# Maximum Area Problem in a Triangle - Not Solved Yet

• Jan 29th 2011, 01:19 PM
oswaldo
Maximum Area Problem in a Triangle - Not Solved Yet
The following extremum problem in a triangle has not been solved. It looks simple, and intuitively "P = mid point of BC" seems to be the answer, but noone was able to prove it yet. The closest solution brings a set of 4 equations which was not able to be solved neither by WolframAlpha nor by Maple.

Thanks.

Creative Unusual Geometry Problems: Fox 322

Attachment 20632

I am walking 8 foxes at the top of my head and trying to keep their tails apart !
• Jan 29th 2011, 10:29 PM
earboth
Quote:

Originally Posted by oswaldo
The following extremum problem in a triangle has not been solved. It looks simple, and intuitively "P = mid point of BC" seems to be the answer, but noone was able to prove it yet. The closest solution brings a set of 4 equations which was not able to be solved neither by WolframAlpha nor by Maple.

Thanks.

This isn't the complete solution, only a first step:

1. $\displaystyle \angle(QPR) = 90^\circ$ because:

$\displaystyle \angle(p_1)=\dfrac{\gamma + \alpha_2}2$

$\displaystyle \angle(p_2)=\dfrac{\beta+\alpha_1}2$

$\displaystyle \angle(PQR)=\angle(p_1) + \angle(p_2) = \dfrac{\gamma + \alpha_2}2 + \dfrac{\beta+\alpha_1}2 = \frac12 \underbrace{((\alpha_1+\alpha_2)+\beta+\gamma)}_{\ text{sum up to 180°}}$

2. Therefore the area of the triangle PQR is calculated by:

$\displaystyle A(\Delta(PQR))=\frac12 \cdot |\overline{PR}| \cdot |\overline{PQ}|$

3. Since you "know" (:D) the angles of the triangles $\displaystyle \Delta(BPR)$ and $\displaystyle \Delta(PCQ)$ and at least one side of these triangles it must be possible to find the optimal place of P on $\displaystyle \overline{BC}$.
• Feb 4th 2011, 08:51 PM
oswaldo
Observing m(RPQ)=90 degrees, is a good start, but then it gets complicated:
Let BP=x, as a variable in the final extremum function, say A(RPQ)=f(x).
sine rule in triangles:
PR / sin(beta) = x / sin(180-beta-p1) => PR = x [sin(beta) / sin(beta+p1)]
Similarlay, PQ = (BC - x) [sin(gamma) / sin(gamma+p2)]

A(RPQ) = PR*PQ/2 = x*(BC-x)*[(sin(beta)*sin(gamma)) / (2*sin(beta+p1)*sin(gamma+p2))] = f(x)

Here is the problem, angles, p1 and p2 also depend on x.

What should be the next step? Any ideas?
• Feb 5th 2011, 01:57 AM
earboth
Quote:

Originally Posted by oswaldo
Observing m(RPQ)=90 degrees, is a good start, but then it gets complicated:
...
What should be the next step? Any ideas?

Quote:

Originally Posted by oswaldo
The following extremum problem in a triangle has not been solved. It looks simple, and intuitively "P = mid point of BC" seems to be the answer, but noone was able to prove it yet. ...

1. I've drawn several examples of the triangle in question if P runs from B to C.

2. Perpendicular to BC over P I have drawn a point whose distance from BC correspond to the value of the area of triangle PQR. How I got the area's value is shown in the box at the upper left part of the image.

3. All these points lie on a curve which pretty much looks like a "simple" parabola. But unfortunately I can't find the equation which belongs to this parabola.
• Feb 8th 2011, 08:01 PM
oswaldo
Thank you earboth for the simulation. Indeed red looks like a parabola. This means that f(x)= K * x(BC-x), and K is a constant. In other words,
(sin(beta)*sin(gamma)) / (2*sin(beta+p1)*sin(gamma+p2)) must be a CONSTANT. The rest is just trigonometry. I worked on it a little, but could not succeed.
We also know that p1+p2=90.
I will try again later..