The six sides of a
cyclic hexagon PQRSTU are PQ, QR, RS, ST, TU and UP. It is known that PQ= RS= TU and the diagonals PS, QT and RU meet at the point V. The lines PS and RT intersect at the point W.
Determine the ratio RW/WT, given that PR = 3*RT.
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. V WS
P T
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. .U
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Unfortunately the reader must visualise the lineal connections.
Let w measure the inscribed angles subtended by
sides PQ, RS, and TU. Let x, y , and z measure
the inscribed angles subtended by sides UP, QR,
and ST respectively.
From triangle RVW,
|RW| |RW| |VW| |VW|
---------- = ---------- = ---------- = ----------
sin(w+x) sin(RVW) sin(VRW) sin(w)
From triangle WVT,
|WT| |WT| |VW| |VW|
---------- = ---------- = ---------- = ----------
sin(w+z) sin(WVT) sin(VTW) sin(y)
Combining these two gives
|RW| sin(y) sin(w+x)
------ = -------- x ---------- (1)
|WT| sin(w) sin(w+z)
From triangle PRT,
3|RT| |PR| |RT| |RT|
---------- = ---------- = ---------- = ----------
sin(w+y) sin(PTR) sin(RPT) sin(w+z)
or
sin(w+y)
---------- = 3 (2)
sin(w+z)
From triangle PRV,
3|RT| |PR| |RV| |RV|
------------- = ---------- = ---------- = --------
sin(2w+y+z) sin(PVR) sin(RPV) sin(w)
From triangle RTV,
|RT| |RT| |RV| |RV|
------------- = ---------- = ---------- = --------
sin(2w+x+z) sin(RVT) sin(RTV) sin(y)
Combining these two gives
sin(y) sin(2w+y+z)
-------- x ------------- = 3 (3)
sin(w) sin(2w+x+z)
If |UP| = |QR|, then x = y and
equation (2) becomes
sin(w+x)
---------- = 3 (4)
sin(w+z)
and equation (3) becomes
sin(y)
-------- = 3 (5)
sin(w)
Plugging (4) and (5) into (1) gives
|RW|
------ = 3*3 = 9
|WT|
Now if someone can just prove |UP| = |QR| or x = y.
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Posted by Bractals
on 2007-12-10 14:01:55 |