170. A nice "slicing" proposed problem for middle school.

by Virgil Nicula, Oct 30, 2010, 7:10 PM

Let $ABCD$ be a aquare and for $E\in (BC)$ consider $F\in (CD)$ so that $\widehat{FAE}\equiv\widehat{FAD}$ . Prove that $AE=BE+DF$ (Mihai Miculita, Oradea).

Method 1 (synthetic). Denote $G\in BC$ so that $B\in (GC)$ , $BG=FD$ and $x=m(\angle GAB)=m(\angle FAD)=m(\angle FAE)$ . Define the point $X\in AD$ so that

$A\in (XD)$ . Observe that $m(\angle EGA)=m(\angle GAX)=(90)^{\circ}-x$ and $m (\angle EAG)=m(\angle EAB)+m(\angle BAG)=\left(90^{\circ}-2x\right)+x=(90)^{\circ}-x$.

In conclusion, the triangle $EGA$ is $E$-isosceles, i.e. $EA=EG=EB+BG=BE+DF$ $\implies$ $AE=BE+DF$ .

Method 2 (trigonometric). Denote $m(\angle FAD)=m(\angle FAE)=x$ . Suppose w.l.o.g. that $AB=1$ . Thus, $m(\angle AEB)=2x$ and $AE=BE+DF$ $\iff$

$\frac {1}{\sin 2x}=\frac {1}{\tan 2x}+\tan x$ $\iff$ $\frac {1-\cos 2x}{\sin 2x}=\tan x$ $\iff$ $\frac {2\cdot\sin^2x}{2\cdot \sin x\cos x}=\frac {\sin x}{\cos x}$ .

Remark. If $E:=C$ , then $AE=BE+DF\iff \sqrt 2=1+\tan \frac {\pi}{8}\iff \tan \frac {\pi}{8}=\sqrt 2-1$ .

Method 3. $AB=1$ , $BE=m$ , $DF=n$ and $m(\angle FAD)=m(\angle FAE)=\phi$ $\implies$ $m(\angle AEB)=2\phi$ and $\tan \phi =\frac {DF}{DA}=n$ , $\tan 2\phi=\frac {BA}{BE}=\frac 1m$ .

Since $\tan 2\phi =\frac {2\tan\phi}{1-\tan^2\phi}$ obtain that $\frac 1m=\frac {2n}{1-n^2}\iff$ $n^2+2mn-1=0\iff$ $\sqrt {1+m^2}=m+n\iff$ $AE=BE+DF$ .

Remark. If $E:=C$ , then $m=1$ and $n=\tan \frac {\pi}{8}$ . Thus, $AE=BE+DF\iff \sqrt 2=1+\tan \frac {\pi}{8}\iff \tan \frac {\pi}{8}=\sqrt 2-1$ .
This post has been edited 22 times. Last edited by Virgil Nicula, Nov 22, 2015, 8:23 PM

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