An Old Sangaku Problem

Method 1:

Let $F$ be the point of tangency between $A$ and $E$ , $G$ be the point of tangency between $E$ and $B$ , $H$ be the point of tangency between $B$ and $C$ , $J$ be the point of tangency between $A$ and $C$ , $I_1$ be the center of the incircle in $\triangle AEC$ , $I_2$ be the center of the incircle in $\triangle BED$ , and $I_3$ be the center of the incircle in $\triangle BDC$ .

Also let $\theta = \angle DBC = \angle DBE$ . Then $\angle ABC = \angle ACB = 2\theta$ and $\angle BED = \angle BCD = 90° - \theta$ , which means $\angle ACE = 3\theta - 90°$ and $\angle CEA = 90° + \theta$ .

By properties of an incircle, $\angle HBI_3 = \frac{1}{2}\theta$ , since the incircle has a radius of $1$ , $HI_3 = 1$ . Then by trigonometry on $\triangle BHI_3, BH = \cot \frac{1}{2}\theta$ . By a similar argument and letting $t = \cot \frac{1}{2} \theta$ , $BG = \cot \frac{1}{2}\theta = t$ , $GE = \cot (45° - \frac{1}{2}\theta) = \frac{t + 1}{t - 1}$ , $EF = \cot (45° + \frac{1}{2}\theta) = \frac{t - 1}{t + 1}$ , $AF = AJ = \cot (90° - 2\theta) = \frac{4t(t^2 - 1)}{(t^2 - 2t - 1)(t^2 + 2t - 1)}$ , and $JC = \cot (\frac{3}{2}\theta - 45°) = \frac{-t^3 - 3t^2 + 3t + 1}{t^3 - 3t^2 - 3t + 1}$ . Since $\triangle ABC$ is an isosceles triangle, $AB = AC$ , so that:

$t + \frac{t + 1}{t - 1} + \frac{t - 1}{t + 1} = \frac{-t^3 - 3t^2 + 3t + 1}{t^3 - 3t^2 - 3t + 1}$

which simplifies to:

$(t - 3)(t^2 + 1)(t^2 + 2t - 1) = 0$

so that $t = 3$ or $t = \sqrt{2} - 1$ for $t > 0$ , but only $t = 3$ leads to a positive solution for $GE = \frac{t - 1}{t + 1}$ .

Therefore $t = 3$ , so $BC = t + \frac{t + 1}{t - 1} = 3 + \frac{3 + 1}{3 - 1} = 5$ and $AC = AB = \frac{4t(t^2 - 1)}{(t^2 - 2t - 1)(t^2 + 2t - 1)} + \frac{-t^3 - 3t^2 + 3t + 1}{t^3 - 3t^2 - 3t + 1} = \frac{4 \cdot 3(3^2 - 1)}{(3^2 - 2 \cdot 3 - 1)(3^2 + 2 \cdot 3 - 1)} + \frac{-3^3 - 3 \cdot 3^2 + 3 \cdot 3 + 1}{3^3 - 3 \cdot 3^2 - 3 \cdot 3 + 1} = \frac{125}{14}$ , so that the area by Heron's Formula is $A_{\triangle ABC} = \frac{150}{7}$ , and $150 + 7 = \boxed{157}$ .

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Method 2:

Let $a = DC = DE$ , $b = BD$ , $c = BC = BE$ , $d = AC = AB$ , and $\theta = \angle DBC = \angle DBE$ .

Since $\triangle ABC$ is an isosceles triangle, $\angle ABC = \angle ACB = 2\theta$ and $\angle BAC = 180° - 4\theta$ .

From $\triangle BDC$ , $\sin \theta = \frac{a}{c}$ , $\cos \theta = \frac{b}{c}$ , and

$a^2 + b^2 = c^2$

From the double angle formulas, $\sin 2\theta = \frac{2ab}{c^2}$ and $\cos 2\theta = \frac{b^2 - a^2}{c^2}$ and $\sin 4\theta = \frac{4ab(b^2 - a^2)}{c^4}$ .

By the law of sines on $\triangle ABC$ , $\frac{\sin (180° - 4\theta)}{c} = \frac{\sin 2\theta}{d}$ , which after substitution and rearranging simplifies to:

$d = \frac{c^3}{2(b^2 - a^2)}$

Since $A = sr$ for the area $A$ , semiperimeter $s$ , and inradius $r$ of a triangle, and the inradius of $\triangle DBC$ and $\triangle AEC$ is $r = 1$ , $A = s$ . Then for $\triangle DBC$ ,

$\frac{1}{2}ab = \frac{1}{2}(a + b + c)$

and for $\triangle AEC$ , $\frac{1}{2}d(d - c) \sin 4\theta = \frac{1}{2}(d - c + d + 2a)$ , which after substitution and rearranging simplifies to:

$2ab(d - c) = c(2a + 2d - c)$

These four equations simplify to:

$(a - 3)(a^2 - 2)(3a^2 - 6a + 2) = 0$

so that $a = 3$ , $a = \sqrt{2}$ , or $a = 1 \pm \frac{\sqrt{3}}{3}$ for $a > 0$ , but only $a = 3$ leads to a positive solution for $b = \frac{2(a - 1)}{a - 2}$ .

Therefore, $a = 3$ , $b = 4$ , $c = 5$ , and $d = \frac{125}{14}$ , and the area of $\triangle ABC$ is $A_{\triangle ABC} = \frac{1}{2}d^2 \sin (180° - 4\theta) = \frac{1}{2} d^2 \cdot \frac{4ab(b^2 - a^2)}{c^4} = \frac{150}{7}$ , and $150 + 7 = \boxed{157}$ .

For the sake of plurality, let’s see a solution that bypasses trigonometry, using exclusively Euclidean geometry tools.

First notice that $\Delta \,BCD = \Delta \,BED$ .

Let $F,\;G,\;H,\;K,\;L,\;Q,\;S,\;T$ be points of tangency and ${I_1},\;{I_2},\;{I_3}$ the centres of the three circles, $CN = h$ $\left( {CN \bot AB} \right)$ , a as seen in the picture.
Label $BF = BH = BG = a$ , $CF = CK = EG = EL = b$ , $AS = AT = c$ , $EQ = ES = x$ (groups of equal tangent segments).

$E{I_1}$ and $E{I_2}$ are angle bisectors of $\angle AEC$ and $\angle BEC$ , thus $\theta + \varphi = 90^\circ$ . Consequently, $\Delta \,E{I_2}L$ and $\Delta \,E{I_1}Q$ are similar right-angled triangles, hence

$\frac{x}{1} = \frac{1}{b} \Rightarrow x = \frac{1}{b} \ \ \ \ \ (1)$ By Pythagorean theorem on $\Delta \,BCD$ , ${\left( {a + b} \right)^2} = {\left( {a + 1} \right)^2} + {\left( {b + 1} \right)^2}$ and this leads to $ab = a + b + 1 \ \ \ \ \ (2)$ Moreover, $AB = AC \Rightarrow BS = CT \Rightarrow BS = CQ \Rightarrow a + b + x = b + 1 + 1 + b - x \Rightarrow$ $2x = b - a + 2 \ \ \ \ \ (3)$ Combining $(1)$ , $(2)$ and $(3)$ , we get to the equation ${b^3} - 5b + 2 = 0$ which can be written as $\left( {b - 2} \right)\left( {b + 1 + \sqrt 2 } \right)\left( {b + 1 - \sqrt 2 } \right) = 0$ Thus, $b = 2$ , or $b = \sqrt 2 - 1$ ( $b$ is positive). But the last solution leads to a negative value for $a$ , therefor it is rejected.

Hence, $\boxed{b = 2}$ , so $(2)$ gives $\boxed{a=3}$ and $\left( 1 \right) \Rightarrow \boxed{x = \frac{1}{2}}$ .

This means that triangles $\Delta \,BCD$ and $\Delta \,BED$ are 3-4-5 right-angled triangles .
(It is so gratifying how often we come across this triangle in Sangaku Problems!)

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Now, we move to the calculations of areas.

By Heron’ s formula, for $\Delta \,BCE$ , which has $BC = BE = 5$ and $CE = 6$ , we get ${\rm{area \ of \ }}\Delta BCE = 12$ .

Alongside, we have ${\rm{area \ of \ }}\Delta BCE = \frac{1}{2}BE \cdot h \Rightarrow 12 = \frac{1}{2} \times 5 \times h \Rightarrow h = \frac{{24}}{5}$ .

For the area of $\Delta \,AEC$ we use both formulae: $A=rs$ , where $r$ is the inradius, $s$ is the semierimeter and $A = \frac{1}{2}AE \cdot h$ .

$s = \frac{{AE + EC + CA}}{2} = \frac{{2AS + 2ES + 2CQ}}{2} = c + x + \left( {CE - x} \right) = c + 6$ .

${\rm{Area \ of \ }}\Delta \,AEC = 1 \cdot \left( {c + 6} \right) = c + 6$

On the other hand, ${\rm{area \ of \ }}\Delta \,AEC = \frac{1}{2}\left( {c + x} \right) \cdot h = \frac{1}{2}\left( {c + \frac{1}{2}} \right) \cdot \frac{{24}}{5} = \frac{{12}}{5}c + \frac{6}{5}$ .

Hence, $\frac{{12}}{5}c + \frac{6}{5} = c + 6 \Rightarrow c = \frac{{24}}{7}$ .

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Finally, ${\rm{area \ of \ }}\Delta \,ABC = \frac{1}{2}AB \cdot h = \frac{1}{2}\left( {a + b + c + x} \right) \cdot h = \frac{150}{7}$ and $150+7=\boxed{157}$

Let $\angle EBD = \angle DBC = \theta$ and $t = \tan \dfrac \theta 2$ . Then $DB = \dfrac 1t + 1 = \dfrac {1+t}t$ ; $ED = DB \times \tan \theta = \dfrac {1+t} \times \dfrac {2t}{1-t^2} = \dfrac 2{1-t}$ ; and $EC = 2 \times ED = \dfrac 4{1-t}$ . But also $EC = \dfrac 1{\tan \frac {\angle AED}2} + \dfrac 1{\tan \frac {\angle ACD}2}$ . Then we have:

$\begin{aligned} \frac 1{\tan \left(45^\circ + \frac \theta 2\right)} + \frac 1{\tan \left(\frac 32 \theta - 45^\circ \right)} & = \frac 4{1-t} \\ \frac {1-t}{1+t} + \frac {\frac {3t-t^3}{1-3t^2}+1}{\frac {3t-t^3}{1-3t^2}-1} & = \frac 4{1-t} \\ \frac {1-t}{1+t} - \frac {1+3t-3t^2-t^3}{1-3t-3t^2+t^3} & = \frac 4{1-t} \\ \frac {1-t}{1+t} - \frac {(1-t)(1+4t+t^2)}{(1+t)(1-4t+t^2)} & = \frac 4{1-t} \\ \frac {8t(t-1)}{(1+t)(1-4t+t^2)} & = \frac 4{1-t} \\ - 2t+4t^2 - 2t^3 & = 1-3t-3t^2+t^3 \\ 3t^3 - 7t^2 - t + 1 & = 0 \\ (3t-1)(t^2-2t-1) & = 0 \\ t & = \frac 13, 1 + \sqrt 2, 1 - \sqrt 2 & \small \blue{\text{For acute }\theta} \\ \implies t = \tan \frac \theta 2 & = \frac 13 \end{aligned}$

Then $BD = \dfrac 1t + 1 = 4$ and $ED=DB = \dfrac 2{1-t} = 3$ . Therefore, $EB=BC = 5$ . The area of $\triangle ABC = \dfrac 12 \times BC \times \dfrac {BC}2 \times \tan (2\theta) = \dfrac {25}4 \times \dfrac {2\tan \theta}{1-\tan^2 \theta} = \dfrac {25}4 \times \dfrac {24}7 = \dfrac {150}7$ . Therefore $a+b = 150 + 7 = \boxed{157}$ .

The (3,4,5) triangle notoriously has inradius $1$ .

Suppose that triangles $DCB$ and $DEB$ are both (3,4,5) triangles. It is a bit of easy trigonometry to deduce that $BA = CA = \tfrac{125}{14}$ , and that $AE = \tfrac{55}{14}$ . Thus triangle $AEC$ has semiperimeter $\tfrac{66}{7}$ .

On the other hand triangle $ABC$ has area $\tfrac{150}{7}$ , so that $AEC$ has area $\tfrac{66}{7}$ . Thus $AEC$ also has inradius $1$ , which means that these are the correct dimensions for this sangaku. Thus the desired answer is $150 + 7 = \boxed{157}$ .

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For a more analytic solution, suppose that the smaller acute angle in $DBC$ is $\alpha$ , so that $\angle DBC = \angle DBE = \alpha$ . Suppose also that $BC = BE = R$ . Then triangle $BDC$ has area $\tfrac12R^2\sin\alpha\cos\alpha= \tfrac14R^2\sin2\alpha$ and semiperimeter $\tfrac12R(1+\sin\alpha\cos\alpha$ , and hence both $DBC$ and $DBE$ have inradius $r_0 \; = \; \frac{R\sin2\alpha}{2(1 + \sin\alpha + \cos\alpha)}$ On the other hand $AB + AC = \tfrac12R\sec2\alpha$ , so that $AE = \tfrac12R(\sec2\alpha - 2)$ . Also $ABC$ has area $\tfrac14R^2\tan2\alpha$ , and hence $AEC$ has area $\tfrac14R^2(\tan2\alpha - 2\sin2\alpha)$ . Moreover $AEC$ has semiperimeter $\tfrac12R(\sec2\alpha - 1 + 2\sin\alpha)$ , and hence $AEC$ has inradius $r_1 \; = \; \frac{R^2\sin2\alpha(1 - 2\cos2\alpha)}{2(1 - \cos2\alpha + 2\sin\alpha\cos2\alpha)}$ Since $r_0=r_1$ we deduce that $\begin{aligned} 1 - \cos2\alpha + 2\sin\alpha\cos2\alpha & = \; (1 - 2\cos2\alpha)(1 + \sin\alpha + \cos\alpha) \\ (4\cos2\alpha - 1)\sin\alpha & = \; (1 - 2\cos2\alpha)(1 + \cos\alpha) + \cos2\alpha - 1 \end{aligned}$ Thus $(4\cos2\alpha - 1)^2\cos^2\alpha + \big[(1 - 2\cos2\alpha)(1 + \cos\alpha) + \cos2\alpha - 1\big]^2 \; = \; (4\cos2\alpha - 1)^2$ which leads us to the sextic equation $\begin{aligned} 80\cos^6\alpha + 16\cos^5\alpha - 164\cos^4\alpha - 12\cos^3\alpha + 114\cos^2\alpha - 24 & = \; 0 \\ 2(4\cos^2\alpha - 3)(2\cos^2\alpha - 1)(5\cos\alpha - 4)(\cos\alpha + 1) & = \; 0 \end{aligned}$ Since $\alpha$ is acute, we see that either $\cos\alpha = \tfrac12\sqrt{3}$ or $\cos\alpha = \tfrac{1}{\sqrt{2}}$ or $\cos\alpha = \tfrac45$ . However, in the first case we must have $\alpha = 30^\circ$ and the above equation relating $\sin\alpha$ and $\cos\alpha$ would imply that $\sin\alpha = -\tfrac12$ , which is impossible. Similarly, in the second case we would have to have $\alpha = 45^\circ$ , and the above equation would again tell us that $\sin\alpha$ had to be negative. Thus the only possiblility is that $\cos\alpha = \tfrac45$ , and so the (3,4,5) solution is the only one.

Write expressions of radius of inscribed circle for right triangle and general upper triangle in term of sides. Further, sum the areas of two right and one general triangle to large isosceles triangle. Solve the system of equations with WolframAlpha to get base of isosceles triangle as 5 and other two sides equal to 5+(55/14) each.

Area=(150/7)

Answer=157

Lots of good solutions here.

I found the best way for me to visualise the problem was to make the line EDC horizontal with the circles below ED and DC fixed. Let's call the circle below ED '1' and the circle below DC '2'.

Then imagine the third inscribed circle as a boulder being pushed along the EDC line. Because the other two circles are fixed the location of the third circle (or boulder) determines the length of AB and AC and ED and DC. With the midpoints of circle 1 and circle 2 fixed therefore the line connecting the midpoint of circle 1 to the midpoint of the boulder is parallel to the line AB. And with the information provided we can use trigonometry to work out the various angles and lengths and where the boulder needs to be located so that ABC is isosceles.