Details

Some fascinating results motivated me to make this Demonstration. For instance, any line through the incenter of a triangle (where the three internal angle bisectors meet) that divides its area in half also divides its perimeter in half; conversely, any line through the incenter that divides the perimeter of the triangle in half also divides its area in half. More generally, we have Haider's theorem: for any triangle and any line , divides the area and perimeter of in the same ratio if and only if it passes through the incenter of .

One of the difficult problems designed to eliminate many targeted applicants in entrance examinations to the Mekh-Mat at Moscow State University during the 1970s and 1980s was to draw a straight line that bisects the area and perimeter of a triangle. Showing the existence of such a line is quite easy; constructing that line is much harder.

References

[1] S. Campo Ruiz. "Solución Problema 138." (Jul 9, 2013) www.aloj.us.es/rbarroso/trianguloscabri/sol/sol138sat.htm.

[2] Math Central. "Problem of the Month, Solution for April 2012." (Jul 9, 2013) mathcentral.uregina.ca/mp/previous2011/apr12sol.php.

[3] I. Vardi. "Mekh-Mat Entrance Examination Problems." (Jul 9, 2013) www.lix.polytechnique.fr/Labo/Ilan.Vardi/mekh-mat.html.