PQR is a right angled triangle with Angle PRQ = 90 degrees. Point N is located inside the triangle in such a manner that the three angles NPQ, NQR and NRP are equal with the common measure denoted by x.
The measure of the acute angle between the medians to PR and QR is denoted by y.
Determine the value of (tan x*cot y)
WOLOG let |PR| = 2 and |QR| = 2a.
<NRQ = <PRQ - <PRN = 90 - <NRP = 90 - <NQR
Therefore, triangle QNR is a right triangle.
|NR| = 2a*sin(<NQR) = 2a*sin(x)
Let z be the measure of angle QPR with
|QR| 2a
tan(z) = ------ = ---- = a.
|PR| 2
Therefore, <NPR = <QPR - <NPR = z - x and
<PNR = 180 - (<NPR + <NRP) = 180 - z.
Applying the law of sines to triangle PNR,
2 |PR| |NR| 2a*sin(x)
-------- = ----------- = ----------- = ------------
sin(x) sin(<PNR) sin(<NPR) sin(z - x)
or
tan(z) a
tan(x) = -------------- = ---------.
a*tan(z) + 1 a^2 + 1
Let w be the measure of angle RPM with
|MR| a
tan(w) = ------ = ---.
|PR| 2
Therefore, <RMQ = <RPM + <MNQ = w + y and
tan(w) + tan(y)
2a = tan(<RMQ) = tan(w + y) = -------------------
1 - tan(w)*tan(y)
(a/2) + tan(y)
= ------------------
1 - (a/2)*tan(y)
or
3a
tan(y) = ------------.
2(a^2 + 1)
Hence,
a
---------
tan(x) a^2 + 1 2
tan(x)*cot(y) = -------- = -------------- = ---.
tan(y) 3a 3
------------
2(a^2 + 1)
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Posted by Bractals
on 2008-01-20 15:42:33 |
Solution
