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- Feb 14, 2012

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Find $\angle MBC$ with proof.

- Thread starter anemone
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- Feb 14, 2012

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Find $\angle MBC$ with proof.

- Jan 25, 2013

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my solution:

Find $\angle MBC$ with proof.

- Nov 29, 2013

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Hello.my solution:

View attachment 1817

from the sketch it is easy to prove that $\angle MBC=17^o=y=\dfrac {\angle ABC}{3}$

Could you show it?.

I have managed to find out, only:

certain angles: 13, 26, 30, 34, 43, 51, 56, 60, 64, 73, 77, 103, 116, 129.

Regards.

Edit: I'm sorry. I had not seen, before, that it was already drawing.

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- #4

- Feb 7, 2012

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In case there are other people like me who took a while to follow Albert's clever proof, here is how it works. You are given the angles of triangle ABC: $103^\circ$ at A, $51^\circ$ at B and therefore $26^\circ$ at C. You are also given the angles of triangle AMC: $30^\circ$ at A, $13^\circ$ at C and therefore $137^\circ$ at M. Since 13 is half of 26 it follows that MC bisects the angle at C. So if we drop perpendiculars E, F from M to AC and BC, the right-angled triangles MEC, MFC will be congruent, with angle $77^\circ$ at M. Now draw the circle centred at M touching AC, BC at E and F, and let BG be the other tangent to the circle from B. Then the right-angled triangles BFM, BGM are congruent. At this stage we know the angles $60^\circ$, $77^\circ$, $77^\circ$ at M. So the remaining angle AMF is $146^\circ$ and this is bisected by BM. Therefore the angle BMF is $73^\circ$; and its complement, the angle MBC, is $17^\circ.$my solution:

View attachment 1817

from the sketch it is easy to prove that $\angle MBC=17^o=y=\dfrac {\angle ABC}{3}$

Opalg, maybe you can enlighten me about Albert's proof. I fail to see why G, the intersection of AM with the circle, must be a tangent point from B. I had played with the problem before looking at the solution and came up with the following:

So an indirect proof follows by showing the given construction must be the triangles as given in the problem. I think this is right.

So an indirect proof follows by showing the given construction must be the triangles as given in the problem. I think this is right.

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- Feb 7, 2012

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That is a very good point, which I had completely overlooked. The only way round it that I can see is to do essentially as you suggest. Namely, suppose that the tangent at G meets the line BC at some point B', and then show that B'=B.Opalg, maybe you can enlighten me about Albert's proof. I fail to see why G, the intersection of AM with the circle, must be a tangent point from B...

- Jan 25, 2013

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MG is a radius , so $\angle BGM=90^o$

MG is a tangent line of circle M

here point M is the**incenter of triangle BHC**

MG is a tangent line of circle M

here point M is the

Last edited:

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- #8

- Feb 7, 2012

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That implies that you are defining G as the point where the tangent from B meets the circle. In that case, you need to explain why G lies on the line AM.MG is a radius , so $\angle BGM=90^o$

MG is a tangent line of circle M

here point M is theincenter of triangle BHC

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- Jan 25, 2013

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$\therefore \angle GBA=17^o$

and $\angle AGB=180-73-17=90^o$

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- #10

- Feb 7, 2012

- 2,792

Yes, that seems to work. You are saying that the angles BAG and BAM are both $73^\circ$. So G (defined as the point where the tangent from B meets the circle) lies on the line AM.

$\therefore \angle GBA=17^o$

and $\angle AGB=180-73-17=90^o$

I think this discussion shows that these "proofs by picture" are inadequate on their own. Unless they are accompanied by a careful explanation of how each constructed point is defined, together with the order in which deductions are made, the structure of the proof is almost impenetrable.

The following sketch shows G' as the second tangency point from B and G the intersection of AM with the circle. I still fail to see why G = G'. (Of course the drawing is not accurate; the actual problem has G = G'.)

- Jan 25, 2013

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146=360-77-77-60

- Nov 29, 2013

- 172

Yes, but the problem is that prove that it are those angles and not others146=360-77-77-60

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- Feb 14, 2012

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I want to also thank to

My solution:

Ceva's theorem tells us the product of the ratios of the pairs of segments formed on each side of the triangle by the intersection point is equal to 1.

Hence we have

$\dfrac{\sin 13^{\circ}}{\sin 13^{\circ}} \cdot \dfrac{\sin 73^{\circ}}{\sin 30^{\circ}} \cdot \dfrac{\sin x}{\sin (51^{\circ}-x)}=1$

$\sin 73^{\circ}\cdot \sin x=\sin 30^{\circ} \cdot \sin (51^{\circ}-x)$

$\sin 73^{\circ}\cdot \sin x=\sin 30^{\circ} \cdot (\sin 51^{\circ} \cos x-\cos 51^{\circ} \sin x)$

$\sin x(\sin 73^{\circ}+\sin 30^{\circ}\cos 51^{\circ})=\sin 30^{\circ} \sin 51^{\circ} \cos x $

$\begin{align*}\tan x&=\dfrac{\sin 30^{\circ} \sin 51^{\circ}}{\sin 73^{\circ}+\sin 30^{\circ}\cos 51^{\circ}}\\&=\dfrac{\dfrac{1}{2} \sin 51^{\circ}}{\sin 73^{\circ}+\dfrac{1}{2}\cos 51^{\circ}}\\&=\dfrac{\sin 51^{\circ}}{2\sin 73^{\circ}+\cos 51^{\circ}}\\&=\dfrac{\sin 3(17)^{\circ}}{2\sin (90-17)^{\circ}+\cos 3(17)^{\circ}}\\&=\dfrac{\sin 17^{\circ}(3-4\sin^2 17^{\circ})}{2\cos 17^{\circ}+\cos 17^{\circ}(4\cos^2 17^{\circ}-3)}\\&=\dfrac{\sin 17^{\circ}(3-4\sin^2 17^{\circ})}{\cos 17^{\circ}(2+4\cos^2 17^{\circ}-3)}\\&=\dfrac{\sin 17^{\circ}(3-4\sin^2 17^{\circ})}{\cos 17^{\circ}(4(1-\sin^2 17^{\circ})-1)}\\&=\tan 17^{\circ}\end{align*}$

$\therefore x=17^{\circ}$

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- #15

- Feb 7, 2012

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Yes, you are right again, and this time I don't see any way to fix it. The distorted diagram showing G and G' as distinct points highlights the difficulty in Albert's argument. Fortunately, anemone has now given us a watertight proof using Ceva's theorem.Albert, Opalg, someone please help. I still don't get it. In Albert's expression for angle BFG, the 77+77+60 is the angle AME, but from where comes 146?? Is he saying triangle AMB is isosceles?

The following sketch shows G' as the second tangency point from B and G the intersection of AM with the circle. I still fail to see why G = G'. (Of course the drawing is not accurate; the actual problem has G = G'.)

View attachment 1826

- Jan 25, 2013

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- Feb 14, 2012

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Ah, I applaud your constant effort to try to show us the reasoning behind your work...but Albert, I only managed to follow up to where you mentioned why $G$ is the midpoint of $AM$, and I don't see how that would imply the two lines $BG$ and $AM$ are perpendicular to one another.

Also, if I could decipher that, what follows immediately is that the two triangles $BGA$ and $BGM$ are congruent triangles and hence $\angle ABM=180^{\circ}-2(73^{\circ})=34^{\circ}$ and that gives $\angle MBC=51^{\circ}-34^{\circ}=17^{\circ}$.

- Jan 25, 2013

- 1,225

please see post #9

Also, if I could decipher that, what follows immediately is that the two triangles $BGA$ and $BGM$ are congruent triangles and hence $\angle ABM=180^{\circ}-2(73^{\circ})=34^{\circ}$ and that gives $\angle MBC=51^{\circ}-34^{\circ}=17^{\circ}$.

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- #19

- Feb 14, 2012

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My solution to determine the $\angle BGA=90^{\circ}$ would be as follow:

Since $AG=GM=r$, therefore we have $AM=2r$. If we consider the triangle $AMC$, using the Sine Rule we get $\dfrac{AC}{\sin (60+77)^{\circ}}=\dfrac{AM}{\sin 13^{\circ}}$, or, $AC=\dfrac{2r\sin 137^{\circ}}{\sin 13^{\circ}}$.

Again, by applying Sine Rule to the triangle $ABC$ gives

$\dfrac{AB}{\sin 16^{\circ}}=\dfrac{AC}{\sin 51^{\circ}}$, or, $AB=\dfrac{2r\sin 137^{\circ}\sin 26^{\circ}}{\sin 13^{\circ}\sin 51^{\circ}}$.

Next, we need to find the length of $BG$ and this could be done by applying Cosine Rule to the triangle $BAG$:

$BG^2=AG^2+AB^2-2(AG)(AB)\cos 73^{\circ}$

$\therefore BG=3.27085r$

Last, we could find the value for $\angle BGA$ by using the Sine Rule again:

$\dfrac{AB}{\sin \angle BGA}=\dfrac{BG}{\sin 73^{\circ}}$

This shows that $\angle BGA$ is indeed, $90^{\circ}$!

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- #20

- Feb 7, 2012

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Post #9 states that $\angle FBG=\dfrac{77+77+60-146}{2}$. I believe that is the step that several of us find incomprehensible. Could you explain it please?please see post #9

- Jan 25, 2013

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