MOSCOW 1952

Geometry Level 4

In △ A B C \triangle{ABC} , AC=BC, ∠ C = 2 0 ∘ \angle C=20^{\circ} , M is on side AC and N is on the side BC, such that ∠ B A N = 5 0 ∘ \angle BAN=50^{\circ} , ∠ A B M = 6 0 ∘ \angle ABM=60^{\circ} . Find ∠ N M B \angle NMB in degrees.


The answer is 30.

Serly Zubir by Serly Zubir , 26, Indonesia

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3 solutions

Let A C = B C . . . . . ∠ AC = BC ..... \angle s at B and C become 8 0 o \ 80^{o}

I n △ A B N . . . . . . . . . . . ∠ N A B = 5 0 o . . . . . ∠ A B N = 8 0 o In \triangle ABN........... \angle NAB =\ 50^{o}..... \angle ABN = \ 80^{o}
∴ ∠ B N A = 5 0 o . . . . . . . . B N = A B . . . . . . . . . . . . . . . . . . ( 1 ) \therefore \angle BNA =\ 50^{o}........BN=AB ................ ..(1)

I n △ M B C . . . . . . . . . . . ∠ M B C = 2 0 o . . . . . ∠ B C M = 2 0 o In \triangle MBC........... \angle MBC =\ 20^{o}..... \angle BCM = \ 20^{o}
∴ ∠ C M B = 14 0 o . . . . . . C M = M B , . . . . . . . . . . . . . . . . ( 2 ) \therefore \angle CMB =\ 140^{o}......CM=MB ,................(2)

I n △ A B C . . . A B S i n 20 = B C S i n 80 . . . . B N = A B = . B C ∗ S i n 20 S i n 80 . . ( 3 ) In \triangle ABC...\frac{ AB}{Sin20} =\frac{ BC}{Sin80}....BN=AB=.\frac{ BC*Sin20}{Sin80}..(3) I n △ M B C . . . C M S i n 20 = B C S i n 140 . . . . M B = C M = . B C ∗ S i n 20 S i n 140 . . ( 4 ) In \triangle MBC...\frac{ CM}{Sin20} =\frac{ BC}{Sin140}....MB=CM=.\frac{BC*Sin20}{Sin140}..(4)

I n △ M B N . . ∠ N M B = X o . . ∠ B N M = 180 − 20 − X o = 160 − X o . In \triangle MBN..\angle NMB = \ X^{o}..\angle BNM =\ 180-20 - X^{o}=\ 160-X^{o}.
M B S i n ( 160 − X ) = B N S i n X . . . . ( 5 ) \frac{ MB}{Sin(160-X)} =\frac{ BN}{SinX}....(5)

. . . M B = C M = . B C ∗ S i n 20 S i n 140 . b y . ( 4 ) . . . . B N = A B = . B C ∗ S i n 20 S i n 80 . b y . ( 3 ) ...MB=CM=.\frac{BC*Sin20}{Sin140}.by.(4) ....BN=AB=.\frac{ BC*Sin20}{Sin80}.by.(3)

By (3),(4),(5) B C ∗ S i n 20 ( S i n 140 ) ∗ S i n ( 160 − X ) = . B C ∗ S i n 20 S i n 80 ∗ ( S i n X ) \frac{BC*Sin20}{(Sin140)*Sin(160-X)}=.\frac{ BC*Sin20}{Sin80*(SinX)} ( S i n 140 ) ∗ S i n ( 160 − X ) = S i n 80 ∗ ( S i n X ) . . . . (Sin140)*Sin(160-X) = Sin80*(SinX).... S i n 140 S i n 80 = S i n X S i n 140 ∗ C o s X − C o s 140 ∗ S i n X = 1 S i n 140 ∗ C o t X − C o s 140 \frac {Sin140}{ Sin80}=\frac{ SinX}{ Sin140*CosX - Cos140*SinX} = \frac{ 1}{ Sin140*CotX - Cos140}

3 0 o \boxed {30^{o}}

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Michael Mendrin
Jul 7, 2014

Problems like this, it's always 20 or 30 degrees, so I picked 30?

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yeah when first attempted the problem i got 20 and then tried 30 which was the right answer.

Mardokay Mosazghi - 6 years, 11 months ago

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I should start posting problems like this where the answer is NOT 20 or 30 degrees. It's time we introduced some variety here.

Michael Mendrin - 6 years, 11 months ago

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yeah look forward to it.

Mardokay Mosazghi - 6 years, 11 months ago

Triangles A C N ACN and M B N MBN are similar, although I don't immediately see how to prove it. But then again, I did not do too deep of an analysis.

Daniel Liu - 6 years, 11 months ago

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The best solution I've seen to these class of problems consisted of drawing a 18 sided polygon with all diameters and a selected number of chords drawn. Then you can find the figure of this problem inside this polygon and the proof becomes straightforward. What I liked about that approach was its generality, without having to resort to trigonometry, which is the usual method.

Michael Mendrin - 6 years, 11 months ago

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hey if you are true then please post the figure of that 18-sided polygon with all constructions that you have conversed above . i will be pleased to see that

Rishabh Jain - 6 years, 11 months ago

I got this problem on my geometry class, sir. I didn't even know about this kind of problem before and I think this problem is really interesting, because I used some congruence triangle to solve it and that was really confused me :D

Serly Zubir - 6 years, 11 months ago
Ajit Athle
Jul 9, 2014

This type of problem is well-known as "Langley's Adventitious Angles." Check this animated solution: http://agutie.homestead.com/files/LangleyProblem.html

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