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So I was playing around GeoGebra and found this thing out, I don't know if this problem has a name or something.

Triangle ABC is inscribed inside a circle, from point D which is located inside the circle, we draw 3 perpendicular lines to each side of the triangle, what is the maximum area of the triangle whose vertices are the intersections of the perpendicular lines and the sides of the triangle? (maximum area of triangle EFG, the red triangle in the picture)

Using Geogebra I found out that this area is always maximal when point D is located at the center of the circle, or in other words, when the perpendiculars divide the sides into 2 equal segments.

If someone could provide a proof/explain why, I would be grateful.

See the diagram below:

enter image description here

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  • $\begingroup$ A synthetic proof here seems like a real bastard, I must say! $\endgroup$ – Noldorin Aug 16 '17 at 17:02
  • $\begingroup$ When $D$ is the circumcentre of $ABC$ the perpendiculars are also the bisector of the sides of $ABC$ and $EFG$ is similar to $ABC$ and it is exactly $\dfrac{1}{4}$ of its area. Left to prove that any other position leads to smaller triangles $\endgroup$ – Raffaele Aug 16 '17 at 17:28
  • $\begingroup$ @Raffaele I think you've left the hard bit! $\endgroup$ – Noldorin Aug 16 '17 at 18:23
  • $\begingroup$ The medial triangle is formed as you specify by connecting midpoints. The concurrence of these perpendicular lines must be important, since a freely chosen set of points on the sides could have area arbitrarily close to the full triangle area. I'm less convinced that the restriction to the circle is important. $\endgroup$ – Joffan Aug 16 '17 at 19:18
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    $\begingroup$ UPDATE 2: Wow, via geogebra, the area of the red triangle depends ONLY on the distance of the point D from the circumcenter, so for the set of all points D which have the same fixed distance from the circumcenter, the red area is equal through all these points! So according to geogebra, let's assume that the distance from D to the circumcenter is X, and the radius of the circle in the picture is r If X increases while X<r, the area decreases and when x=r the area is 0 If you increase X while X>r, the area keeps increasing (up to infinity) $\endgroup$ – Wajd Aug 16 '17 at 21:43
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$+1$ for re-discovering this neat property of pedal triangles: the area is proportional to the power of the point with respect to the circumcircle, in other words it only depends on the distance from the point to the circumcenter of the original triangle:

$$ S_{EFG}=\frac{R^2-OD^2}{4 R^2} \,S_{ABC} $$

Mathworld quotes on this Johnson, R. A. Modern Geometry: An Elementary Treatise on the Geometry of the Triangle and the Circle. Boston, MA: Houghton Mifflin, 1929.

A proof can be found for example on cut-the-knot.

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    $\begingroup$ Wow, this explains everything! Thank you! The problem looks difficult to solve if one did not know that the area depends only on the distance from D to the circumcenter and the area of the original triangle ABC $\endgroup$ – Wajd Aug 17 '17 at 9:52

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