As said in the title, I'm looking for the maximum area of a isosceles triangle in a circle with a radius $r$.
I've split the isosceles triangle in two, and I solve for the area $A=\frac{bh}{2}$*. I have made my base $x$, and solve for the height by using the Pythagorean theorem of the smaller triangle (seen in picture).
$h=r+\sqrt{r^2-x^2}$
So my the formula, I think, for both triangles should be $A=x(r+\sqrt{r^2-x^2})$
But after I solved for the derivative, when put "$= 0$", and checked on my calculator, I got the maximum to be about $4.3301r$, which differs a lot from my book's answer of $\frac{3\sqrt{3}}{4}r^2$*. Is my formula for the area right? Am I going about this the wrong way, or is it just my derivative that is wrong? Thanks in advance
* Edited from original post
$A=rx+x\sqrt{r^2-x^2}$
$A'=r+x(\frac{1}{2})(r^2-x^2)^{-\frac{1}{2}}(-2x)+\sqrt{r^2-x^2}$
$r+\sqrt{r^2-x^2}=\frac{x^2}{\sqrt{r^2-x^2}}$
$r\sqrt{r^2-x^2}+(r^2-x^2)=x^2$
$r^2+r\sqrt{r^2-x^2}=2x^2$
$r^2(r^2-x^2)=(2x^2-r^2)^2$
$r^4-r^2x^2=4x^2-4x^2r^2+r^4$
$4x^4=3x^2r^2$
$x=\frac{\sqrt{3}}{2}r$
$\endgroup$ 34 Answers
$\begingroup$Let $\theta$ be one-half of the vertex angle (less than a right angle) of the isosceles triangle.
Exercise: Show that the area of the inscribed triangle is
$A(\theta) = \dfrac{h b}{2} = \dfrac{(r + r \cos\theta)}{2} (2 r \sin\theta) = r^2 \,(1+\cos\theta) \, \sin\theta $
Differentiating $A$ and setting it to $0$, you will be left with the problem of solving
$\cos\theta = -\cos2 \theta.$
Exercise: Show that the only $\theta$ in $\left(0, \dfrac{\pi}{2}\right)$ that works is $\dfrac{\pi}{3}$.
Plugging these into $A$ we get the answer: $\dfrac{3\sqrt 3 r^2}{4}.$
$\endgroup$ $\begingroup$We can also split the triangle into three smaller triangles using $\frac{1}{2} ab \sin C$. Therefore $\Delta ABC$ equals:
$$\frac{1}{2} r^2 \big(\sin \theta + \sin \theta + \sin(360º - 2 \theta ) \big)$$$$\frac{1}{2} r^2 \big(2 \sin \theta + \sin(- 2 \theta ) \big)$$
Since the area is only dependent on $\theta$, we can take the derivative of:
$$f(\theta) = 2 \sin \theta + \sin(-2 \theta) \implies f' (\theta) = 2 \cos \theta - 2 \cos(-2\theta)$$
and setting $f'(\theta)$ equal to $0$:
$$\cos \theta - \cos(-2 \theta) = 0$$$$\Rightarrow \cos \theta - (2 \cos^2 \theta - 1) = 0$$$$\Rightarrow 2 \cos^2 \theta - \cos \theta - 1 = 0$$$$\Rightarrow \cos \theta = -\frac{1}{2}, \cos \theta = 1$$
and the only solution that makes sense in the range $\theta = (0º, 180º)$ is $\theta = 120º$.
Therefore the maximum possible area is:$$\frac{1}{2} r^2 \big(2 \sin 120º + \sin(- 2 \times 120º ) \big) = \frac{3 \sqrt3}{4} r^2.$$
$\endgroup$ $\begingroup$A slightly different perspective:
Use above drawing:
Label apex of isosceles triangle $A$, centre of circle $O$, extend $AO$ to intersect the circle in $C$. Length $AC = 2r$. Pick any point, say, on the circle's left part, call $ B$. Look at $\triangle ABC$.
$\triangle ABC$ is a right triangle (Thales circle)
Draw the height to side $AC$. Let length of height = $x$.
Let footpoint of height on $AC$ be $X$.
Height $x$ divides length $AC$ into $q$ and $ p$, where $q $ the upper part.
Area of $\triangle ABX$ is half the area of the isosceles triangle we are originally looking at.
Altitude rule for $\triangle ABC$ : $x^2 = q×p$.
With $p= 2r - q$ we get: $x^2 = q(2r - q)$.
Area $ \triangle ABX = (1/2) qx = (1/2)q(q(2r - q))^{1/2}$.
Area = $(1/2)(q^3(2r - q)^{1/2}$.
To find extremal points we look at the square of the area function above, factor 1/2 has been dropped, I.e.:
$f(q) = q^3(2r -q)$.
$f'(q) = 6rq^2 - 4q^3 = 0$.
1)$q = 0$, Area = $0$, or
2) $4q = 6r$, $q = (3/2)r$.
Maximum area of original triangle :
AREA = $2(1/2)(q^3(2r-q))^{1/2} $=
$[((3/2)r)^3(2r -(3/2)r)]^{1/2}$ =
$(3/2)r[(3/2)r(r/2)]^{1/2}$ =
$(3/4)r^2√3$.
AREA = $(3/4)√3r^2$.
$\endgroup$ $\begingroup$Could it be possible that the 2 rs in the drawing are not equal and the other r (diagonal) could be longer than the vertical r in the drawing (or vice versa). In such case, the pythagorean equation h=r + sqrt(r^2 -x^2) is not applicable.
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