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NOTE: This very difficult problem of elementary geometry has an ancient Japanese source (See “Sacred Mathematics: Japanese Temple Geometry”. Princeton University Press, 2008, by F. Hidetoshi & T. Rothman). It was given by F. Hidetoshi to the Spanish international journal “Revista de la O. I. M.” for publication. I was the only one to submit a complete solution, while two Spaniards and a Chilean gave partial solutions. I would love to see another answer but do not know if you can post it here with that precedent.

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    $\begingroup$ Not necessarily, user3491648. $\endgroup$
    – Piquito
    Apr 14 '15 at 22:11
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    $\begingroup$ As I indicated in my answer, it seems to me that there is some information missing in your problem statement. Can you please double-check that? It would also be interesting to know some more precise references. Which page of the book on Temple Geometry? Which issue of the journal? Is it problem 165 from number 35? The solution in there is only relative to the dimensions (height $h$ and half baseline $a$) of the triangle, and also some variable $d$ which I don't know where it came from. It should have stated as much in the problem statement! $\endgroup$
    – MvG
    Apr 15 '15 at 12:22
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    $\begingroup$ This PDF of the book seems to have this as Problem 3 of Chapter 6 (pg 194), except that the book includes one important addition: $\overline{CH}\perp\overline{AP}$ (in the notation of the problem as given above). $\endgroup$
    – Blue
    Apr 15 '15 at 12:50
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    $\begingroup$ I remember that the Cut the Knot website had this one on its Sangaku section. $\endgroup$
    – Sawarnik
    Apr 15 '15 at 17:55
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    $\begingroup$ @Sawarnik, are you referring to this? $\endgroup$ Apr 16 '15 at 2:11
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Let's define some coordinates.

\begin{align*} A &= (-1,0) & P &= C+a(B-C) & C_1 &= (c,r) \\ C &= (1,0) & H &= A+b(P-A) & C_2 &= (d,e) \\ B &= (0,y) &&& C_3 &= (0,f) \end{align*}

Here $C_{1,2,3}$ are the centers of the three circles, from bottom to top. The above coordinates contain $9$ variables: $a\dots f,r,x,y$. You can express the tangentialities based on these:

\begin{align*} AP,C_1: 0 &= -ac^2y+ar^2y-2acr-2acy-2ar+4cr-ay+4r \\ CH,C_1: 0 &= -abc^2y+abr^2y-2abcr+2abcy+2abr+4bcr-aby-4br-4cr+4r \\ AP,C_2: 0 &= -a^2d^2y^2+a^2r^2y^2-2a^2dey-2a^2dy^2-a^2e^2+a^2r^2-2a^2ey\\&\phantom=+4adey-a^2y^2+4ae^2-4ar^2+4aey-4e^2+4r^2 \\ CH,C_2: 0 &= -a^2b^2d^2y^2+a^2b^2r^2y^2-2a^2b^2dey+2a^2b^2dy^2-a^2b^2e^2+a^2b^2r^2\\&\phantom=+2a^2b^2ey+4ab^2dey-a^2b^2y^2+4ab^2e^2-4ab^2r^2-4ab^2ey-4abdey\\&\phantom=-4abe^2-4b^2e^2+4abr^2+4b^2r^2+4abey+8be^2-8br^2-4e^2+4r^2 \\ CB,C_2: 0 &= -d^2y^2+r^2y^2-2dey+2dy^2-e^2+r^2+2ey-y^2\\ AB,C_3: 0 &= r^2y^2-f^2+r^2+2fy-y^2\\ AP,C_3: 0 &= a^2r^2y^2-a^2f^2+a^2r^2-2a^2fy-a^2y^2+4af^2-4ar^2+4afy-4f^2+4r^2 \end{align*}

There is one more equation, which relates $X$ to these variables:

$$X^2 = a^2b^2y^2+a^2b^2-4ab^2+4ab+4b^2-8b+4$$

I obtained these equations in the following way: I homogenized the two points defining a line, and computed their cross product to represent the line joining them. I also expressed the circles in a projective way, as symmetric $3\times 3$ matrices representing the quadratic form of a conic section in homogeneous coordinates. Taking the adjunct of that matrix I obtained a description of the dual conic, in terms of tangent lines instead of incident points. So I multiplied that matrix with the vector of the line from both sides, let my CAS handle all the gory details, and pasted the result above. The final equation is simply the expanded form of $X^2=\langle C-H,C-H\rangle$.

It's also good to have one approximate solution at hand. I constructed one using Cinderella, using a small script to adjust one parameter using bisection. I obtained a configuration with

\begin{align*} a&\approx\phantom+0.60827488013287\\ b&\approx\phantom+0.5392413189553\\ c&\approx-0.199319828889346\\ d&\approx\phantom+0.284316034733135\\ e&\approx\phantom+0.803339370375978\\ f&\approx\phantom+1.443050448525532\\ r&\approx\phantom+0.327733253018358\\ X&\approx\phantom+1.451198842282192\\ y&\approx\phantom+2.25 \end{align*}

Example configuration

Now one can try to eliminate variables $a$ through $f$ and $y$ using standard elimination techniques. I used resultants, and furthermore, after each resultant computation I factorized the result, and kept only the single factor which almost vanished for my approximate example data. In the end, I obtained the solution

$$r^6 + 3r^4X^2 + r^2X^4 + 4r^4X - 2r^2X^3 + 5r^4 - 8r^2X^2 - X^4 - 4r^2X + 4X^3 + 4r^2 - 4X = 0$$

Obviously it would have ben impossible to follow this approach without massive support from a CAS, Sage in my case.

The result is not homogeneous, though, which means that it makes implicit reference to the fact that I fixed the length of edge $AC$ as $2$. The way I see it, the problem can't be answered without knowing either the length of one of the edges, or the ratio $AB/AC$. Is there some information missing in the question?

Fixing one length appears slightly inelegant. If we have to introduce an extra parameter, it would be better to choose a dimensionless ratio for this. The ratio $AB/AC$ mentioned above would lead to a somewhat longer expression, so I prefer to use $s$ to denote the ratio between height and base length of the triangle. $s$ stands for “shape” since that ratio only depends on the shape of the triangle, not its size. In the above coordinates, $s$ would be $y/2$. Using this parameter, and following the same approach outlined above, you'd end up with the condition

$$ 16s^5r^{10} - (16s^5 + 28s^3)r^8X^2 - (8s^4 + 4s^2)r^7X^3 + (4s^5 + s^3 + 12s)r^6X^4 + (12s^4 + 11s^2 + 4)r^5X^5 + (25s^3 + 15s)r^4X^6 - (4s^4 - 10s^2 - 5)r^3X^7 - (11s^3 + s)r^2X^8 - 3s^2rX^9 + s^3X^{10} = 0 $$

This condition is homogeneous in $r,X$ since their powers always add up to $10$, as is to be expected since now everything is invariant under scaling.

With right angle

It appears that the original statement of this problem had $AP\perp CH$. In my world, this translates to $\langle A-P, C-H\rangle=0$ or

$$a^2by^2 + a^2b - 4ab + 2a + 4b - 4=0$$

Adding this to my system of equations, one can solve for all variables. It turns out that all the variables are even rational numbers:

\begin{align*} a &= \frac{14}{25} & b &= \frac{1}{2} & c &= -\frac{1}{5} \\[2ex] d &= \frac{7}{25} & e &= \frac{26}{25} & f &= 2 \\[2ex] r &= \frac{2}{5} & X &= \frac{8}{5} & y &= \frac{24}{7} \end{align*}

You can verify that this satisfies all the equations stated above. So if you choose your coordinate system in such a way that the points $A$ and $C$ are at the given coordinates, then the other points will have the coordinates I just stated. It is trivial to read $X = 4r$ from this. You may notice that for $s=12/7$ the equation given above factors, and $X-4r$ is one of its factors.

If you scale everything by $175$ you even get integer coordinates:

\begin{align*} A &= (-175,0) & P &= (77,336) & C_1 &= (-35,70) \\ C &= (175,0) & H &= (-49,168) & C_2 &= (49,182) \\ B &= (0,600) &&& C_3 &= (0,350) \end{align*}

Right-angled situation with integer coordinates

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    $\begingroup$ So we conclude that Edo-period Japan secretly had electronic computers! $\endgroup$
    – GEdgar
    Apr 15 '15 at 12:05
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    $\begingroup$ @GEdgar: No, we conclude that if you can't have a computer do some (massive but stupid) thinking for you, you actually have to start thinking yourself to come up with a more reasonable approach suitable for manual solution. $\endgroup$
    – MvG
    Apr 15 '15 at 12:07
  • $\begingroup$ Interesting. I have been helped by a friend of mine but just to verify with computer my algebraic calculations at hand $\endgroup$
    – Piquito
    Apr 15 '15 at 15:24
  • $\begingroup$ By the way, the right angle you consider now is exactly as the original problem was. $\endgroup$
    – Piquito
    Apr 15 '15 at 16:13
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In this diagram, $$|\overline{AB}| = |\overline{CB}| = a \qquad |\overline{AC}| = 2b \qquad |\overline{AD}| = d \qquad |\overline{CE}| = e$$ Also, $$x = \frac{|\overline{BD}|}{|\overline{BC}|} \qquad y = \frac{|\overline{AE}|}{|\overline{AD}|} \qquad \angle AEC = 2\theta$$ (where $\angle AEC$ may or may not be a right angle). Writing $T$ for the area of $\triangle ABC$, we see that $$|\triangle ABD| = Tx \qquad |\triangle CDE| = T(1-x)(1-y) \qquad |\triangle ACE| = T(1-x)y$$


The relation $$\text{inradius}\cdot\text{perimeter} = 2\cdot\text{area}$$ implies $$\begin{align} r\;\left(\; a+ax+d \;\right) &= 2\;|\triangle ABC| = 2 T x \tag{1}\\ r\;\left(\; a (1-x) + d (1-y) + e \;\right) &= 2\;|\triangle CDE| = 2T(1-x)(1-y) \tag{2}\\ r\;\left(\; e + 2 b + d y \;\right) &= 2\;|\triangle ACE| = 2T (1-x)y \tag{3} \end{align}$$

We also have $$T^2 = b^2 ( a^2 - b^2 ) \tag{4}$$

Stewart's Theorem applied to Cevians $d$ and $e$ of $\triangle ABC$ and $\triangle ACD$, ultimately gives us that $$\begin{align} d^2 &= a^2 ( 1 - x )^2\phantom{y} + 4 b^2 x \tag{5}\\ e^2 + d^2 y ( 1 - y ) &= a^2 ( 1 - x )^2 y + 4 b^2 ( 1 - y ) \tag{6} \end{align}$$ Finally, the Law of Cosines provides this relation $$4 b^2 = d^2 y^2 + e^2 - 2 d e y \cos2\theta \tag{7}$$

Thus, we have seven equations in seven parameters: $a$, $b$, $d$, $e$, $x$, $y$, $T$. There is hope of eliminating six parameters to achieve a relation $F(r,e,\theta) = 0$.


The elimination process is tedious, and I haven't found a particularly "good" path through it. However, I'll note that we can solve $(1)$, $(2)$, $(3)$ that are linear in $a$, $b$, $T$ to get:

$$a = \frac{ex - d(1-y)(1-2x)}{(1 - x)(1 - y - x y )} \qquad b = \frac{2 e (1+x)y - e - d (1-2x)y}{2(1 - y - x y)} \tag{$\star$}$$ $$T = r\;\frac{e (1+x) + 2 d x - d (1+x) y}{2( 1 - x)( 1 - y - x y)}$$

This immediately reduces our burden a bit. Also, we can solve $(5)$ and $(6)$ for $a^2$ and $b^2$ $$a^2 =\frac{d^2 (1-y-xy+xy^2) - e^2 x}{(1 - x)^2(1 - y - x y)} \qquad b^2 = \frac{e^2 - d^2 y^2}{4(1 - y - x y)} \tag{$\star\star$}$$ which helps in quickly establishing further reductions. (For instance, one can equation $a^2$ from $(\star)$ with $a^2$ from $(\star\star)$.)

With the help of Mathematica, I was able to wade through a river of resultants to get to this relation

$$\begin{align} \phantom{\cdot} &\left(\;2 \hat{r}^3 + \hat{r}^2 \hat{e}\;(1-3\sin^2\theta) - 2 \hat{r} \hat{e}^2 \sin^2\theta + \hat{e}^3 \sin^4\theta\;\right) \\ \cdot &\left(\;2 \hat{r}^3 - \hat{r}^2 \hat{e}\;(1-3\sin^2\theta)- 2 \hat{r} \hat{e}^2 \sin^2\theta - \hat{e}^3 \sin^4\theta\;\right) = 0 \end{align}$$

where $\hat{r} := r\sin\theta$ and $\hat{e} := e\cos\theta$.

(Of course, since the factors are merely cubics, we can solve for roots $\hat{r}$ explicitly. This is left as an exercise for the reader.)

When $\angle AEC$ is a right angle, $\theta = \pi/4$ and the equation reduces to

$$(4r - e) (2r^2 - e^2)\cdot(4r + e) (2r^2 - e^2) = 0$$

Since $r$ and $e$ are non-negative (and $r \leq e/2$), we conclude $r = e/4$.

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J. Marshall Unger (Department of East Asian Languages and Literatures, Ohio State University) has a new book:

Sangaku Proofs: A Japanese Mathematician at Work (January, 2015, Cornell East Asia)

Elsewhere, Unger has discussed other Sangaku problems, including this one. Unger analyzes Kitagawa's proof, noting certain unjustified leaps in the reasoning. Then he gives a new elementary proof of his own. (Cut-The-Knot LINK for Unger's solution.)

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  • $\begingroup$ Following the link to Unger's solution, he starts by stating that $BC=CD$ which would be $AC=CP$ in the notation used here. The question gives no indication to this effect, and the illustration there contradicts such an assumption. This PDF however has the same right angle, on page 195. So perhaps that's indeed the bit of information I had been missing. $\endgroup$
    – MvG
    Apr 15 '15 at 15:01
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    $\begingroup$ OK, now we are coming to understand what the discussion is. The original problem in Japan (1806) had a right angle specified at $H$. But Luis claims to have solved it without that assumption? And MvG claims it cannot be solved without that assumption... $\endgroup$
    – GEdgar
    Apr 15 '15 at 15:19
  • $\begingroup$ All I want to say about is my algebraic calculations were correct $\endgroup$
    – Piquito
    Apr 19 '15 at 14:38
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Oh! You can see my solution here: http://www.oei.es/oim/revistaoim/numero35/165.pdf Clearly, If this solution is correct which I hope and believe, one has F(X, r) = (X -4r)^2 where F(X, r) = 0 is my answer. But I did not even think about this desirable simplification (work to go to F was quite hard).

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  • $\begingroup$ I have some trouble following you. In the link you post, where does it say $F(X,r)=(X-4r)^2$? Or is that not what you're referring to when you write “this solution”? You know that $(X-4r)^2=0$ is equivalent to $X=4r$, right? Is your work, too, based on an assumed right angle at $H$, or is it trying to be more general? In the latter case, I'd be really surprised by this result. In the former case, I wonder why the illustration in your question didn't include that right angle. $\endgroup$
    – MvG
    Apr 15 '15 at 19:12
  • $\begingroup$ The problem I had resolved was without the restriction of the original problem with the perpendicularity but so was the statement cited in the Journal in which it was previously warned of this. Concerning the square of my assumption (perhaps light) was because my answer was also quadratic (not simplified because doing so was very laborious). Given the surprising (for me) simplicity of the answer, I was perhaps wrong though frankly I do not think so.Regards.(my english is bad I know). $\endgroup$
    – Piquito
    Apr 15 '15 at 23:53
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Given the isosceles triangle $ABC$ with unit legs $|AC|=|BC|=1$ and the base $|AB|=c\in(0,2)$, $|AD|=v$, $|BE|=u$, $|BD|=t$, inscribed circles of $\triangle ADC$, $\triangle ABE$ and $\triangle BDE$ has the same radius $r_c$.

Obviously, the value of $c$ uniquely defines all the other parameters of this sangaku configuration, that is, the corresponding values of $t,v,u$ and $r_c$, but to find an explicit expressions for $t(c),v(c),u(c)$ and $r_c(c)$ is not an easy task.

However, $v,u,r_c$ can be easily expressed in terms of $c$ and $t$:

\begin{align} v^2&=t^2+c^2(1-t) \tag{1}\label{1} ,\\ 4u^2 &= t c(2 +c) \tag{2}\label{2} ,\\ r_c &= \frac{c(1-t)\sqrt{4-c^2}}{2\Big(2-t+\sqrt{t^2+c^2-t\,c^2}\Big)} \tag{3}\label{3} . \end{align}

Using the two other questions, Is there a way to reduce a specific quintic to cubic? and Approximating function for the root of quintic polynomial, the relationship between $r$ and $u$ can be represented by a quarter of a nice $\infty$-shaped curve:

enter image description here

Interestingly, the solution $u=4r_c$ to the original sangaku construction, when $\angle DEB=90^\circ$ and hence $c=t=0.56$, is not unique. There is another configuration, when $u=4r_c$:

\begin{align} c & \approx 0.13935311959635 ,\\ t & \approx 0.38561654555749 ,\\ u& \approx 0.16953033464425 ,\\ r_c&\approx 0.04238258366106 ,\\ \angle DEB&\approx 128^\circ . \end{align}

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