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a difficult sangaku

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Home » Forum » Olympiad Section » Geometry » Geometry Unsolved Problems

Moderators: High School Olympiad Moderators, darij grinberg, N.T.TUAN, orl, pbornsztein, pohoatza, yetti

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alexb
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a difficult sangaku
A problem said to be useful for other numerous sangaku

Here is a problem with no elegant solution in sight.

http://www.geocities.com/gregm112358/Hard_sudoku.pdf

Many thanks for trying.

Posted: Thu Jan 31, 2008 11:44 pm

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armpist
Riemann Hypothesis

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Dear Alexb,

Are you going to package a solution into an applet and

try to sell it later?

T.Y.
M.T.

Posted: Fri Feb 01, 2008 5:54 am

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alexb
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a difficult sangaku
solving problems and seling applets

This would depend on the solution. For example, long trigonometric derivations are not very amenable to Java implementation.

But I wonder, do you have a problem with applets being sold?

That's probably OK with you that, say, Mathforum lives off tax payer's money and the AoPS organization sells their books. I do not charge a penny of site visitors. I do not force them to buy an applet. On the other hand, I do maintain my own server and I do spend of my own time. And, between us, if I could I would spend more time at the site. The trouble is I have to make my living somehow. So, perhaps, if you relent somewhat and purchase your next book from my amazon store, you are sure to make a contribution to a good cause. This is of cause in addition to supporting the book's author and the publishing house.

I often receive requests from non-profit organizations to contribute this or that resource to a worthy cause. I always ask this question: do you fellow receive a salary from your dot org? If you do, would you please pay for the resource you request; for I do not receive a salary.

Does the above answer your question?

Best regards,
Alexander Bogomolny
http://www.cut-the-knot.org

Posted: Fri Feb 01, 2008 6:22 am

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armpist
Riemann Hypothesis

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Even the worst sweat-shops in South-East Asia
using under-aged labor at least pay those
unfortunate kids something out of their future profits.

T.Y.
M.T.

Posted: Fri Feb 01, 2008 9:16 pm

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alexb
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metaphor
the meaning is quite unclear

Well, I am not sure about your metaphor. Seems to me it's neither here nor there.

I am not a kid and no one pays me for my efforts. As to you, no one forces you to sweat out a solution and I doubt in a sweat shop (not necessarily of the worst kind) dissension would be tolerated.

Posted: Fri Feb 01, 2008 9:47 pm

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yetti
Navier-Stokes Equations

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alexb wrote:

Problem: Consider a triangle with vertices $A, B, C$ and opposite sides $a, b, c.$ Suppos that vertices $A$ and $B$ lie on a circle, with $C$ lying within the circle. Let $M$ be the midpoint of $AB$ and let $N$ be chosen on the arc $AB$ such that $MN$ is perpendicular to $AB.$ Let $R_C$ be radius of the circle inscribed in the curvilinear triangle with sides $A, B,$ and arc $AB.$ Prove that

$R_c = r + \frac {2 (MN) (s - a) (s - b)}{cs},$

where $s = \frac {_{a + b + c}}{^2}$ is the semiperimeter of $ABC$ and $r$ is radius of the incircle of $ABC.$

Extend $BC$ or $AC$ (it does not matter which one, say $BC$ ) to cut the given circle $(O)$ again at a point $P.$ The result is Thebault theorem setup for the $\triangle ABP.$ The original notation is not handy for calculations; re-label $C$ as $D$ and then $P$ as $C.$ Use the following notation:

Sides of the $\triangle ABC, \triangle ABD, \triangle ADC$ : $a = BC, b = CA, c = AB, d = AD, u = BD, v = DC,$ where $u + v = a.$
Semiperimeters: $s = \frac {_{a + b + c}}{^2}, s_b = \frac {_{u + c + d}}{^2}, s_c = \frac {_{v + b + d}}{^2},$ where $s = s_b + s_c - d.$
Areas: $\triangle, \triangle_b, \triangle_c,$ where $\triangle = \triangle_b + \triangle_c.$

Incenters: $I, I_b, I_c,$ radii of the incircles $(I), (I_b), (I_c)$ : $r, r_b, r_c.$
Tangency points of the incircles $(I), (I_b), (I_c)$ with $BC$ : $T, T_b, T_c.$

Incenters of the curvilinear $\triangle ABD, \triangle ADC$ : $J_b, J_c,$ radii of the incircles $(J_b), (J_c)$ : $\varrho_b, \varrho_c.$
Tangency points of the incircles $(J_b), (J_c)$ with $BC$ : $S_b, S_c$ and with $AD$ : $Q_b, Q_c.$

Midpoints of the segments $b = CA, c = AB$ : $M_b, M_c.$
Midpoints of the circumcircle arcs $CA, AB$ opposite to $B, C,$ respectively: $N_b, N_c.$
Angles: $\phi = \angle ADB, \psi = \angle ADC,$ where $\phi + \psi = \pi.$

In this notation, you want to show:

$(?)\ \ \frac {\varrho_b}{r_b} = 1 + \frac {2(M_cN_c) (s_b - u) (s_b - d)}{c r_b s_b},$

or (using Heron's formula)

$(?)\ \ \frac {\varrho_b}{r_b} - 1 = \frac {2(M_cN_c) (s_b - u) (s_b - d)}{c \triangle_b} = \frac {2(M_cN_c) \triangle_b}{c s_...$

By Thebault theorem, the incenters $I, J_b, J_c$ are collinear. An important step in the proof shows that the lines $S_bQ_b, S_cQ_c$ intersect at the incenter $I.$ This can be used to calculate the desired ratios $\frac {\varrho_b}{r_b}, \frac {\varrho_c}{r_c}$ (one ratio is sufficient).

$\overline{DT_b} = s_b - c$

$\overline{DS_b} = \overline{DT} + \overline{TS_b} = \overline{DB} - \overline{TB} + \overline{TS_b} = u - (s - b) + r \tan \f...$

$\frac {\varrho_b}{r_b} = \frac {\overline{DS_b}}{\overline{DT_b}} = \frac {u - (s - b) + r \tan \frac {\phi}{2}}{s_b - c}$

Using $u - (s - c) = u - (s_b + s_c - d - b) = (s_b - c) - (s_c - b),$

$\frac {\varrho_b}{r_b} - 1 = \frac {r \tan \frac {\phi}{2} - (s_c - b)}{s_b - c}$

We have to show equality of this and the above expressions for $\frac {\varrho_b}{r_b} - 1.$

$(?)\ \ \frac {r \tan \frac {\phi}{2} - (s_c - b)}{s_b - c} = \frac {2(M_cN_c) r_b}{c (s_b - c)}$

$(?)\ \ r \tan \frac {\phi}{2} = \frac {2(M_cN_c) r_b}{c} + s_c - b$

$\angle M_cAN_c \equiv \angle BAN_c = \angle BCN_c = \frac {\angle C}{2}.$ Substituting $\frac {2(M_cN_c)}{c} = \frac {M_cN_c}{M_cA} = \tan \frac {C}{2} = \frac {r}{s - c}$ and $\tan \frac {\phi}{2} = \frac {r_b}{s_b - c},$

$(?)\ \ \frac {r r_b}{s_b - c} = \frac {r r_b}{s - c} + s_c - b$

Using $(s - c) - (s_b - c) = s - s_b = s_c - d,$

$(?)\ \ \frac {r r_b (s_c - d)}{(s_b - c)(s - c)} = s_c - b$

Using $\frac {r}{s - c} = \tan \frac {C}{2} = \frac {r_c}{s_c - d},$

$(?)\ \ \tan \frac {\phi}{2} \tan \frac {\psi}{2} = \frac {r_b}{s_b - c} \cdot \frac {r_c}{s_c - b} = 1$

which is an obvious identity, because $\phi + \psi = \pi.$

_________________
Carthage must be destroyed.

Posted: Sun Feb 03, 2008 10:30 am
Last edited by yetti on Mon Feb 04, 2008 2:26 am; edited 1 time in total

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alexb
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Re: difficult sangaku
appreciate the solution

It's a pleasure getting a relevant response. Thank you.

Posted: Sun Feb 03, 2008 2:51 pm

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7 Posts • Page 1 of 1

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