A Problem Involving a median, altitude and bisector

[Sean McGrath]

Hi, I am a high school math team coach and I have been collecting some
harder problems than I am used to. I need some serious help on one of
them. Here it is:

In triangle ABC, N, L and M are (in that order) on side AC, N nearer
A, so that BN is an altitude, BL bisects angle ABC, and BM is a
median. Also, angle ABN = angle NBL = angle LBM = angle MBC. Find
the measures of the angles of triangle ABC.

Help.

Thanks, Sean McGrath
Algonquin RHS
Northboro, MA
mathte...@aol.com

[Ken Pledger]

In article <goz8p7...@forum.mathforum.com>, mathte...@aol.com (Sean
McGrath) wrote:

> ....

> In triangle ABC, N, L and M are (in that order) on side AC, N nearer
> A, so that BN is an altitude, BL bisects angle ABC, and BM is a
> median. Also, angle ABN = angle NBL = angle LBM = angle MBC. Find

> the measures of the angles of triangle ABC....

Angle A = 3(pi)/8, B = pi/2, C = pi/8.

Proof.
Let each of your four equal angles be X (although on paper I used
alpha). Using the right angle at N shows that

A = (pi/2 - X), B = 4X, C = pi/2 - 3X.

Apply the sine rule to triangles ABM and BCM to get

sin(A)/sin(3X) = BM/AM = BM/MC = sin(C)/sin(X).

Substituting the above values for A and C leads to

cos(X)/sin(3X) = cos(3X)/sin(X).

Cross-multiply and double both sides to get

sin(6X) = sin(2X).

If 6X = an even multiple of pi plus 2X,
then B = 4X = 6X - 2X = an even multiple of pi
which is impossible for an angle of a triangle.

Therefore 6X = an odd multiple of pi minus 2X,
so B = 4X = (6X + 2X)/2 = an odd multiple of pi/2
which can only be pi/2. Hence X = B/4 = pi/8, etc.

Ken Pledger.

[Martin Lukarevski]

Sean McGrath wrote:

>In triangle ABC, N, L and M are (in that order) on side AC, N nearer
>A, so that BN is an altitude, BL bisects angle ABC, and BM is a
>median. Also, angle ABN = angle NBL = angle LBM = angle MBC. Find
>the measures of the angles of triangle ABC.

We will make use of this:

ang(NBL) = ang(LBO) , where O is circumcenter of ABC

Since ang(NBL) = ang(LBM) by assumption, it follows that
O lies on BM.On the other hand O lies on the perpendicular of AC at M.
From this we conclude that O = M.Now it follows immediately that

B = pi/2

Now,

ang(ABN) = pi/2 - A = B/4 and

A - C
ang(NBL) = ----- = B/4
2


From this two equations and B = pi/2 one easily finds that

A = 3pi/8 , B = pi/2 , C = pi/8


Martin Lukarevski
from Skopje,Macedonia