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Find the magnitude of the vertex angle $\alpha$ of an isosceles triangle with the given area $A$ such that the radius $r$ of the circle inscribed into the triangle is maximal.


My attempt:
enter image description here

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  • $\begingroup$ By the way, I think your first area is wrong. You may also want to check en.wikipedia.org/wiki/…. Complete answer to follow $\endgroup$ – David Winton Jun 30 '17 at 14:51
  • $\begingroup$ I've taken the freedom to rotate and crop your image. Hope you appreciate. If not, you can do a rollback of the previous version. The original is quite annoying on an iPad, because it rotates the image every time you try to take a look at it. $\endgroup$ – Han de Bruijn Jul 1 '17 at 14:53
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+100
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We'll prove that for all triangle $$r\leq\sqrt{\frac{A}{3\sqrt3}}.$$

Indeed, let $AB=c$, $AC=b$ and $BC=a$.

Hence, $A=\frac{1}{2}ra+\frac{1}{2}rb+\frac{1}{2}rc$, which gives $r=\frac{2A}{a+b+c}$.

Thus, we need to prove that $$\frac{2A}{a+b+c}\leq\sqrt{\frac{A}{3\sqrt3}}$$ or $$12\sqrt3A\leq(a+b+c)^2.$$ But by Heron formula we have $$A=\sqrt{p(p-a)(p-b)(p-c)},$$ where $p=\frac{a+b+c}{2}$.

Thus, we need to prove that $$(a+b+c)^2\geq12\sqrt3\sqrt{\frac{a+b+c}{2}\cdot\frac{a+b-c}{2}\cdot\frac{a+c-b}{2}\cdot\frac{b+c-a}{2}}$$ or $$(a+b+c)^3\geq27(a+b-c)(a+c-b)(b+c-a)$$ or $$\frac{(a+b-c)+(a+c-b)+(b+c-a)}{3}\geq\sqrt[3]{(a+b-c)(a+c-b)(b+c-a)},$$ which is AM-GM.

The equality occurs for $$a+b-c=a+c-b=b+c-a$$ or $$a=b=c,$$ which says that $r$ gets a maximal value, when our triangle is an equilateral triangle,

which says that $\measuredangle BAC=60^{\circ}$.

Done!

If the following is obvious for you, then I am ready to delete it.

The equality occurring in our AM-GM just for $a+b-c=a+c-b=b+c-a$ we can understand by the following way.

Let $a+b-c=x^3$, $a+c-b=y^3$ and $b+c-a=z^3$.

Hence, $x$, $y$ and $z$ are positives and the equality case gives $$\frac{x^3+y^3+z^3}{3}=xyz$$ or $$x^3+y^3+z^3-3xyz=0$$ or $$x^3+3x^2y+3xy^2+y^3+z^3-3x^2y-3xy^2-3xyz=0$$ or $$(x+y)^3+z^3-3xy(x+y+z)=0$$ or $$(x+y+z)((x+y)^2-(x+y)z+z^2)-3xy(x+y+z)$$ or $$(x+y)^2-(x+y)z+z^2-3xy=0$$ or $$x^2+y^2+z^2-xy-xz-yz=0$$ or $$2x^2+2y^2+2z^2-2xy-2xz-2yz=0$$ or $$(x-y)^2+(x-z)^2+(y-z)^2=0$$ or $$x=y=z,$$ which gives $a=b=c$.

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Suppose that the base of our trangle is $2a$ then the coresponding height is $A/a$ and the length of the side is $b=\sqrt{a^2+A^2/a^2}$, and finally the semi perimeter of the triangle is $$p= a+\sqrt{a^2+\frac{A^2}{a^2}}$$ Now because $pr=A$ the maximal value of $r$ corresponds to the smallest value of $p$. So, we need to find the minimum of $p$ as function of $a$.

To simpify things we introduce a new variable $x=a/\sqrt{A}$ so that $$f(x)=\frac{p}{\sqrt{A}}= x+\sqrt{x^2+\frac{1}{x^2}}$$ Clearly,

$$\eqalign{f'(x)&=1+\frac{x^2-1/x^2}{\sqrt{x^4+1}}=\frac{\sqrt{x^4+1}+x^2-1/x^2}{\sqrt{x^4+1}}\cr &=\frac{3x^4-1}{x^4\sqrt{x^4+1}(\sqrt{x^4+1}-x^2+1/x^2)} }$$ Thus $f(x)$ is decreasing on $(0,1/\sqrt[4]{3})$ and increasing on $(1/\sqrt[4]{3},\infty)$. Thus $f$ attains its minimum when $x=1/\sqrt[4]{3}$. The minimum corresponds to $$ 2a=2\sqrt{\frac{A}{\sqrt{3}}},\qquad b=\sqrt{a^2+A^2/a^2}=2a.$$ So, our triangle when $r$ is maximum is equilateral, and the desired apex is equal to $\frac{\pi}{3}$.

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  • $\begingroup$ If $h$ is the height of the isosceles triangle, then your last part can be replaced by: $$ A = h\, a \quad \Longrightarrow \quad h = \frac{A}{a^2}a = \frac{1}{x^2} a = \sqrt{3}\, a $$ From which it is clear with Pythagoras that $b = 2a$ and the triangle is equilateral. Good (+1) answer though. $\endgroup$ – Han de Bruijn Jul 1 '17 at 19:36
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We know that every triangle has a unique incircle; we also have the following 2 theorems:

Theorem 1: Among all triangles of given perimeter, the equilateral one has the largest area.

Theorem 2: The radius $r$ of the incircle for a triangle $\triangle ABC$ is given by
$ r = 2 \frac{Area(\triangle ABC) } {Perimeter(\triangle ABC)}$

Let $A > 0$ and restrict our focus to all triangles satisfying $Area(\triangle ABC) = A$.

By Theorem 2, if we have any two such triangles, the one with the smaller perimeter will give us a larger incircle radius $r$.

Let $\mathcal T$ be any triangle with area $A$ and suppose it is not equilateral. By Theorem 1, the equilateral triangle with the same perimeter has a greater area. We can scale this equilateral triangle down to a triangle $\mathcal E$ where $Area(\mathcal E) = A$ and $Perimeter(\mathcal E) < Perimeter(\mathcal T)$.

Since this equilateral triangle $\mathcal E$ is an isosceles triangle, we arrive at the answer:

$\alpha = 60 °$.

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0
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If you draw segments from the center of the circle to each vertex, you cut the triangle into 3 smaller triangles. Since the radii are perpendicular to the sides, the heights of all 3 triangles is $r$. Let the length of the sides adjacent to $\alpha$ be $b$. Then 2 of the triangles have area $br/2$. The side opposite $\alpha$ has length $2b\sin(\alpha/2)$ so the area of the 3rd triangle is $br\sin(\alpha/2)$. The total area, then is

$$A=\frac{br}{2}+\frac{br}{2}+br\sin\left(\frac{\alpha}{2}\right)=br\left((1+\sin\left(\frac{\alpha}{2}\right)\right).$$

We need to express $b$ in terms of $r$. Another espression for the area is $(1/2) b^2\sin(\alpha/2)\cos(\alpha/2).$ Solving for $b$ and plugging in the above expression and solving for $r$ gives

$$r=\frac{A}{\sqrt{\frac{2A}{\sin \alpha}}\left(1+\sin(\alpha/2)\right)}.$$

So now we "simply" maximize $r$. This is a computational mess. Take the derivative of $r$ with respect to $\alpha$. Set the junk equal to zero and you get $\pi/3$. But I did this last bit with Maple. It looks do-able by hand by a very patient person.

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As a primer, there's a section in https://en.wikipedia.org/wiki/Incircle_and_excircles_of_a_triangle#Incircle_and_its_radius_properties with the answer to this question. For the rest of the question I use $\alpha$ to be $\frac{\alpha}{2}$ for simplicity of reading.

We start by considering the area in terms of $\alpha$. $$A = \frac{base * height}{2}= \frac{a*sin(\alpha) * a * cos(\alpha)}{2} = \frac{a^2sin(2\alpha)}{4}$$ where the last step uses the sine double angle formula.

Now, we consider the area in terms of the radius of the incircle. You can get better details in the wikipedia page, but we're going to decompose the triangle into three smaller triangles by drawing lines from their edges to the center of the circle. Because the circle is tangent to each of the edges, we know the radius that connects the center to each edge meets the edge at a right angle and consequently the height of each of these three subtriangles is r. Therefore, the over all area is $$A = \sum_{subtriangles} A_{subtriangle} = (\frac{a*r}{2} + \frac{a*r}{2} + \frac{base*r}{2}) = \frac{r}{2}(2a + 2a*sin(\alpha))$$

So we have $$\frac{a^2sin(2\alpha)}{4} = \frac{r}{2}(2a + 2a*sin(\alpha))$$ which implies $$ r = \frac{a^2sin(2\alpha)}{4a(1+sin(\alpha))}$$ We set a = 1 arbitrarily because scale is irrelevant in the problem and take a derivative with respect to $\alpha$. Unfortunately at this point I get stumped and turn to W|A, but it seems other answers use maple so I feel less bad at this point.

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  • $\begingroup$ I see I conflated alpha and alpha over 2 which is why I get half the answer I should. $\endgroup$ – David Winton Jun 30 '17 at 15:17
-1
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It is well known (and easy to prove via the convexity of $\tan$) that a triangle of smallest area circumscribed to a circle $C$ of radius $r>0$ has to be equilateral. A fortiori an isosceles triangle of smallest area circumscribed to $C$ has to be equilateral. It follows that ${{\rm area}(\triangle_{\rm iso})\over r^2}$ is minimal for an equilateral triangle; hence ${r^2\over {\rm area}(\triangle_{\rm iso})}$ is maximal when the angle at the vertex is $60^{\circ}$.

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