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?


Video about the problem

  • $\begingroup$ 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

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.


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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