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Question:

Given the base and vertical angle of a triangle show that its area is greatest when the triangle is isosceles.

My attempt:

For isosceles triangle (with base given 2x, and vertical angle let z):

Note: $y$ is got by $(180 - z)/2$

I calculated area to be: $$CD^2 * tan(y)$$ by using trigonometry and the fact that the altitude in isosceles triangle from vertex angle bisects the base.

I don't know how I can prove this area to be greatest.

Please help me. Thank you!

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  • $\begingroup$ Do you have any background in calculus? It can easily be proved with derivatives. $\endgroup$ – Ari Dec 3 '14 at 16:13
  • $\begingroup$ @Ari Sorry. I do not have experience with calculus. The book in which this question is given doesn't teach about calculus. I guess the question has to be solved without it. Thank you! $\endgroup$ – Gaurang Tandon Dec 3 '14 at 16:14
  • $\begingroup$ Use Heron formula: $A=\sqrt{s(s-a)+(s-b)+(s-c)}$ and $2s=a+b+c$. $\endgroup$ – ulead86 Dec 3 '14 at 16:22
  • $\begingroup$ Could you clarify which is the vertical angle? $\endgroup$ – ghosts_in_the_code Dec 3 '14 at 16:22
  • $\begingroup$ @user45195 The one opposite the base i.e. $\angle BAC$ $\endgroup$ – Gaurang Tandon Dec 3 '14 at 16:25
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The circle is the locus of all triangles with a given base and a given angle opposite to it. If a line is drawn parallel to the base there are two points of intersection. Altitude of these two scalene triangle drawn between intersection points can be increased by drawing the parallel line progressively farther away until it cuts at two coincident points as a tangent when the triangle vertex is situated at apex as coalesced tangent point making it isosceles by symmetry.

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The inscribed angle theorem gives a circle as the locus of all triangles with a given base and a given angle opposite that base. Among all possible corners on that circle, the one which maximizes height is the one leading to an isosceles triangle.

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  • $\begingroup$ Yes, you are right. Thank you! $\endgroup$ – Gaurang Tandon Dec 3 '14 at 16:26
  • $\begingroup$ But how do we know that the height is maximum in that case ? (theoretically, with math) $\endgroup$ – Gaurang Tandon Dec 3 '14 at 16:27
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    $\begingroup$ @GaurangTandon: Depends on what tools you have. one central fact is that at the point of maximal height, the tangent has to be parallel to the base. Otherwise there would be a direction where you could increase the height. You can either prove that parallel tangent with some geometric arguments, or fire up calculus as Ari suggested. $\endgroup$ – MvG Dec 3 '14 at 16:29

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