Take the 2-minute tour ×

Two lines $l_1$ and $l_2$ intersects at point $A$ such that the angle they intersect is $\alpha$. A line segment has endpoints $B$ and $C$ in the lines $l_1$ and $l_2$, respectively, and $|BC|=l$. What is the maximal area of $ABC$ in terms of $\alpha$ and $l$ if those two variables are fixed?

share|improve this question
Hint: Maximum area is achieved when the other two sides are equal. –  ronno Nov 3 '12 at 16:12
Another hint: Area is length of base, $l$, times the altitude. So consider things relative to the base $l$ and where the vertex $A$ should end up to get maximum altitude... –  coffeemath Nov 3 '12 at 16:36
Thanks for the edit. So the task is to determine where the points of intersection B and C should lie on $\ell_1$ and $\ell_2$ to give you the maximum area. I think @ronno 's hint is "right on the money" so to speak! –  amWhy Nov 3 '12 at 17:16
I was wondering what is the trick to show that the triangle is isosceles. –  student Nov 3 '12 at 17:56
So is the trick to put the figure to coordinate system where BC is in the $y$-axis, from (0,-l/2) to (0,l/2)? Then point where given line segment can be seen from a given angle forms two circles. Then the points in the circle where $|x|$ is maximal has $y$ coordinate zero. Then by symmetry, the triangle is isosceles. –  student Nov 3 '12 at 19:21

2 Answers 2

up vote 1 down vote accepted

There are several ways to parametrize this problem and therefore several expressions for the target function. I will express it in terms of a variable angle. enter image description here

Denote one of the variable sides, say $AC$, $x$ and $\angle ACB=\beta$. The area of $\triangle ABC$ is half the product of two sides times the sine of the enclosed angle:


From the sine rule we deduce the following relation: $$\frac{l}{\sin\alpha}=\frac{x}{\sin\gamma}$$ Substituting in the expression for $S$: $$S=\frac{l^{2}}{2\sin\alpha}\sin\beta\sin\gamma$$ So the problem reduces to the following: $$f=\sin\beta\sin\gamma\to\max$$ subject to the constraint: $$\beta+\gamma=\pi-\alpha$$ Simplify using complimentary formula: $$f\left(\beta\right)=\sin\beta\sin\left[\pi-\left(\alpha+\beta\right)\right]=\sin\beta\sin\left(\alpha+\beta\right)$$ $$f'\left(\beta\right)=\sin\left(\alpha+\beta\right)\cos\beta+\sin\beta\cos\left(\alpha+\beta\right)=\sin\left(\alpha+2\beta\right)$$ Find stationary points: $$f'\left(\beta\right)=0$$ $$\sin\left(\alpha+2\beta\right)=0$$ From the graph of sine and the consideration that $0\le\alpha,\beta\le \pi$ we obtain: $$\alpha+2\beta=\pi$$ $$\beta=\frac{\pi-\alpha}{2}$$ $$\gamma=\pi-\alpha-\frac{\pi-\alpha}{2}=\frac{\pi-\alpha}{2}$$

Hence the triangle is isosceles. Now the area can be easily computed.

share|improve this answer

Consider any such triangle $\triangle A'BC$, not necessary with maximal area. Then any other such triangle must have $A$ lying on the circumcircle of $\triangle A'BC$, and on the arc $BA'C$. Now, consider the line through $A$ parallel to $BC$. For the area to be maximal, this line must be as far away from $BC$ as possible. This obviously happens when it is tangent to the circle. In this case, The diameter through $A$ is perpendicular to $BC$, and hence is the perpendicular bisector of $BC$. So $AB = AC$.

share|improve this answer

Your Answer


