Minimizing Isosceles triangle with a circle inscribed

[Dethrone]

Find the smallest possible area of an isosceles triangle that has a circle of radius $r$ inside it.

I cannot seem to find the relationship between the circle and triangle. Any hints?
I'm thinking similar triangles, but I want to know if they're any other approaches before I try that.

 

[MarkFL]

The approach I used involves coordinate geometry in the first quadrant. I considered the top-right quarter of the circle having radius $r$ and center $(0,r)$, and the line through $(0,h)$ and $(b,0)$.

If we observe there is an isosceles triangle with vertices $(-b,0),\,(0,h),\,(b,0)$, then our objective function is:

\(\displaystyle A(b,h)=bh\)

Now, in order to obtain a constraint, we may consider the line and the circle I mentioned before will only have one point of intersection, and so equating them and setting the discriminant of the resulting quadratic to zero will give you a constraint. Then you may express the area of the triangle as a function of one variable, and minimize.

 

[Dethrone]

Before I hit the hay and have sweet dreams about this problem ;), I just want to make sure my next step is correct:

Assuming the line you are referring to is one of the sides of the triangle, from $(0, h)$ to $(b, 0)$, then the slope of the line is:

$$m=-\frac{h}{b}$$

The equation of the line:

$$y=-\frac{hx}{b}+h$$

Am I correct so far?

 

[MarkFL]

If you mean:

\(\displaystyle y=-\frac{h}{b}x+h\)

(which I am certain you do), then yes, you are correct. :D

 

[Dethrone]

Yes! I've changed my answer to reflect that. (Whew)

 

[Dethrone]

Is this the equation of the circle?

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

 

[MarkFL]

Is this the equation of the circle?

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

Yes...good! Since the tangent between the line and the circle must be on the upper half of the circle, can you write this upper half as a function, i.e., solve for $y$?

 

[Dethrone]

I'm not sure how to do that...

$$x^2-2yr+y^2+r^2=r^2$$
$$x^2-2yr+y^2=0$$

 

[MarkFL]

Using your original equation of the circle, subtract through by $x^2$, and then take the square root of both sides, and discard the negative root...:D

 

[Dethrone]

So we have:

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

Equating:

$$\frac{h}{b}x + \sqrt{r^2-x^2} + (r-h) =0$$

How do I proceed?

EDIT: I think I have an idea...never mind, it seems too difficult that way.

$$\sqrt{r^2-x^2}=-\frac{h}{b}x+(h-r)$$
$$r^2-x^2=\left(-\frac{h}{b}x+(h-r)\right)^2$$
$$r^2-x^2=\frac{h^2}{b^2}x^2+2(h-r)(-\frac{h}{b}x)+(h-r)^2$$

I do not think this is the quadratic you're looking for...

 

[MarkFL]

You are almost there...expand $(h-r)^2$, and then write the quadratic in standard form.

 

[Dethrone]

So it was the quadratic you were looking for :D

$$0=\left(\frac{h^2}{b^2}+1\right)x^2+2(h-r)(-\frac{h}{b}x)+h^2-2hr$$

Am I suppose to solve the quadratic?

 

[MarkFL]

No need to solve...what can we say must be true of the discriminant and why?

 

[Dethrone]

The discriminant must be equal to 0 for there to be one point of intersection.

 

[MarkFL]

The discriminant must be equal to 0 for there to be one point of intersection.

Exactly! (Yes)

This requirement will give you a relationship between $h$ and $b$ (our constraint), and allow you to express the area of the triangle as a function in one variable. :D

What do you find the constraint to be?

 

[Dethrone]

Am I to set $b^2-4ac=0$?

 

[MarkFL]

Am I to set $b^2-4ac=0$?

Yes, and I chose to solve for $b$...(our $b$, not the $b$ in the formula for the discriminant).

 

[Dethrone]

This is very difficult to isolate, is there a trick? I've been asking you to confirm every step because of this nasty algebra.

$$b^2-4ac=0$$
$$\left((2)(h-r)(-\frac{h}{b})\right)^2-4\left(\frac{h^2}{b^2}+1\right)\left(h^2-2hr\right)=0$$

 

[MarkFL]

Here is the expression I wrote for the discriminant being zero and the next step:

\(\displaystyle \left(\frac{2h(h-r)}{b}\right)^2-4\left(\frac{h^2+b^2}{b^2}\right)\left(h^2-2hr\right)=0\)

Multiply through by \(\displaystyle \frac{b^2}{4h}\):

\(\displaystyle h(h-r)^2-\left(h^2+b^2\right)(h-2r)=0\)

 

[Dethrone]

I forgot that was a legal move, but it works because the other side is 0. But, I guess we are allowed to divide by $h$ because we're trying to isolate for $b$ and missing solutions of $h$ isn't an issue. (dividing by a variable can cause missing solutions, right?)

Would isolating for $h$ work too? I guess for obtaining a constraint equation, solving the quadratic equation is unnecessary, since we don't need the know where they intersect.

So far, I have:

$$b^2=\frac{(h)(h-r)^2}{h^2-2hr}-h^2$$

We are only dealing with $b>0$
$$b=\sqrt{\frac{(h)(h-r)^2}{h^2-2hr}-h^2}$$

 

[MarkFL]

I forgot that was a legal move, but it works because the other side is 0. But, I guess we are allowed to divide by $h$ because we're trying to isolate for $b$ and missing solutions of $h$ isn't an issue. (dividing by a variable can cause missing solutions, right?)

We know we must have $0<2r\le h$, so as long as we know $h$ cannot be zero, it is fine to divide though by it.

Would isolating for $h$ work too? I guess for obtaining a constraint equation, solving the quadratic equation is unnecessary, since we don't need the know where they intersect.

Yes, we could solve for $h$, but I chose $b$ for simplicity.

So far, I have:

$$b^2=\frac{(h)(h-r)^2}{h^2-2hr}-h^2$$

We are only dealing with $b>0$
$$b=\sqrt{\frac{(h)(h-r)^2}{h^2-2hr}-h^2}$$

You don't quite have $b^2$ correct...here's what I did:

Spoiler

\(\displaystyle b^2=\frac{h(h-r)^2}{h-2r}-h^2\)

\(\displaystyle b^2=\frac{h(h-r)^2-h^2(h-2r)}{h-2r}\)

\(\displaystyle b^2=\frac{h\left(h^2-2hr+r^2\right)-h^2(h-2r)}{h-2r}\)

\(\displaystyle b^2=\frac{h^3-2h^2r+hr^2-h^3+2h^2r}{h-2r}\)

\(\displaystyle b^2=\frac{hr^2}{h-2r}\)

Since $0<b$, we obtain:

\(\displaystyle b=\sqrt{\frac{hr^2}{h-2r}}\)

 

[Dethrone]

But if you had $x^2+2x=0$, and you divided both sides by $x$, you would end up missing a solution to the equation, right? In our case though, we were trying to find $b$, and so missing solutions of $h$ wasn't an issue, right? Dividing by $h$, then $h$ cannot be equal to 0, but the missing solutions is still a problem if we were to isolate for $h$ instead...

Back to the problem:

You're right, I forgot to divide an $h$ from the second part.

$$\displaystyle b=\sqrt{\frac{hr^2}{h-2r}}$$

$$A(b,h)=\frac{bh}{2}$$
$$A(h) = \frac{h\sqrt{\frac{hr^2}{h-2r}}}{2}$$
$$A'(h)=0=3h^2r^2(h^3-2rh^2)-(3h^2-4rh)(h^3r^2)$$

Am I correct up this point?

 

[MarkFL]

But if you had $x^2+2x=0$, and you divided both sides by $x$, you would end up missing a solution to the equation, right? In our case though, we were trying to find $b$, and so missing solutions of $h$ wasn't an issue, right? Dividing by $h$, then $h$ cannot be equal to 0, but the missing solutions is still a problem if we were to isolate for $h$ instead...

The only solution we could potentially lose is $h=0$, but we know $h\ne0$, so there is no issue here.

Back to the problem:

You're right, I forgot to divide an $h$ from the second part.

$$\displaystyle b=\sqrt{\frac{hr^2}{h-2r}}$$

$$A(b,h)=\frac{bh}{2}$$
$$A(h) = \frac{h\sqrt{\frac{hr^2}{h-2r}}}{2}$$
$$A'(h)=0=3h^2r^2(h^3-2rh^2)-(3h^2-4rh)(h^3r^2)$$

Am I correct up this point?

Recall the area of the isosceles triangle is twice that of the first quadrant right triangle, so the objective function is:

\(\displaystyle A(b,h)=bh\)

I would look at optimizing the square of the objective function for simplicity, since they must have the same critical value. I'm not certain what you are doing with your differentiation.

 

[Dethrone]

Right...so...

$$A^2(h)=\frac{hr^2}{h-2r}h^2=\frac{h^3r^2}{h-2r}$$
$$\left(A^{2}(h)\right)^{'}=\frac{(2r^2h^2)(h-3r)}{\left(h-2r\right)^2}$$
$$h=3r$$

Sub this back into the Area of the isosceles triangle:
$$A=\sqrt{\frac{hr^2}{h-2r}h^2}h$$
$$A=3\sqrt{3}r^2$$

I'm pretty sure this is the right answer!

I just looked this question up, and someone does this question with another method:
https://ca.answers.yahoo.com/question/index?qid=20071123210841AAjodSe

I don't follow it, can you explain how he worked out the formula for the area?

 

[MarkFL]

Yes, that's the correct answer. Now, a couple of follow up questions...

a) What is the significance of the critical value (in the denominator) of $h=2r$.

b) For the result you cite, what type of triangle do we have? (never mind, I see this was stated in the yahoo link)

As far as what the poster at Yahoo did, they used the fact that the line through the center of the circle and the tangent point must be perpendicular to the line on which the hypotenuse of the right triangle with which we worked in the first quadrant. I have never heard of Prognathous, but I assume he meant Pythagoras.

So, let's look at a diagram:

View attachment 2990

Now the post at yahoo should make more sense. :D

 

[Dethrone]

What I didn't understand about the post was why the perpendicular line from the center of the circle cuts the furthest side of the triangle into $b$ and $s-b$. How do we know that it is length $b$?

I'm not sure about the significant of $h=2r$, but I do know that is impossible since the height of the triangle can't be equal to the height of the circle. That would be the case where the triangle is inside the circle.

https://www.physicsforums.com/attachments/2990

 

[MarkFL]

What I didn't understand about the post was why the perpendicular line from the center of the circle cuts the furthest side of the triangle into $b$ and $s-b$. How do we know that it is length $b$?

Look at the pair of acute angles in the two triangles I have asserted are congruent. How can we know the corresponding angles are the same?

I'm not sure about the significant of $h=2r$, but I do know that is impossible since the height of the triangle can't be equal to the height of the circle. That would be the case where the triangle is inside the circle.

Picture $h$ getting smaller and smaller, approaching $2r$...what happens to $b$? What conclusion then can you draw about the area of the triangle?

 

[Dethrone]

The two acute triangles are right triangles and they both share one side, which we call $h$, the hypotenuse. Then we know, by Pythagoras, that $h^2=r^2+b^2$ for the bottom triangle, and for the top triangle we know $h^2-r^2=a^2$. $a=b$. This also means that the angles are equal. I'm not sure if there's any other way to think of it.

I'm going to give a rash answer that when $h$ approaches $2r$, $b$ (by my imagination and by the formula) decreases until the triangle becomes a line! Hence, the area of the triangle becomes the area of the line (Dull)

On second though, $b$ increases to infinity, so we have an infinity area.

 

[MarkFL]

The two acute triangles are right triangles and they both share one side, which we call $h$, the hypotenuse. Then we know, by Pythagoras, that $h^2=r^2+b^2$ for the bottom triangle, and for the top triangle we know $h^2-r^2=a^2$. $a=b$. This also means that the angles are equal. I'm not sure if there's any other way to think of it.

Actually, all we need to observe is that yes, they are both right triangles, they both share two sides, so the third side in both must also be the same. And so we know the two triangles are congruent.

I'm going to give a rash answer that when $h$ approaches $2r$, $b$ (by my imagination and by the formula) decreases until the triangle becomes a line! Hence, the area of the triangle becomes the area of the line (Dull)

No, the triangle doesn't become a line, it's height approaches $2r$, which is presumably non-zero. As $h$ shrinks, $b$ grows, and by the time $h$ approaches $2r$, $b$ is growing without bound. Thus, the significance of $h=2r$ as a critical value is that the maximal area of the triangle is unbounded. We may place $h$ arbitrarily close to $2r$ to make $b$ as large as we want.

 

[Dethrone]

So how would we compute the maximum area of the triangle?

 

[MarkFL]

So how would we compute the maximum area of the triangle?

If $b$ is unbounded, then so is the area of the triangle. It has no real maximum, we can make it as large as we want. :D

 

[Dethrone]

I found another solution to this problem, but very similar.

View attachment 3007

Can someone explain how that's similar triangles? I can kind-of visualize it if you flip it...

 

[MarkFL]

They are both right triangles, and they share an acute angle, so we know they have the same 3 internal angles, and so they must be similar. :D

 

[I like Serena]

I found another solution to this problem, but very similar.

View attachment 3007

Can someone explain how that's similar triangles? I can kind-of visualize it if you flip it...

Yep. Flip it and you can visualize it! (Mmm)
It means that the smaller triangle can be scaled to the bigger triangle (after flipping).
And that means that the ratio of the sides next to the right angle in both triangles is the same.
In the big triangle the ratio is $y:x$, while in the small triangle the ratio is $r : \sqrt{x^2-2rx}$.Alternatively, an equilateral triangle will have an extreme area due to its symmetry. Break the symmetry in any direction and the area will change.
Since you can also make the triangle as big as you want, by choosing a basis that is either large enough, or small enough, it follows that the equilateral triangle will have minimum area. (Nerd)

 

[Dethrone]

I see now that the acute angle in both right triangles are the same...but it wasn't immediately obvious. I had to work out (arbitrary) angles in my head. Anyway to quickly notice that they're similar without working out angles? I know the clue is that they're both right triangles.