3
$\begingroup$

The problem is:

If one side and the opposite angle of a triangle are fixed, and the other two sides are variable, use the law of cosines to show that the area is a maximum when the triangle is isosceles.

I know that, if I let $c$ denote the fixed side and $\theta$ the angle opposite to $c$, then area $A=1/2 ab \sin{\theta}$, where $a$ and $b$ are the variable sides. I've tried substituting $b$ by the expression for it determined by the law of cosines, but that didn't seem to get me anywhere.

$\endgroup$
3
$\begingroup$

Fix the side of length $c$. Let $C$ be the vertex of the triangle opposite to $c$. The set of points that can be chosen for $C$, forming a fixed angle $\theta$, forms a circle, such that $c$ is a chord of that circle. (Actually it forms two circles, one on each side, but we can just take one WLOG.)

The area of the triangle is $\frac12 c h$. So the question is what point $C$ on this circle maximizes the height $h$ perpendicular to $c$. Somewhat clearly, that would be the point $C$ such that the line tangent to the circle at $C$ is parallel to $c$. It follows from this that the triangle is isosceles.

Alternate approach: If you want to make your idea work, you would be trying to maximize $$ \left(\frac12 \sin \theta \right) ab $$ subject to the constraint $a^2 + b^2 - 2ab \cos \theta = c$, for a fixed $c$ and $\theta$. This is a Lagrange multipliers problem, with two variables $a$ and $b$.

$\endgroup$
  • $\begingroup$ I'm sorry, but why does angle $\theta$ remain constant as we move its vertex $C$ along the circle? That's the only thing I didn't understand. I looked it up on Google but could only find it stated as an Euclid's theorem here: math.stackexchange.com/q/1586946. Does that theorem have a name? Or can you link me to a proof? Thank you. $\endgroup$ – Sasaki Jan 10 '18 at 17:13
  • 1
    $\begingroup$ @Sasaki Try "the inscribed angle theorem" (see Wikipedia). Angle ACB is called an "inscribed angle" -- the theorem is that the measure of an inscribed angle is half of the measure of the central angle, no matter where the point C is located on the circle. The proof is on Wikipedia, or on this ProofWiki page, or you could search "inscribed angle theorem proof" for a number of other hits. $\endgroup$ – 6005 Jan 10 '18 at 20:10
  • $\begingroup$ Also, Jack's image is nice. $\endgroup$ – 6005 Jan 10 '18 at 20:10
3
$\begingroup$

If $AB$ is fixed and $\widehat{ACB}$ is fixed, the circumradius of $ABC$ is fixed as well. enter image description here

If $C$ travels on the (major, or minor) $AB$-arc of a fixed circle $\Gamma$, the distance of $C$ from $AB$, hence the area of $ABC$, is clearly maximized when $C$ lies on the perpendicular bisector of $AB$, hence when $ABC$ is isosceles with respect to the base $AB$.

$\endgroup$
1
$\begingroup$

Let $a$ be a fixed side and $\alpha$ be a fixed angle of $\Delta ABC$.

Thus, in the standard notation we obtain: $$S_{\Delta ABC}=\frac{1}{2}bc\sin\alpha=2R^2\sin\alpha\sin\beta\sin\gamma=$$ $$=R^2\sin\alpha(\cos(\beta-\gamma)-\cos(\beta+\gamma))\leq R^2\sin\alpha(1+\cos\alpha).$$ The equality occurs for $\cos(\beta-\gamma)=1$, which gives $\beta=\gamma$ and we are done!

$\endgroup$
  • $\begingroup$ I understand it up to the inequality sign. Why is the LHS less than or equal to the RHS? Besides, I understand that them being equal implies an isosceles triangle, but I can't quite see how that proves what is to be proven. Maybe there's some geometric interpretation to those quantities that I'm missing out on? Thanks very much. $\endgroup$ – Sasaki Jan 9 '18 at 22:52
  • $\begingroup$ Because $\cos(\beta-\gamma)\leq1$ and $-\cos(\beta+\gamma)=-\cos(180^{\circ}-\alpha)=\cos\alpha$. Also, we got that $R^2\sin\alpha(1+\cos\alpha)$ depends on $\alpha$ and $a$. $\endgroup$ – Michael Rozenberg Jan 10 '18 at 2:21
1
$\begingroup$

hint

Do not forget the useful equality

$$\frac {a}{\sin (\alpha)}=\frac {b}{\sin (\beta)}=\frac {c}{\sin (\gamma)} $$

$\endgroup$
1
$\begingroup$

I you use the extended version of the sine rule, with $a$ the side, $A$ the opposite angle and $R$ the circumradius, you find that $2R=\frac a{\sin A}$, so that the third vertex of the triangle lies on a fixed circle. The centre of the circle lies on the perpendicular bisector of the side you have been given (the circle passes through the two vertices you already have) and the greatest possible area is when the vertex also lies on the bisector of this side (greatest height on fixed base).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.