a sangaku problem

alexb

Chords KN and ST are perpendicular to diameter CP of circle (O) at points Q and R. SQ intersects the circle in V. (K, S are on one side of CP, N and T on the other.) Let r be the radius of the circle (M) inscribed into the curvilinear triangle TQV. Prove that

1/r = 1/PQ + 1/QR.

In addition, it would be nice to have a short independent proof of the fact that r is independent of the radius of (O).

**Posted:** Sat Aug 12, 2006 5:02 pm

yetti

The points P, Q, R, C have to follow on the line PC in this order. Let r > q be radius of the circle (O),

$SR^{2} = PR \cdot RC = PR (2r - PR),\ \ r = \frac {SR^{2} + PR^{2}}{2PR}$

The triangle $\triangle STQ$ is isosceles and $QN \parallel ST,$ hence QN bisects the angle $\angle TQV.$ Let (M) be a circle centered on the ray (QN with radius $q = \frac {PQ \cdot QR}{PR}$ and tangent to both QT and QV at X, Y. The right triangles $\triangle QXM \sim \triangle SRQ$ are similar, $QX = \frac{MX \cdot SR}{QR} = \frac{PQ \cdot SR}{PR}.$ Using Pythagorean theorem for the right triangle $\triangle OQM,$

$OM^{2} = QM^{2} + QO^{2} = MX^2 + QX^2 + (PQ - r)^{2} =$

$= q^{2} + \frac {PQ^{2}\cdot SR^{2}}{PR^{2}} + PQ^{2} - 2r \cdot PQ + r^{2} =$

$= q^{2} + PQ^{2}\cdot \frac {SR^{2} + PR^{2}}{PR^{2}} - 2r \cdot PQ + r^{2} =$

$= q^{2} + PQ^{2}\cdot \frac {2r}{PR} - 2r \cdot PQ + r^{2} =$

$= q^{2} - 2r \cdot \frac {PQ \cdot (PR - PQ)}{PR} + r^{2} =$

$= q^{2} - 2rq + r^{2} = (r - q)^{2}$

As a result, OM = r - q and the circle (M) is internally tangent to the circle (O). Given collinear points P, Q, R, the point S on a normal to PR at R is arbitrary, i.e., radius r of (O) varies, while $q = \frac {PQ \cdot QR}{PR}$ does not.

**Posted:** Sun Aug 13, 2006 2:57 am

alexb

Impressive. I saw all the right and similar triangles there are, but tried to get from OM = r - q through a chain of applications of the Pythagorean theorem to "q = something" and could not.

Thank you.

**Posted:** Mon Aug 14, 2006 8:50 am

yetti

Starting with $OM = r - q,$ where $q = MX,$

$(r - q)^2 = OM^2 = QM^2 + QO^2 = MX^2 + QX^2 + (PQ - r)^2\ \Longleftrightarrow$

$QX^2 + PQ^2 - 2r PQ + 2rq = 0.$

Substituting $QX = \frac {MX \cdot SR}{QR},$

$q^2 \frac {SR^2}{QR^2} + 2rq - PQ (2r - PQ) = 0.$

Substituting $SR^2 = PR \cdot RC,$ $2r - PQ = QC,$

$q^2 + 2rq\ \frac{QR^2}{PR \cdot RC} - \frac{PQ \cdot QR}{PR} \cdot \frac{QC \cdot QR}{RC} = 0$

Substituting $2r QR = (PR + RC)(PR - PQ) = PR \cdot QC - RC \cdot PQ,$

$q^2 - q\ \left(\frac{PQ \cdot QR}{PR} - \frac{QC \cdot QR}{RC}\right) - \frac{PQ \cdot QR}{PR} \cdot \frac{QC \cdot QR}{RC} =...$

The two roots of the quadratic are $q = \frac {PQ \cdot QR}{PR} > 0$ and $q' = - \frac {QC \cdot QR}{RC} < 0.$ If $(M), (M')$ are directed incircle and directed excircle of the curvilinear $\triangle TQV$ in the angle $\angle TQV,$ internally and externally tangent to the directed $(O),$ then $(M)$ has the same direction as $(O)$ and $(M')$ has the opposite direction. $q > 0, q' < 0$ are their radii.

**Posted:** Sun Apr 12, 2009 2:52 pm

_________________
Carthage must be destroyed.

armpist · Riemann Hypothesis Offline Joined: 04 Sep 2005 Posts: 266

Using the two chords as bases of two triangles and completing it with the Thebault

circles, the circle in question is the common T-circle of both triangles.

The other T-circles have their centers on the diameter in known points.

The triangles incenters are on a line parallel to this diameter.

I conjecture that after applying T- formulas that inradius is a linear combination of

T-radii and incenter divides the distance between T-centers in the ratio

of ( tan X )^2, see

http://forumgeom.fau.edu/FG2002volume2/FG200223.pdf,

and after a couple of similar triangles and a lot of complicated multiplication

and division, the T-radii will be replaced by diameter segments of the big circle

and the result as required will be obtained.

I just do not have time now for it, and those SINs and COSs scare me a bit, sorry.

T.Y.
M.T.

**Posted:** Mon Apr 13, 2009 6:26 am

yetti

Rescue ? If your life depended on the answer and you tried this:

**Posted:** Wed Apr 15, 2009 9:29 pm

_________________
Carthage must be destroyed.