Thank you

Starting from n pcs of equal areas of triangle pairs let's assume slopes are

s_1 and s_2 in absoulte value while base lengths are dx_1 and dx_2 respectively.

Then there has to be: dx_2 = dx_1 * s_1 / s_2.

For a single pair of trianges area is:

A_s = s_1 * dx_1 * dx_1 /2 + s_2 *dx_2 * dx_2 / 2 =

= dx_1 * dx_1 * ( s_1 + s_1 * s_1 / s_2 ) /2,

total area is: A_t = n * A_s.

In case any of the n area together with height and base length is shrinked

another needs to be enlarged with the same amount,

so dx-1 for one triangle pair becomes dx_1 - e, for another dx_1 + e.

Since s_1 and s_2 are constants area depends only dx_1.

Especially area goes quadratically with dx_1.

Which means one of the n pcs of A_s become (dx_1 - e / dx_1)^2 times samller,

another become (dx_1 + e / dx_1)^2 times larger.

Their sum must be larger than the sum of original A_s pair,

since (dx_1 - e )^2 + ( dx_1 + e )^2 =

= ( (dx_1)^2 + e^2 - 2* dx_1 * e ) + ( (dx_1)^2 + e^2 + 2* dx_1 * e ) =

= 2*(dx_1)^2 + 2e^2 thus always larger than 2*(dx_1)^2.

Extending for any further changes of triangle pairs sizes is streightforward.

Dear Zlotan,

Thank you for your reply. However, could you explain further, or rather draw, what is s_1 , s_2, dx_1 , dx_2?

Hello,

Sorry for the different notation that seemed easier to me.

Each triangle pairs on your figure consists of 2 rectangular triangles.

For each of these have one vertical and one horizontal side and one with a particular slope on the left and another one with different slope on the right. One of the slopes is s_1 another is s_2, maybe a and c on your figure.

Within each each pair of triangles vertical sides have equal length, thus the base lengths dx_1 and dx_2, might have been t_1 and t_2 on your figure, respectively, depend on the slopes as described above.

Thanks a lot. You help me very much.