Diamond Sangaku - What a Gem!

Geometry Level 3

In the convex pentagon $A_1A_2B_2CB_1$ ,

$|\overline{A_1B_1}| = |\overline{A_1A_2}| = |\overline{A_2B_2}|$
$\overline{A_1A_3} \perp \overline{A_1B_1}$ and $\overline{A_2A_3} \perp \overline{A_2B_2}$
The isosceles right triangle $\Delta A_1A_2A_3$ contains the incircle of radius $R_1$
The concave pentagon contains two incircles both of maximum radius $R_2$ .
The isosceles right triangle $B_1B_2C$ contains the incircle of radius $R_3$
$\overline{B_1B_2}$ bisect both right angles $\angle A_1B_1 C$ and $\angle A_2B_2 C$

Which of the following is larger? $(R_1 + R_2)$ or $R_3$ ?

(R_1 + R_2)

R_3

Both are the same. Not enough information.

1 solution

Brian Moehring
Jul 26, 2018

Let $\overleftrightarrow{A_1A_3}$ and $\overleftrightarrow{A_2A_3}$ intersect $\overline{B_1B_2}$ in points $P_1$ and $P_2$ respectively.

Then $\triangle A_1A_2A_3, \triangle B_1P_1A_1, \triangle B_2P_2A_2, \triangle B_1B_2C$ are all isosceles right triangles with incircles of radii $R_1, R_2, R_2, R_3$ , respectively. Then by similarity, $\frac{R_1}{A_1A_2} = \frac{R_2}{B_1P_1} = \frac{R_3}{B_1B_2}$

If we let $x := A_1A_2 = A_1B_1 = A_2B_2$ , then $B_1P_1 = x\sqrt{2}$ $B_1B_2 = x + 2x\frac{\sqrt{2}}{2} = x(1+\sqrt{2})$

Therefore $\begin{aligned} R_1+R_2 &= \frac{R_1}{A_1A_2}(A_1A_2) + \frac{R_2}{B_1P_1}(B_1P_1) \\ &= \frac{R_3}{B_1B_2}\cdot x + \frac{R_3}{B_1B_2}\cdot(x\sqrt{2}) \\ &= \frac{R_3}{B_1B_2}(x(1+\sqrt{2})) \\ &= \frac{R_3}{B_1B_2}(B_1B_2) \\ &= R_3 \end{aligned}$

That is, $R_1+R_2$ and $R_3$ have $\boxed{\text{the same value}}$ .

Diamond Sangaku - What a Gem!

1 solution

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