# A Modified Sangaku

### Source

### Problem

In rectangle $ABCD,$ $E\in AB,$ $F\in CD,$ $EF\parallel BC;$ $P\in AD;$ $Q=CP\cap EF.$

Each of the quadrilaterals $AEQP,$ $DPQF,$ and $BCQE$ is inscriptible, the former two are equal.

If $r$ is the common radius of the incircles of $AEQP$ and $DPQF,$ and $R$ the inradius of $BCQE,$ prove that $\displaystyle \frac{R}{r}=\varphi,$ the Golden Ratio.

### Solution

Extend $CP$ and $BA$ to the cross-point $K:

Obviously, line $PQ$ passes through the center of the rectangle $AEFD.$ That point projects to the midpoint of $BC,$ implying that it is also the midpoint of $KC.$ Thus, $BE=AK=2R$ and $AE=2r.$ The incircle of $BCQE$ serves as the incircle of $\Delta BCK$ while the incircle of $AEQP$ is the incircle of $\Delta EQK.$ We thus have the proportion:

$\displaystyle \frac{BK}{EK}=\frac{4R+2r}{2R+2r}=\frac{R}{r},$

giving an equality $\displaystyle (2R+r)r=(R+r)R,$ or, $R^2-Rr-r^2=0,$ from which $\displaystyle \frac{R}{r}=\frac{\sqrt{5}+1}{2}.$

### Extra

As the consequence of the above, it is easily seen that the configuration sports multiple occurrences of the Golden Ratio, as is illustrated below.

### Acknowledgment

The problem has been posted by Kadir Altintas at the Peru Geometrico facebook group, with a reference to Géry Huvent note Sangaku.

Several additional solutions can be found by following the link above to the Peru Geometrico.

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