American Mathematical Monthly Problem 11393 by Cosmin Pohoata, Buharest, Romania.
In triangle ABC,letM andQbe points on the segmentAB,letN andRbe points on the segment CA,and let P and S be points on the segment BC. Denote
m= |BM|
|M A|, n= |CN|
|N A|, p= |CP|
|P B|, q= |AQ|
|QB|, r = |AR|
|RC|, s= |BS|
|SC|. Prove that the lines M N, P Qand RS are concurrent if and only ifmpr+nqs+mq+ nr+ps = 1.
Solution by Darij Grinberg.
First, we reformulate (and, to some extent, generalize) the problem using directed lengths1:
Problem 11393’. In triangle ABC, let M and Q be points on the line AB, let N and R be points on the lineCA, and let P and S be points on the lineBC. Denote
m= BM
M A, n = CN
N A, p= CP
P B, q= AQ
QB, r = AR
RC, s= BS SC. Prove that the lines M N, P QandRS are concurrent2 if and only if mpr+
nqs+mq+nr+ps= 1.
Solution. LetD=M N ∩BC, E =P Q∩CA and X =M N ∩P Q.The Menelaos theorem for triangle ABC and the collinear pointsD, N, M on its sidelinesBC, CA, AB yields
BD DC · CN
N A · AM
M B =−1.
Since AM M B = 1
m (as immediately follows from BM
M A =m) and CN
N A =n, this becomes BD
DC ·n· 1
m =−1, so that BD
DC =−m
n. Similarly, CE
EA =−p
q. Thus, EA CE =−q
p, hence CA
CE = CE+EA
CE = 1 + EA
CE = 1− q
p = p−q
p , yielding CE= p p−qCA.
On the other hand, CN
N A = n leads to N A CN = 1
n, therefore CA
CN = CN +N A
CN =
1 + N A
CN = 1 + 1
n = n+ 1
n , henceCN = n
n+ 1CA. Thus, EN
CE = CN −CE
CE = CN
CE −1 = n n+ 1CA
p p−qCA
−1 = n(p−q)
(n+ 1)p−1 = − p+nq (n+ 1)p.
1We will denote the directed length of any segmentXY byXY .Of course, this directed length is well-defined only if the pointsX andY lie on some directed line.
2”Concurrent” means ”concurrent or parallel” here (parallel lines are said to concur at an infinite point).
1
Similarly,
DP
CD =− n+pm (p+ 1)n. Also, BM
M A =m yields BA
M A = BM +M A
M A = BM
M A + 1 = m+ 1, what rewrites as AB
AM =m+ 1. Thus,AM = 1
m+ 1AB. Similarly, BQ= 1
q+ 1BA, so that M Q=AQ−AM =AB+BQ−AM =AB+ 1
q+ 1 BA
|{z}
=−AB
− 1
m+ 1AB= mq−1
(q+ 1) (m+ 1)AB.
Consequently,
AM M Q =
1 m+ 1AB mq−1
(q+ 1) (m+ 1)AB
= q+ 1 mq−1.
The Menelaos theorem for triangle CP E and the collinear points X, N, D on its sidelinesP E, EC, CP yields3
P X XE ·EN
N C · CD
DP =−1, thus P X
XE =−N C EN ·DP
CD. (11393.1)
Then, the Menelaos theorem for triangle AEQ and the collinear points X, M, N on its sidelines EQ, QA, AE yields
QX XE ·EN
N A · AM
M Q =−1, thus QX
XE =−N A EN ·M Q
AM. (11393.2)
Now, P X
QX = P X XEQX
XE =
−N C EN ·DP
CD
−N A EN · M Q
AM
(by (11393.1) and (11393.2))
= N C N A · DP
CD · AM
M Q =−CN N A
|{z}=n
· DP CD
|{z}
=− n+pm (p+ 1)n
· AM M Q
| {z }
= q+ 1 mq−1
= (n+pm) (q+ 1) (mq−1) (p+ 1).
Similarly, if we denote Y =RS∩P Q,we have QY
P Y = (r+qs) (p+ 1)
(sp−1) (q+ 1), so that P Y
QY = (sp−1) (q+ 1) (r+qs) (p+ 1).
3We silently disregard the casesM N kBC, P Qk CAandRSk ABas limiting cases in which a part of the proof continues to apply while the rest of the proof simplifies.
2
Now, the lines M N, P Q and RS are concurrent if and only if X = Y (since we definedX asM N∩P Qand Y as RS∩P Q), what is equivalent to P X
QX = P Y
QY ,what, in turn, rewrites as
(n+pm) (q+ 1)
(mq−1) (p+ 1) = (sp−1) (q+ 1) (r+qs) (p+ 1).
But this equation is equivalent to mpr+nqs+mq+nr+ps= 1,since (n+pm) (q+ 1)
(mq−1) (p+ 1) − (sp−1) (q+ 1)
(r+qs) (p+ 1) = q+ 1
p+ 1 · mpr+nqs+mq+nr+ps−1 (mq−1) (r+qs)
(the q+ 1 factor is never zero, unless we allow Qto be a point at infinity). This solves the problem.
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