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Let $ABC$ be a triangle and $O$ the center of its enscribed circle.

Let $M = BO \cap AC$ and $N=CO \cap AB$ such that $\measuredangle NMB = 30°, \measuredangle MNC = 50°$.

Find $\angle ABC, \angle BCA$ and $\angle CAB$.

I also posted this here at the Art of Problem Solving.

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At first, I took $O$ to denote the center of the incircle inscribed into the triangle. Only a comment below clarified that you were actually taling about the center of the circumcircle circumscribed around the triangle. Therefore, I have two solutions, one for each interpretation.


Angles without a proof

I constructed your situation using Cinderella. At first I had one point chosen freely on a line, which I later moved into a position where everything fit the way it should. From the resulting figure, I obtained the following measurements:

\begin{align*} \measuredangle ABC &= 60° \\ \measuredangle BCA &= 70° \\ \measuredangle CAB &= 50° \end{align*}

Warning: I accidentially swapped $M$ and $N$ as well as $B$ and $C$ in the following image. So take care to look more at the actual angles than the denoted point names.

image of desired triangle with circumcircle

I guess that once you know these angles at the corners, you can choose suitable coordinates and using these proove that the angles specified in your question are indeed as required.

Oriented angles

In the above, I interpreted the angles in your question as unoriented, and measured one of the clockwise and the other counter-clockwise. I furthermore assumed $M$ and $N$ to lie between the corresponding corners of the triangle. Strictly speaking, your angles are given in an oriented fashion, and as both of them are positive, $B$ and $C$ must lie on different sides of $MN$. this leads to rather ugly triangles. Of the two possible solutions I found, the better one is the following:

Circumcenter outside the triangle

The angles for this triangle are rather ugly, compared to the nice numbers resulting from the interpretation above.


Angles without a proof

This result I obtained in a similar way to the one outlined above, using Cinderella with one free point on a line adjusted till things line up as intended.

\begin{align*} \measuredangle ABC &= 40° \\ \measuredangle BCA &= 120° \\ \measuredangle CAB &= 20° \end{align*}

Image of triangle with incircle


Let us call the half corner angles like this:

\begin{align*} \alpha &= \angle BAO = \angle OAC \\ \beta &= \angle CBO = \angle OBA \\ \gamma &= \angle ACO = \angle OCB \end{align*}

From $\triangle BCO$ you see that $\angle BCO = 180° - \beta - \gamma$. That is the same as $\angle MON$, so from $\triangle MNO$ you can conclude that $\beta+\gamma = 30° + 50°$. From $\triangle ABC$ you know that $2\alpha + 2\beta + 2\gamma = 180°$, so $\alpha = 90° - (\beta + \gamma) = 10°$.

Based on this angle, you can construct the triangle even without knowing the corner angles up front:

  1. Choose $M$ and $N$ arbitrarily
  2. Draw the rays $MO$ and $NO$ using the given angles to obtain $O$
  3. Draw lines at $90° - \alpha = 80°$ from $MO$ through $M$ and $O$ as indicated in the figure below (green). Around their intersection, draw a circle containing all points that see $M$ and $O$ under an angle of $10°$.
  4. Repeat the previous step for $NO$. The intersection of these two circles is $A$.
  5. Now $C = AM \cap NO$ and $B = AN \cap MO$.

Illustration of construction for incircle

Measuring the angles in this figure will give you the desired result up to the precision with which you performed your drawings and measurements.

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I'm sorry, I think I used the wrong notion. I mean that $O$ is the circumcenter, the center of the circle passing through the three vertices of the triangle – Andrei Oct 22 '12 at 12:30
@Andrei: If you're talking about circumcircle instead of incircle, that changes the problem fundamentally. I didn't know the word “enscribed” in this context, and so assumed “inscribed”. Will think about the other task. – MvG Oct 22 '12 at 14:47
@Andrei, please have a look at my revised answer. – MvG Oct 23 '12 at 12:04
I guess the values are corect, but can we find them without the software? – Andrei Oct 30 '12 at 7:57

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