endgroup$ – Takahiro Waki Jan 8 '17 at 6:05
  • $\begingroup$ @TakahiroWaki Why do you think it is wrong? Are you getting something else? I have added some explanation, hope it helps. $\endgroup$ – user1131274 Jan 8 '17 at 7:58
  • $\begingroup$ I don't know from second to third line. I think that's wrong. $\endgroup$ – Takahiro Waki Jan 8 '17 at 8:28

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