# Optimization—Finding the Area of the Largest Isoceles Triangle

I managed to solve $(a)$. Since the area of a triangle is determined by $\frac{1}{2}$ base $\times$ height, and we already know the height, we just have to solve for the base. Using Pythagorean theorem, we can deduce base as $\sqrt{36-h^2}$. Therefore, the area is $(6+h)(\sqrt{36-h^2})$. To find $h$, we will take the derivative of the area equation and set that equal to zero. It turns out $h = 3$. Plugging $h$ into the area equation gives us an area of $27\sqrt{3}$.

Can you guys help me do $(b)$? I believe for $(c)$, the largest type of triangle of maximum area is an equilateral.

Thanks.

• I am answering the question for you right now, however do we want to incorporate $h$ as a variable for $(b)$ and $(c)$ as well or apply the constant? – Julian Rachman Jan 7 '15 at 6:38
• And what is $(c)$ really asking for? – Julian Rachman Jan 7 '15 at 6:44
• I dont't believe we can incorporate h as a variable of (b) as it asks for area in terms of just alpha, and not alpha and $h$. (c) is referring to the shape of the triangle (isosceles, equilateral, etc). – Sentient Jan 7 '15 at 6:46

For solving $(b)$, we can find the total height $(h_T)$ of the isosceles triangle by adding the value for $h$ found in $(a)$ to the other height $6$:

$$h_T=6+3=9.$$

So, we can say that the total height of the triangle is $9$ units. Then we can find the base:

$$b=\sqrt{36-3^2}=\sqrt{36-9}=\sqrt{27}=3\sqrt{3}.$$

We can then divide the base by $2$ to get the length of the base for half the triangle (by the $AAS$ Similar Triangles Theorem) to get the base to be $\frac{3\sqrt{3}}{2}$. We can now focus on the right half of the triangle divided by the dotted line going through the vertex point of the circle. We can tell that the dotted line through the vertex makes a right angle. Thus, we can use Trigonometry to solve for $\alpha$. So,

$$\alpha=\tan\frac{\text{opposite}}{\text{adjacent}}=\tan\frac{\frac{3\sqrt{3}}{2}}{9}=\tan\frac{27\sqrt{3}}{2}.$$

Since we have found what $\alpha$ is, we can find the function of it with respect to $\alpha$:

$$\boxed{\alpha=\tan\left(\frac{(6+h)(\sqrt{36-h^2})}{2}\right)},$$

or as an incorporated function:

$$\frac12bh=\frac12(6\sin\alpha)(12\cos\alpha)=\boxed{36\sin\alpha\cos\alpha}.$$

And we are done with $(b)$!

For $(c)$, By the Isoperimetric Theorem, it states

Theorem: Among all triangles inscribed in a given circle, the equilateral one has the largest area.

Therefore, the equilateral triangle has the maximum area.

Part b looks to be the tough part and c would likely quickly follow from it. It is possible, however, to verify your guess from what you've already done. One of the legs of that lower right triangle is $h$ which is equal to $3$. The hypoteneuse has length $6$. Therefore, the angle in between is $60$ degrees. The supplementary angle is then $120$ and the other $2$ angles of the isosceles triangle on the right are $30$ each. That makes the angles of the large triangle $60$ degrees each.

The only thing I can think of for part b is to cut the larger triangle into several smaller ones. If you draw an altitude from the center of the circle to the base of the isosceles triangle right of center, you can calculate the lengths of the base and height through trigonometric formulae given $\alpha$ and the length of the radii. So the isosceles portion at the right has area of

$$\frac12bh=\frac12(6\sin\alpha)(12\cos\alpha)=36\sin\alpha\cos\alpha$$

That just leaves the area of the lower right triangle. Once that's calculated, double everything to account for the congruent left half of the big triangle. For this, either use the inscribed angle theorem (had to look up the name) or just note that the isosceles triangle on the right has angles of $\alpha,\alpha$ and $180-2\alpha$, making the supplementary angle $2\alpha$.