angle MBC is equal to angle FC iff ABC is equilateral

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angle MBC is equal to angle FC iff ABC is equilateral

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Source: Moldova TST 2005 for JBMO, day 2, problem 1

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#1 h Apr 11, 2005, 5:53 PM • 2 Y

Y by Adventure10, Mango247

Let $ABC$ be an acute-angled triangle, and let $F$ be the foot of its altitude from the vertex $C$ . Let $M$ be the midpoint of the segment $CA$ . Assume that $CF=BM$ . Then the angle $MBC$ is equal to angle $FCA$ if and only if the triangle $ABC$ is equilateral.

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#2 h Apr 11, 2005, 7:30 PM • 2 Y

Y by Adventure10, Mango247

One direction is very easy. I will prove the other direction.

Let $m\angle{FCA} = \alpha = m\angle{CBM}$ . Let $m\angle{CBA} = \beta$ , and $x = m\angle{BMA}$ . Now, by the Law of Sines, $\frac{\sin{90-\beta+\alpha}}{\sin{90-\alpha}} = \frac{AB}{BC} = \frac{\frac{AB}{AM}}{\frac{BC}{MC}} = \frac{\frac{\sin{x}}{\sin{\beta-\alpha}}}{\frac{\sin{180-x}}{\sin{\alpha}}} = \frac{\sin{\alpha}}{\sin{\beta-\alpha}}$ . That is, $\cos{(\beta-\alpha)}\sin{(\beta-\alpha)} = \cos{\alpha}\sin{\alpha}$ . This gives $\beta = 2\alpha$ or $\beta = 90$ . Only the first is valid. But then the angle bisector bisects the opposite side, giving $AB = BC$ . Then basic trigonometry completes the problem.

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#3 h Jul 24, 2005, 2:28 PM • 2 Y

Y by Adventure10, Mango247

Why use trigonometry? There is a short synthetic solution:

We have to show that < MBC = < FCA holds if and only if triangle ABC is equilateral. Well, one direction is obvious: If triangle ABC is equilateral, then both the median BM and the altitude CF are actually medians, so that M and F are the midpoints of the sides CA and AB, respectively, and then < MBC = < FCA = 30° is clear from symmetry. So, it remains to prove the other direction: If < MBC = < FCA, then we have to show that triangle ABC is equilateral.

In the right-angled triangle AFC, we have < FCA = 90° - < FAC, so that < MBC = < FCA = 90° - < FAC = 90° - A.

If O is the circumcenter of triangle ABC, then BO = CO on the one hand, so that triangle BOC is isosceles, while on the other hand, the central angle theorem yields < BOC = 2 < BAC = 2A; thus, the base angle of the isosceles triangle BOC is $\measuredangle OBC=\frac{180^{\circ}-\measuredangle BOC}{2}=\frac{180^{\circ}-2A}{2}=90^{\circ}-A$ . Thus, < OBC = MBC. Hence, the points B, O and M lie on one line l.

Now, if the points O and M would coincide, then it would follow that the circumcenter O of triangle ABC lies on its side CA, so that triangle ABC is right-angled at B; this would contradict the assumption that triangle ABC is acute-angled. So the points O and M cannot coincide. Now, the points O and M both lie on the perpendicular bisector of the segment CA (since O is the circumcenter of triangle ABC and M is the midpoint of the segment CA). Hence, the line l, passing through the points O and M, must be the perpendicular bisector of the segment CA. And thus, since the point B lies on this line l, we have BC = BA, so that a = c. So the triangle ABC is isosceles, and since M is the midpoint of its base CA, we have < BMC = 90°. On the other hand, < CFA = 90°. Thus, < BMC = < CFA. Also, we know that < MBC = < FCA and BM = CF. Hence, the triangles BMC and CFA are congruent, and thus BC = CA. In other words, a = b. Together with a = c, this yields a = b = c, and thus the triangle ABC is equilateral. Proof complete.

EDIT: See also http://www.mathlinks.ro/Forum/viewtopic.php?t=1118 for the second direction.

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