Prove that the radius of the incircle of one triangle has two times the radius of the incircle of another triangle with Euclidean geometry.

Question

"Let ABCD be a square. Let CDP be an equilateral triangle inside the square. Let Q be the intersection of the line drawn through points A and P and the side BC. Let R be a point such that the triangle AQR is equilateral. Let S be the intersection of the lines QR and PC. Let T be the intersection of the lines AR and PD. Prove that the radius of the incircle of the triangle CRS is two times the radius of the incircle of the triangle ADT."

I honestly don't even know where to start this. I know how to prove it using trigonometry but it is supposed to be done with pure Euclidean geometry: similar triangles, adding points, etc. I see no connection whatsoever between the triangles CRS and ADT.

Thank you in advance for any insight.

Answer 1

Here is a possible approach w/o trigonometry. Let $\overline{AB} = 1$. Let $U$ be the point symmetric to $T$ w.r.t. $AD$, and $V$ be the point symmetric to $D$ w.r.t. $TU$.

1. By angle chasing $\triangle DTU$ and $\triangle VTU$ are equilateral, and $\triangle TVA$ is isosceles. Since $$\overline{AV} +\sqrt 3 \cdot \overline{TV} = 1,$$ therefore, we get $$\overline{TD} = \frac{\sqrt 3 -1}2.$$
2. Angle chasing shows that $\triangle CRQ$ is isosceles. Distance from $P$ to $AB$ and Thales Theorem give $BQ = 2-\sqrt 3$. Therefore $$\overline{CR} = \sqrt 3-1.$$
3. We have that $$\triangle CRS \sim \triangle RDT,$$ and, thanks to 1. and 2. we deduce that the similarity ratio is $2$. So our thesis is reached once we show that the inradii of $\triangle RDT$ and $\triangle ADT$ are equal.
4. The ratio between the areas of $\triangle RDT$ and $\triangle ATD$ is equal to $$\frac{\mathcal A_1}{\mathcal A_1} = \frac{\overline{RT}}{\overline{AT}}=\frac{1+\frac{\sqrt 3}2}{\frac{\sqrt 3}2}=\frac{\sqrt 3+2}{\sqrt 3},$$using 1. and Thales Theorem. We can use the above ratio plus $\overline{AT} = \sqrt 2(\sqrt 3-1)$ to find $$\overline{AT} = \frac{\sqrt 2}2$$ and $$\overline{RT} = \sqrt 6-\frac{3\sqrt 2}2.$$ We now have all the information to compute the ratio bewteen the perimeters of the two triangles, i.e. $$\frac{\mathcal P_2}{\mathcal P_1} = \frac{\overline{AT}+\overline{TD}+1}{\overline{RT}+\overline{TD}+\overline{RD}} =\frac{\sqrt 3}{\sqrt 3+2}.$$
5. From 4., the fact that the ratio between the inradii is equal to $$\frac{r_1}{r_2} = \frac{\mathcal A_1}{\mathcal A_2}\cdot \frac{\mathcal P_2}{\mathcal P_1}=1,$$ and recalling the observation in 3., we have our result.