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the question

Consider the triangle $ABC$ and a point $M$ inside the triangle such that $\angle MAB = 10 ,\angle MAC = 40 ,\angle MCA = 30 $ and $\angle MBA = 20 $ . Show that triangle $ABC$ is isosceles.

my idea

the drawing

enter image description here

As you can see I calculated the other angles we have in the triangle...I'm preatty sure we need an auxiliar construction, but I can't figure out which one to do. I dont know how to start. Hope one of you can help me! Thank you!

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    $\begingroup$ Extending $BM$ you get an exterior angle of triangle $BMC$ equal to $80^\circ$. This means $\angle MBC + \angle MCB = 80^\circ$. This and the boomerang theorem on $MACB$ gives $\angle AMB = \angle MAC + \angle ACB + \angle MBC$. Try using this to your advantage. $\endgroup$
    – Rusurano
    Apr 18 at 13:41
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    $\begingroup$ Later today, I will return here to leave a complete proof of this problem without trigonometry. $\endgroup$
    – Rusurano
    Apr 18 at 13:57
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    $\begingroup$ @Rusurano sadly properties of triangles was left out of our syllabus, hence I didn't get the chance to look into it in depth. My answer was nothing but trigo. I am looking forward to your answer :) $\endgroup$
    – Gwen
    Apr 18 at 14:32
  • $\begingroup$ Are you familiar with the trigonometric form of ceva's theorem? $\endgroup$
    – Calvin Lin
    Apr 18 at 16:28
  • $\begingroup$ @Gwen Posted, enjoy Euclidean geometry! :) $\endgroup$
    – Rusurano
    Apr 19 at 6:47

6 Answers 6

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By law of sines, $$\frac{AM}{\sin 30°}=\frac{AC}{\sin 110°} \tag{1}\label{1} $$ $$\frac{AB}{\sin 150°}=\frac{AM}{\sin 20°} \tag{2}\label{2}$$ We can combine ($\ref{1}$) and ($\ref{2}$) to get $$\frac{AC\sin 30°}{\sin 110°}=\frac{AB\sin 20°}{\sin 150°}$$ Now we can apply values of $\sin 30°$, $\sin 150°$: \begin{align} AC& =2AB \left(2\sin 20°\sin 110°\right) \\ AC & =2AB(\cos(-90°)-\cos(130°)) \\ AC & =-2AB\cos(180°-50°) \\ AC & =2AB\cos 50° \tag{3} \label{3} \end{align} Again by law of cosines, $$BC^2=AC^2+AB^2-2ABAC\cos 50°$$ Now substituting from $(\ref{3})$, \begin{align} BC^2& =4(AB)^2\cos^2 50° + AB^2 - 4(AB)^2\cos^2 50° \\ BC^2 & =AB^2 \\ BC & =AB \end{align} Hence $\triangle ABC$ is isoceles.

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  • $\begingroup$ How did you know that $2*sin(20)*sin(110)=cos(-90)-cos(130)$ $\endgroup$ Apr 18 at 14:23
  • $\begingroup$ That follows an identity $$\cos(A-B)-\cos(A+B)=2\sin A\sin B$$ It can be easily proven by expanding them $\endgroup$
    – Gwen
    Apr 18 at 14:25
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Here is a more general hint towards solving these kinds of problems. When you see a problem where degrees are given in $10x$ increments (this is sometimes disguised a bit), the problem is likely constructed with the following schematic in mind, where an equilateral triangle is divided by rays subtending $10$ degrees each. Note the emergence of many right angles, and several $80-20-80$ isosceles triangles. These latter triangles are special, for, wherever two congruent $80-20-80$ triangles share an $80$-degree angle, embedded within them are several equilateral and isosceles triangles with vertices on the intersection of the sides of one and altitude of the other triangle. It's a good exercise to prove this yourself. Note that this only works for congruent, not similar, pairs of such triangles (see the blue vs gray equilateral triangles I marked).

In your particular version, you are given a situation with side $AB$ and point $M$ (in red) and the ray $BC'$ (also in red). You only do not know where the final side is, but it is instantly fixed once you find the triangle at $M$ with an angle of $150$, for this is part of the $80-20-80$ in the lower right corner of the diagram with $MF$ as its upper side.

enter image description here

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    $\begingroup$ This is actually beautiful approach and definitely deserves an upvote. $\endgroup$
    – Rusurano
    Apr 19 at 11:14
  • $\begingroup$ just a typo: "and the ray $AC′$ (also in red)" - ray $BC'$ is in red. $\endgroup$ Apr 25 at 16:16
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    $\begingroup$ @PiotrWasilewicz - thanks, fixed $\endgroup$ Apr 25 at 17:00
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This solution was given by user vineet on AoPS.

enter image description here

Construct an equilateral triangle $\triangle AMD$ with base $AM$, on the opposite side of $B$. Let $E = DB\cap CM$.

Claim 1: $BM$ is the perpendicular bisector of $AD$.
Proof: $\angle AMB = 150^\circ$, so $BM$ is angle bisector of $\angle DMA$, hence the perpendicular bisector of $AD$.

Claim 2: $BD \perp AC$
Proof: Due to claim 1, $A$ and $D$ are symmetric about $BM$. Particularly, $\angle BDM = \angle BAM = 10^\circ$. Also $\angle CAD = 20^\circ$ (as $\angle DAM = 60^\circ$) and $\angle ADM = 60^\circ$. Summing them, we get the required claim.

Claim: $AMED$ is cyclic.
Proof: Due to claim 2, $\angle DEC=60^\circ = \angle DAM$.

Then, note that, $\angle EAM = \angle EDM = \angle BDM = 10^\circ$, so that $$\angle EAC = 30^\circ = \angle ECA \implies \triangle ECA \text{ is isosceles}$$This, coupled with claim 2, gives that $BD$ is the perpendicular bisector of $AC$, which proves the required statement. $\blacksquare$

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Here is an elementary proof of this statement that does not involve trigonometry.

The drawing

By plane separation axiom, line $AC$ separates a plane into two half-planes. Construct an equilateral triangle $CAK$ on side $AC$, so that point $K$ lies in the same half-plane with point $B$.

Extend a segment $CM$ to intersection with $AK$ at a point $H$. As $CH$ is the angle bisector of $\angle ACK$, and $_\Delta ACK$ is equilateral by construction, then $CH$ is also a median and an altitude.

Drop a perpendicular $KL$ on $AC$. As $_\Delta ACK$ is equilateral, it will also be the median and the angle bisector, so $\angle AKL = \angle CKL = \frac{1}{2}\cdot 60^\circ = 30^\circ$.

In $_\Delta AMB$, $\angle AMB = 180^\circ - (\angle MAB + \angle ABM) = 180^\circ - (10^\circ + 20^\circ) = 180^\circ - 30^\circ = 150^\circ$.

In right triangle $_\Delta AMH$, $\angle AMH = 90^\circ - \angle MAH = 90^\circ - 20^\circ = 70^\circ$. Therefore, $\angle BMH = \angle AMB - \angle AMH = 150^\circ - 70^\circ = 80^\circ$. $\angle OMB$ is supplementary to $\angle BMH$, therefore $\angle OMB = 180^\circ - \angle BMH = 180^\circ - 80^\circ = 100^\circ$.

In right triangle $_\Delta KOH$, $\angle KOH = 90^\circ - \angle OKH = 90^\circ - 30^\circ = 60^\circ$.


Warning: the following part contains a circular argument. If you know how to fix it, please do!

Assume that point $B$ does not belong to $KL$. Then let $B_1$ be the intersection point of $MB$ and $KL$.
Consider the triangle $_\Delta AB_1K$. There, $\angle B_1AK = 10^\circ$, $\angle AKB_1 = 30^\circ$. Therefore, $\angle AB_1K = 180^\circ - (\angle B_1AK + \angle AKB_1) = 180^\circ - 40^\circ = 140^\circ$.
Now, consider the triangle $_\Delta MOB_1$. There, $\angle B_1OM = 60^\circ$, $\angle B_1MO = 100^\circ$. By exterior angle theorem, an exterior angle should be equal to the sum of two opposite interior angles. $\angle MB_1K$ is exterior to $_\Delta MOB_1$. Therefore, $\angle MB_1K = \angle B_1OM + \angle B_1MO = 60^\circ + 100^\circ = 160^\circ$.
Since $AB_1K = 140^\circ$ and $\angle MB_1K = 160^\circ$, we also must have $\angle AB_1M = \angle MB_1K - \angle AB_1K = 160^\circ - 140^\circ = 20^\circ$.
$\angle ABM = 20^\circ$. Then we have $\angle AB_1M = \angle ABM$. But by angle construction axiom, there is only one way to construct an angle less than $180^\circ$ on a given half-plane. So we have a contradiction. Therefore, points $B$ and $B_1$ are coincident.


As point $B$ belongs to $KL$, and $KL$ is a perpendicular bisector of $AC$, then, by the properties of perpendicular bisector, point $B$ is equidistant from the points $A$ and $C$. Therefore, $AB = AC$, which leads to the conclusion that $_\Delta ABC$ is isosceles by definition, which was to be proven.


By the way, this is Problem 5 from USAMO 1996, and there are pretty trigonometric solutions published on Art of Problem Solving.

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    $\begingroup$ This is wrong. $\angle B_1AK$ is not $10^\circ$. $\angle BAK$ is. Here, you have assumed $\angle B_1AK=10^\circ$ and hence your argument is circular. $\endgroup$
    – D S
    Apr 19 at 7:19
  • $\begingroup$ @DS Ah, true. Though, similar reasoning must absolutely work if I define $B_1$ correctly. Would you suggest a fix to that reasoning? $\endgroup$
    – Rusurano
    Apr 19 at 8:16
  • $\begingroup$ Your proof uses a circular argument and yet it received four votes. Standards are deteriorating. $\endgroup$
    – Ted Black
    Apr 26 at 20:56
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We can construct an isosceles triangle $AFC$ by drawing a segment $AF$ at a $10^\circ$ angle to $AM$. $FF_\perp$ is the perpendicular bisector of this triangle with $\angle F_\perp F C = 60^\circ$. This means that the angle formed between $F F'$ (the extension of $F_\perp F$) and $FM$ is also $60^\circ$. Since $\angle AMC=110^\circ$ it is also the case that $\angle AFM=60^\circ$ and so segment $FM$ bisects $\angle AFF'$.

If we can prove that the point where the extension of $F_\perp F$ and $AB$ intersect is vertex $B$ then $BF_\perp$ is a perpendicular bisector of $AC$ and therefore $\triangle ABC$ is isosceles with $|AB|=|BC|$.

enter image description here

Lets call the point of intersection of lines $AB$ and $F_\perp F$, $B_F$. $\triangle AFB_F$ has $M$ as a concurrent point for segments $AM$, $FM$ and $B_FM$. We know that $AM$ (resp. $FM$) bisects $\angle B_FAF$ (resp. $\angle AFB_F$). The only point of concurrency with this property is the incenter which means that $B_FM$ also bisects $\angle AB_FF$. Since $\angle AB_FF=40^\circ$ it follows that $\angle AB_FM=20^\circ$. But we know from the conditions of the problem that $\angle ABM=20^\circ$. This is only possible if $B \equiv B_F$ and this completes the proof.

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HINT.-Let $a,b,c$ the sides. The angles to determine are $\alpha=\angle{MCB}$ and $\beta=\angle{MBC}$. These are solution of the system

$$\alpha+\beta=80^{\circ}\\ac\sin 50^{\circ}=bc\sin (20+\beta)=ab\sin(30+\alpha)$$ there are two solutions $(\alpha, \beta)=(20,60)$ which yields to $b=c$ and $(\alpha, \beta)=(50,30)$ which yields to $b=a$

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    $\begingroup$ Can you elaborate on how you solved this system? There seems to be way too many unknowns, like why can't we let $ \alpha = 10^\circ, \beta = 70^\circ$, and set $a, b, c$ accordingly? $\endgroup$
    – Calvin Lin
    Apr 19 at 15:16

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