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Take any equilateral triangle and pick a random point inside the triangle.

Draw from each vertex a line to the random point. Two of the three angles at the point are known let's say $x$,$y$.

If the three line segments from each vertex to the random point were removed out of the original triangle to form a new triangle , what would the new triangle's angles be?

triangle

Video about the problem

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  • $\begingroup$ I'm trying to parse your final sentence, but it reads to me like you want to undo what you did initially. Which would be trivial. Or am I misreading? $\endgroup$ – Raskolnikov Jan 29 '18 at 12:57
  • $\begingroup$ @Raskolnikov, the three line segments were used(removed) to form a new triangle. $\endgroup$ – prog_SAHIL Jan 29 '18 at 12:59
  • $\begingroup$ My guess is that it means, "If we create a new triangle whose side lengths are equal to the distances from the given point to the vertices of the original triangle..." $\endgroup$ – Taneli Huuskonen Jan 29 '18 at 13:00
  • $\begingroup$ @quasi, We need to find the angles in relations with x,y and the information is sufficient. I was told that this problem has a very elegant solution. $\endgroup$ – prog_SAHIL Jan 29 '18 at 13:02
  • $\begingroup$ So there are three triangles in the end? $\endgroup$ – TheSimpliFire Jan 29 '18 at 13:02
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enter image description here As in the attached diagram, let $ABC$ be the original equilateral triangle and let $D$ be a point in $\triangle ABC$.

We let point $E$ be on the opposite side of $BC$ as $D$ such that $\triangle BDE$ is equilateral. Then $BD=BE$, $BA=BC$ and $\angle DBA=\angle EBC=60^{\circ}-\angle DBC$. And therefore $\triangle DBA$ and $\triangle EBC$ are congruent. This implies that $EC=DA$ and since $DE=BD$, we now have $\triangle CDE$ as the triangle we want.

Let $\angle ADB=x$ and $\angle BDC=y$. Then $\angle EDC=y-60^{\circ}$, $\angle DEC=x-60^{\circ}$ and $\angle DCE=300^{\circ}-x-y$ are our desired angles.

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Geometrical solution In the attached figure, EC, EA and AG are parallel to and of equal length as AD, CD and DB respectively. The angles opposite to them are $120^{\circ}$ since $\triangle EFG$ is exactly inverted to $\triangle ABC$. Let's call these points I, J and K. Hence EI = EJ, AJ = AK and CI = GK due to the same reason. Thus the green triangles can be perfectly joined to form the red triangle and the angles will be $60^{\circ}$ less than the given angles i.e. $x-60^{\circ}$, $y-60^{\circ}$ and $300^{\circ}-x-y$ since EADC and the rest (not shown in diagram for simplicity) are parallelepipeds.

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