Overlapping triangles

[onta...@hotmail.com]

Two identical isoscelean triangles with angles A,(180-A)/2 and (180-A)/2 have their vertices on the circumference of the same circle. They have an angular displacement of 125 degrees to each other. If the overlap of the triangle areas is equal to one half of the area of one triangle find angle A. Consider that there may be more than one answer.

[Narasimham]

On May 8, 4:58 pm, "ontad...@hotmail.com" <ontad...@hotmail.com>
wrote:

> Two identical isoscelean triangles with angles A,(180-A)/2 and (180-A)/2 have their vertices on the circumference of the same circle. They have an angular displacement of 125 degrees to each other.

When they are rotated with respect to circum-center (by implication)?

[onta...@hotmail.com]

Yes.

[Philippe 92]

onta...@hotmail.com wrote :

Seems it is again calculation intensive...
I got two solutions A ~ 39° and A ~ 77°
by playing with JavaSketchpad.

Also why 125° ? It is not compass and straightedge constructible...
neither have simple trigonometric values.
Why not 120° ? with solutions about 40° and 78°

You may play with the applet at
http://mathafou.free.fr/pbw_en/pb402.html

Interresting and even harder to calculate is displacement = 60°
then the overlapping area is an hexagon, with A ~ 34° and A ~ 102°
(to get area ratio = 1/2).

My calculation method for the 'simpler' case :
Area of a triangle, given base c and adjacent angles A and B
is (1/2) * c^2 * sin(A)*sin(B)/sin(A+B)

Then calculate all necessary angles at vertex A, and deduce area of
suitable triangles, then add/substract to get overlapping area.
I gave up after two pages of sin and cos expressions...
(I'm unable to write more than two lines of algebra without any error)

Regards.

--
Philippe C., mail : chephi...@free.fr
site : http://chephip.free.fr/ (recreational mathematics)

[onta...@hotmail.com]

Hi Phillipe,
Thank you for a very nice graphical representation of the problem. You ask "Why 125 degrees ?". Well my only answer is "Why not, it makes it more difficult !!". I think your answers are fairly close as I have 38.778873 degrees and 76.913213 degrees after lots of math.
This problem is a variation of an earlier one of mine, with two triangles (with 30,75,75 angle vertices) and requiring the angular displacement for a half area overlap, and which was easier to solve.
Regards
Bill.