A Sangaku: Two Unrelated Circles
What if applet does not run? 
Chords KN and ST are perpendicular to diameter CP of a circle with center 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. Q is between P and R.) Let q be the radius of the circle inscribed into the curvilinear triangle TQV. Prove that
(1)
1/q = 1/PQ + 1/QR.
Activities Contact Front page Contents Geometry
Copyright © 19962018 Alexander Bogomolny
Solution
The solution comes from an Art of Problem Solving Forum.
For (1) to hold, the radius q must satisfy
(2)
q = PQ·QR / (PQ + QR) = PQ·QR / PR.
So assume a number q is defined by (2). Denote
(3)
OM = r  q,
where r is the radius of the given circle.
What if applet does not run? 
Start by applying the Pythagorean theorem in ΔOQM:
OM^{2}  = MQ^{2} + OQ^{2} 
= q^{2}/(sinθ)^{2} + (PQ  r)^{2}  
= q^{2} + q^{2}/(tanθ)^{2} + (PQ  r)^{2}  
= q^{2} + PQ^{2}·QR^{2}/(PR·tanθ)^{2} + (PQ  r)^{2}  
= q^{2} + PQ^{2}·SR^{2}/PR^{2} + PQ^{2}  2r·PQ + r^{2}  
= q^{2} + PQ^{2}·(SR^{2} + PR^{2})/PR^{2}  2r·PQ + r^{2}. 
By the Intersecting Chords Theorem,
SR^{2} = PR·(2r  PR) = 2r·PR  PR^{2}. 
Using this we continue:
OM^{2}  = MQ^{2} + OQ^{2} 
= q^{2} + PQ^{2}·(SR^{2} + PR^{2})/PR^{2}  2r·PQ + r^{2}  
= q^{2} + PQ^{2}·2r·/PR  2r·PQ + r^{2}  
= q^{2}  2r·PQ·(PR  PQ)/PR + r^{2}  
= q^{2}  2r·PQ·QR/PR + r^{2}  
= q^{2}  2r·q + r^{2}  
= (r  q)^{2}, 
from which OM = r  q, as expected.
It is noteworthy to observe that the radius in (1) does not depend on the radius r of the given circle, and thus not on point S on the perpendicular PC at R.
Michel Cabart has observed that this problem can be solved by a similar calculation but without first assuming the formula to be true. He proceeds as follows:
We can suppose the circle is of radius 1. Say
OQ = 1  x,
OR = x + y  1,
QT^{2} = y^{2} + [1  (x + y  1)^{2}] = 2(x + y)  2xy  x^{2},
(sinθ)^{2} = y^{2}/[ 2(x + y)  2xy  x^{2}].
By hypothesis OM^{2} = (1  q)^{2} = (1  x)^{2} + q^{2}/(sinθ)^{2}.
Move (1  x)^{2} to left and divide all by x^{2}q^{2}.
[(1  q)^{2}  (1  x)^{2}]/x^{2}q^{2}  = [x  q][2  (q + x)]/ x^{2}q^{2} (first member) 
= [2(x + y)  2xy  x^{2}]/x^{2}y^{2} (2nd member) 
Replace X = 1/x, Y = 1/y, Q = 1/q. First member becomes
(t + Y) [t(2X  1) + 2X(X + Y)  2X  Y] = f(t).
Second member becomes Y[2X(X + Y)  2X  Y] = f(0).
The trinom f(t) has all its coefficients positive, because
2X  1 = (2  x)/x > 0, and
2X(X + Y)  2X  Y = QT^{2}/x^{2}y^{2}
thus its roots are negative and f(t) is growing when
A third approach is by Nathan Bowler who starts with the smaller circle:
Take coordinates such that it is the unit circle
 QV has equation xg + yh = 1.
 Q is at (1/g, 0).
 O is at (1/g, v/ug) = W/ug.
 V satisfies xg + yh = 1 and (x  1/g)^{2} + (y  v/gu)^{2} = (1  1/ug)^{2}.
Further y^{2}  2ygv/u + 2g/u  g^{2}  1 = 0 so that
QR  = y 
= gv/u  sqrt(g^{2}/u^{2}  2g/u + 1)  
= gv/u + 1  g/u  
= 1  g(1+v)/u. 
and
PQ = 1  1/ug + v/ug = 1  (1v)/ug.
So (PQ  1)(QR  1) = (1v)(1+v)/u^{2} = 1. Equivalently,
Sangaku

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Activities Contact Front page Contents Geometry
Copyright © 19962018 Alexander Bogomolny