Nice exercise include all levels for high school

Exercise:

Show that, the ratio of AB over BC is golden ratio if an only

if the quadrilateral of side $AB$ is square in the right figure.


Solution: Suppose that, the circle is unit circle then $r=1$

On the other hand we have the two triangles $NCM$ and $MBC$

are similar by A.A due to the following reasons

We have $\angle M=90$ (inscribed angle property), then $\angle M=\angle B$ and $\angle C$ is common angle

thus, the similar ratio with $BC=x\And BM=y$

 $\frac{NCM}{MCB}=\frac{NC}{MC}=\frac{CM}{CB}=\frac{NM}{MB}=\frac{2r}{MC}=\frac{MC}{y}=\frac{NM}{x}\Leftrightarrow M{{C}^{2}}=2ry=2y$

From Pythagoras theorem in triangle$MBC$,$M{{C}^{2}}={{x}^{2}}+{{y}^{2}}=2y$    (*)

Notice that, $\angle MKC=\angle MNC=\frac{\overset\frown{MC}}{2}$ and x$\angle KMN=\angle KCN=\frac{\overset\frown{NK}}{2}$

But $\left( KM \right)\parallel \left( AB \right)$ as $KMBA$ is rectangle, then by alternate interior

Angles property we have $\angle MKC=\angle KCN$ and $\angle KMN=\angle MNC$

Thus $\angle MNC=\angle KCN$ i.e  $\vartriangle QNC$ is isosceles triangle of vertex $Q$      (**)

Also we have $\angle NKC=\angle NMC=90$ (inscribed angle property) and $NC$ is common side

Thus $\vartriangle NKC$ and $\vartriangle NMC$ are equal by $H.R.A$ therefore, $NK=MC\And NM=KC$

In addition to that, $\angle KNA=\frac{KM+MC}{2}\And \angle MCB=\frac{KM+NK}{2}\Leftrightarrow \angle MCB=\angle KNA$
Also $AK=MB=x$ hence $\vartriangle NKA$ and $\vartriangle MBC$ are equal by $H.R.A$

therefore, $NA=BC$ and $NC=2y+AB\Leftrightarrow AB=2\left( 1-y \right)$             (***)

Now if $KMBA$ is square, then $AB=BM\Leftrightarrow x=2\left( 1-y \right)\Leftrightarrow M{{C}^{2}}={{\left( 2\left( 1-y \right) \right)}^{2}}+{{y}^{2}}=2y$

$\Rightarrow 4\left( 1-2y+{{y}^{2}} \right)+{{y}^{2}}-2y=0\Leftrightarrow 5{{y}^{2}}-10y+4=0\Leftrightarrow y=\frac{5\pm \sqrt{5}}{5}$

$\Leftrightarrow \frac{AB}{BC}=\frac{2\left( 1-y \right)}{y}=\frac{\mp \frac{2}{\sqrt{5}}}{\frac{5\pm \sqrt{5}}{5}}=\frac{\mp 10}{5\sqrt{5}\pm 5}=\frac{\mp 2}{\sqrt{5}\pm 1}\times \frac{\sqrt{5}\mp 1}{\sqrt{5}\mp 1}=\frac{\mp 2\left( \sqrt{5}\mp 1 \right)}{4}=\frac{1\mp \sqrt{5}}{2}$

But $AB>0\And BC>0$ thus $\frac{AB}{BC}=\frac{1+\sqrt{5}}{2}>0$ is only the acceptable ratio

However, if $KMBA$ is rectangle then $AB\ne BM$ as result we have on the other hand

$\vartriangle NCW$ is isosceles triangle of vertex $W$ hence, $\left( OW \right)$ is the perpendicular

bisector of $\left[ NC \right]$ this means that, $Q$ is the orthocenter of this triangle $NCW$ and

$QA=QB$ hence $O$ is the midpoint of $\left[ AB \right]$.from orthocenter property we have

$QM=\frac{NM}{3}\And NQ=\frac{2NM}{3}$

Hence, From Thales’s we have $\frac{MQ}{QN}=\frac{KQ}{QC}=\frac{KM}{NC}$ with $QN=QC\And QK=QM$ then

$\Leftrightarrow \frac{QM}{QN}=\frac{KM}{NC}\Leftrightarrow \frac{1}{2}=\frac{KM}{2}\Leftrightarrow $ hence, $AB=KM=1$ and $2y+1=2\Leftrightarrow y=0.5$

Therefore, $\frac{AB}{BC}=\frac{1}{y}=2$ impossible as $AB<NC$ therefore the accepted case for $ABKM$

must be square inorder to obtain golden ratio.                          Q.E.D
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