A Sangaku: Two Unrelated Circles
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
Solution
Copyright © 1996-2008 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.
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So assume a number q is defined by (2). Denote  QST = QTS = θ. Then QN bisects TQV and NQV = θ. Place point M on QN at distance q/sinθ from Q. A circle with center M and radius q will touch QV in G, such that MG  QV. We wish to show that this circle is also tangent to the given one, which, for example, will follow from
where r is the radius of the given circle.
Start by applying the Pythagorean theorem in ΔOQM:
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| OM2 | = MQ2 + OQ2 |
| | = q2/(sinθ)2 + (PQ - r)2 |
| | = q2 + q2/(tanθ)2 + (PQ - r)2 |
| | = q2 + PQ2·QR2/(PR·tanθ)2 + (PQ - r)2 |
| | = q2 + PQ2·SR2/PR2 + PQ2 - 2r·PQ + r2 |
| | = q2 + PQ2·(SR2 + PR2)/PR2 - 2r·PQ + r2. |
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By the Intersecting Chords Theorem,
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SR2 = PR·(2r - PR) = 2r·PR - PR2.
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Using this we continue:
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| OM2 | = MQ2 + OQ2 |
| | = q2 + PQ2·(SR2 + PR2)/PR2 - 2r·PQ + r2 |
| | = q2 + PQ2·2r·/PR - 2r·PQ + r2 |
| | = q2 - 2r·PQ·(PR - PQ)/PR + r2 |
| | = q2 - 2r·PQ·QR/PR + r2 |
| | = q2 - 2r·q + r2 |
| | = (r - q)2, |
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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 QP = x, QR = y. Then successively
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OQ = 1 - x,
OR = x + y - 1,
QT2 = y2 + [1 - (x + y - 1)2] = 2(x + y) - 2xy - x2,
(sinθ)2 = y2/[ 2(x + y) - 2xy - x2].
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By hypothesis OM2 = (1 - q)2 = (1 - x)2 + q2/(sinθ)2.
Move (1 - x)2 to left and divide all by x2q2.
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| [(1 - q)2 - (1 - x)2]/x2q2 | = [x - q][2 - (q + x)]/ x2q2 (first member) |
| | = [2(x + y) - 2xy - x2]/x2y2 (2nd member) |
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Replace X = 1/x, Y = 1/y, Q = 1/q. First member becomes (Q - X)[2QX - (X + Q)], which by posing Q = X + Y + t becomes
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(t + Y) [t(2X - 1) + 2X(X + Y) - 2X - Y] = f(t).
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Second member becomes Y[2X(X + Y) - 2X - Y] = f(0).
The trinom f(t) has all its coefficients positive, because
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2X - 1 = (2 - x)/x > 0, and
2X(X + Y) - 2X - Y = QT2/x2y2
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thus its roots are negative and f(t) is growing when t > 0. So f(t) = f(0) iff t = 0.
A third approach is by Nathan Bowler who starts with the smaller circle:
Take coordinates such that it is the unit circle (q = 1), with Q on the x axis. Let G = (g, h), W = (u, v). Then:
- 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 y2 - 2ygv/u + 2g/u - g2 - 1 = 0 so that
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| QR | = -y |
| | = -gv/u - sqrt(g2/u2 - 2g/u + 1) |
| | = -gv/u + 1 - g/u |
| | = 1 - g(1+v)/u. |
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and
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PQ = 1 - 1/ug + v/ug = 1 - (1-v)/ug.
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So (PQ - 1)(QR - 1) = (1-v)(1+v)/u2 = 1. Equivalently, 1/PQ + 1/QR = 1 = 1/q.
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Copyright © 1996-2008 Alexander Bogomolny
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