5
$\begingroup$

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).

$h=r+\sqrt{r^2-x^2}$

enter image description here

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

$A=rx+x\sqrt{r^2-x^2}$

$A'=r+x(\frac{1}{2})(r^2-x^2)^{-\frac{1}{2}}(-2x)+\sqrt{r^2-x^2}$

$r+\sqrt{r^2-x^2}=\frac{x^2}{\sqrt{r^2-x^2}}$

$r\sqrt{r^2-x^2}+(r^2-x^2)=x^2$

$r^2+r\sqrt{r^2-x^2}=2x^2$

$r^2(r^2-x^2)=(2x^2-r^2)^2$

$r^4-r^2x^2=4x^2-4x^2r^2+r^4$

$4x^4=3x^2r^2$

$x=\frac{\sqrt{3}}{2}r$

$\endgroup$
3
  • 1
    $\begingroup$ Your formula for the area of the isosceles triangle in terms of $x$ and $r$ is correct, but the value you got for the maximum area is definitely incorrect (you can immediately tell that it must be wrong because the area should be proportional to $r^2$, not $r$). It would be easier to diagnose where you made a mistake if you posted your work. $\endgroup$
    – David H
    Dec 7, 2014 at 18:27
  • 2
    $\begingroup$ That looks good. Substitute that value in $A$ to get the answer. $\endgroup$
    – Macavity
    Dec 7, 2014 at 19:00
  • 2
    $\begingroup$ In general, the polygon with the greatest area inscribed in a circle is a regular polygon. In this case, we are dealing with an equilateral triangle. Is such a triangle isosceles ? $\endgroup$
    – Lucian
    Dec 7, 2014 at 22:21

4 Answers 4

1
$\begingroup$

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: $\dfrac{3\sqrt 3 r^2}{4}.$

$\endgroup$
1
$\begingroup$

enter image description here

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: $$\frac{1}{2} r^2 \big(2 \sin 120º + \sin(- 2 \times 120º ) \big) = \frac{3 \sqrt3}{4} r^2.$$

$\endgroup$
0
$\begingroup$

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 = $(3/4)√3r^2$.

$\endgroup$
-1
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ $r$ is just the radius of the circle. $\endgroup$ Nov 30, 2019 at 2:30

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .