I created a Sangaku-style geometry problem involving an equilateral triangle and three circles. Can you solve it without a computer?

Question

Inspired by this difficult Sangaku problem, I created the following Sangaku-style problem of my own.

In equilateral $\triangle ABC$, $D$ is on $AB$, $E$ is on $AC$, and the incircles of $\triangle ADE$, $\triangle DBE$ and $\triangle EBC$ have equal radii. Prove that $BD=DE$.

I have a solution that requires a computer. I am looking for a solution that does not require a computer.

My solution

Assume the side length of $\triangle ABC$ is $1$.

Let
$x=BD$
$y=DE$
$\theta=\angle BDE$

Use the sine rule and cosine rule to express $\theta$, $AE$ and $BE$ in terms of $x$ and $y$:
$\theta=\arcsin{\left(\frac{\sqrt3 (1-x)}{2y}\right)}+\frac{\pi}{3}$
$AE=\sqrt{(1-x)^2+y^2+2y(1-x)\cos{\theta}}$
$BE=\sqrt{x^2+y^2-2xy\cos{\theta}}$

The inradius of a triangle is $\frac{2\times \text{area}}{\text{perimeter}}$.

$r_1=$ inradius of $\triangle ADE=\dfrac{y(1-x)\sin{\theta}}{1-x+y+AE}$

$r_2=$ inradius of $\triangle DBE=\dfrac{xy\sin{\theta}}{x+y+BE}$

$r_3=$ inradius of $\triangle EBC=\dfrac{(1-AE)\frac{\sqrt3}{2}}{2-AE+BE}$

Setting $r_1=r_2$ and $r_1=r_3$ gives two equations in $x$ and $y$. Desmos shows they are satisfied only when $x=y$, both approximately $0.5502282106$.

Answer 1

Here is a solution from one of my coworkers.

Assume $AD=DE$ and $OF=KG=r$. We will prove that $HL=r$.

Define $\alpha=\angle{DAE}$. Simple angle chasing gives the other angles shown in the diagram.

$$AT=CO+OE+EK$$ $$\frac1r(OE+CO)=\frac1r(AT-EK)$$ $$\cot{(60^0-\alpha)}+\sqrt3=\cot{\left(30^0-\frac{\alpha}{2}\right)}-\cot{\left(30^0+\frac{\alpha}{2}\right)}$$ $$\frac{\cos{\alpha+\sqrt3 \sin{\alpha}}}{\sqrt3 \cos{\alpha}-\sin{\alpha}}+\sqrt3=\frac{\sin{(30^0+\alpha/2)\cos{(30^0-\alpha/2)-\cos{(30^0+\alpha/2)}\sin{(30^0-\alpha/2)}}}}{\sin{(30^0-\alpha/2)\sin{(30^0+\alpha/2)}}}$$ $$\frac{\cos{\alpha}+\sqrt3 \sin{\alpha}}{\sqrt3\cos{\alpha}-\sin{\alpha}}+\sqrt3=\frac{2\sin{\alpha}}{\cos{\alpha}-\cos{60^0}}$$ $$2\cos^2{\alpha}-\cos{\alpha}=\sqrt3 \sin{\alpha}\cos{\alpha}-\sin^2{\alpha}$$ $$\cos^2{\alpha}-\cos{\alpha}+1=\sqrt3\sin{\alpha}\cos{\alpha}$$ $$(\cos^2{\alpha}-\cos{\alpha}+1)^2=3(1-\cos^2{\alpha})\cos^2{\alpha}$$ $$(2\cos{\alpha}-1)(2\cos^3{\alpha}-1)=0$$ Obviously, $\cos{\alpha}\ne1/2$.

Let $m=2^{1/3}$.

$$\cos{\alpha}=\frac1m=\frac{1-\tan^2{(\alpha/2)}}{1+\tan^2{(\alpha/2)}}$$ $$\tan{\alpha}=\sqrt{m^2-1}$$ $$\tan{\frac{\alpha}{2}}=\sqrt{\frac{m-1}{m+1}}$$ $$\begin{align} HL&=DE\cos{\alpha}\tan{\frac{\alpha}{2}}\\ &=(DM+ME)\cos{\alpha}\tan{\frac{\alpha}{2}}\\ &=r(\cot{\alpha}+\cot{(60^0-\alpha)})\cos\alpha\tan{\frac{\alpha}{2}}\\ &=r\left(\frac{1}{\sqrt{m^2-1}}+\frac{1+\sqrt3\sqrt{m^2-1}}{\sqrt3-\sqrt{m^2-1}}\right)\frac1m\sqrt{\frac{m-1}{m+1}}\\ &=\frac{\sqrt3mr}{(m+1)(\sqrt3-\sqrt{m^2-1})}\\ &=\frac{\sqrt3mr}{\sqrt3m+\sqrt3-(m+1)\sqrt{m^2-1}}\\ &=\frac{\sqrt3mr}{\sqrt3m+\sqrt3-\sqrt{\color{red}{(m^2-1)(m+1)^2}}}\end{align}$$

$$\color{red}{(m^2-1)(m+1)^2}=(m+2)(m^3-2)+3=0+3=3$$

$$\therefore HL=r$$

Answer 2

The following does not use a computer, except for GeoGebra, and argues not from equal inradii to $BD=DE$, but rather the converse: if $BDE$ is a certain constructible isosceles triangle, then the inradii are equal.

In equilateral triangle $ABC$, with side $AB=1$, construct $\angle CBE=22.5^o$ and $\angle AED=45^o$. Since $\angle BEC=180^o-60^o-22.5^o=97.5^o$, then $\angle DEB=\angle EBD=37.5^o$ and $BD=DE=x$ in triangle $BDE$.

Since the inradius is twice the area divided by the perimeter, in triangle $ADE$,$$FJ=\frac{x(1-x)\sin 75^o}{x+(1-x)+AE}$$Relying on GeoGebra, that $x\approx\frac {11}{20}=.55$, then by the law of cosines$$AE=\sqrt{x^2+(1-x)^2-2x(1-x)\cos75^o}\approx .614$$and$$FJ\approx\frac{.239}{1.614}=.148$$ Next, in triangle $DBE$, since $\angle BDE=105^o$, then $$GK=\frac{x\sin 37.5^o\cdot BE}{2x+BE}$$and$$BE=\sqrt {2x^2-2x^2\cos 105^o}\approx.873$$so that$$GK\approx\frac{.292}{1.973}=.148$$Finally, in triangle BEC, again$$HL=\frac{BC\cdot BE\sin 22.5^o}{BE+EC+BC}=\frac{.873\cdot\sin 22.5^o}{.873+(1-.614)+1}\approx\frac{.334}{2.259}=.148$$Thus it seems possible, with ruler and compass, in an equilateral triangle to construct three triangles, the middle one isosceles, such that their incircles are equal.

Triangle $ADE$ is oddly reminiscent of the first Pythagorean triple, having its angles rather than its sides in the $3-4-5$ ratio. Moreover, all angles are integral multiples of $7.5^o$, as indicated in the figure below.

Answer 3

Here is a response to OP's Bonus Question.

Let $ABC$ be a $30^o-60^o-90^o$ right triangle. Construct $DE$ perpendicular bisector of $AB$, and join $EB$.

Since, by SAS, $\triangle ADE\cong\triangle DBE$, then$$\frac{FH}{GI}=\frac{\triangle ADE}{\triangle DBE}=\frac{AE+ED+DA}{BE+ED+DB}=\frac{1}{1}$$[More briefly: Congruent triangles have equal incircles.]

Next, since $\angle EBD=\angle DAE=30^o$, then $\angle CBE=\angle EBD$, and by AAS, $\triangle DBE\cong\triangle CBE$, and again$$\frac{GI}{JK}=\frac{\triangle DBE}{\triangle EBC}=\frac{BE+ED+DB}{BE+EC+CB}=\frac{1}{1}$$

Therefore, a $30^o-60^o-90^o$ right triangle is divisible into three equal triangles having equal inradii, and$$\frac{AB}{BC}=\frac {2}{1}$$Is this the only triangle to fulfill the conditions? It seems clear that since equal triangles under the same height must have equal bases, then triangles $ADE$, $DBE$ will have equal perimeters, making $FH=GI$, only if $ED\perp AB$ so that $AE=BE$.

Likewise, equal triangles $DBE$, $CBE$ on the same base $BE$ have equal perimeters, making $GI=JK$, only if the triangles are congruent (if two equal triangles on the same base are non-congruent, the more nearly isosceles triangle has a greater incircle), i.e. only if $\angle ECB=\angle BDE=90^o$, and hence $\angle CBA=2\angle EBA$. Therefore, since $\angle CBA+\angle CAB=90^o$, then $\angle CBA=60^o$ and $\angle CAB=30^o$.

Finally, and by the way, as in the solution for equilateral $ABC$, here too we find $\angle AED=2\angle CBE$.