Golden Ratio In a 3x3 Square II

The following has been posted by Tran Quang Hung at the CutTheKnotMath facebook page.

In a $3\times 3\;$ let $P\;$ be the incenter of the inscribed isosceles triangle as shown below; $S\;$ and $T\;$ the two opposite corners of the middle square:

.

Golden  Ratio In a 3x3 Square II

Then $P\;$ divides $ST\;$ in the Golden Ratio: $\displaystyle\frac{PT}{PS}=\varphi.$

Assuming the side length of the small squares is $1,\;$ the two sides of the triangle are equal to $\sqrt{10}\;$ while its base is $2\sqrt{2},\;$ making its semiperimeter $p=\sqrt{10}+\sqrt{2}.\;$ Its area is obviously $S=\frac{1}{2}2\sqrt{2}\cdot 2\sqrt{2}=4.\;$ We may thus compute its inradius: $\displaystyle r=PT=\frac{S}{p}=\frac{2\sqrt{2}}{\sqrt{5}+1}.\;$ Further, $\displaystyle PS=\sqrt{2}-r=\sqrt{2}\frac{\sqrt{5}-1}{\sqrt{5}+1}\;$ so that, finally,

$\displaystyle\frac{PT}{PS}=\frac{2\sqrt{2}}{\sqrt{5}+1}\cdot \frac{1}{\sqrt{2}}\cdot \frac{\sqrt{5}+1}{\sqrt{5}-1}=\frac{2}{\sqrt{5}-1}=\frac{\sqrt{5}+1}{2}.$

Here's an alternative proof by Quang Duong. Let $D\;$ be the incenter of $\Delta ABT:$

Golden  Ratio In a 3x3 Square II, proof 2

Apply Menelaus' Theorem in $\Delta BPS\;$ with the transversal $GDT:$ $\displaystyle\frac{ST}{TP}\cdot\frac{PD}{DB}\cdot\frac{BG}{GS}=1\;$ which reduces to $\displaystyle\frac{ST}{TP}=\frac{DB}{PD}.$

Since $D\;$ is the incenter of $\Delta ABT,\;$ $\displaystyle\frac{DB}{DP}=\frac{AB+BT}{AT}.\;$ But $AB=\sqrt{10},\;$ $BT=\sqrt{2},\;$ $AT=2\sqrt{2}.\;$ Thus, $\displaystyle\frac{DB}{PD}=\frac{AB+BT}{AT}=\frac{\sqrt{5}+1}{2}\;$ and, consequently, $\displaystyle\frac{ST}{TP}=\frac{\sqrt{5}+1}{2}\;$ from which, too, $\displaystyle\frac{PT}{SP}=\frac{\sqrt{5}+1}{2}.$

Grégoire Nicollier has pointed to the shortest proof: by the Angle Bisector theorem, $AP:PT=\sqrt{10}:\sqrt{2},\;$ and the result follows from here.

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