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What's the radius and area of circle of max area that can be inscribed in a isoceles triangle with $2$ equal sides of length $1$? Radius formula is given, $r = \dfrac{2A}{P}$, where $A$ is area of triangle and $P$ is perimeter of triangle. I have no idea how to do this. I'm guessing the triangle might be equilateral because equilateral is also isosceles.

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  • $\begingroup$ i give up. i'm going to bed. i will look again in the morning. $\endgroup$ – Lying Cat Nov 3 '14 at 6:12
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    $\begingroup$ The one that you think helped you the most. $\endgroup$ – Ali Caglayan Nov 3 '14 at 18:02
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Computing the area and perimeter, we get $A=x\sqrt{1-x^2}$ and $P=2+2x$.

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Therefore, $$ \begin{align} r^2 &=\frac{4A^2}{P^2}\\ &=x^2\frac{1-x}{1+x}\tag{1} \end{align} $$ This reaches maximum when the derivative of $(1)$ is $0$; that is, when $$ x^3+x^2-x=0\require{cancel}\tag{2} $$ The only root of $(2)$ within the allowable range of $x\in(0,1)$ is $x=\frac{-1+\sqrt5}2=\frac1\phi$. Thus, the maximum area of the inscribed circle is $$ \begin{align} \pi r^2 &=\frac\pi{\phi^5}\\ &\doteq0.283277232857953\tag{3} \end{align} $$ Divide the area by $\pi$ and take the square root to get the radius of the circle.

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We need to optimize the area of the triangle to optimize the radius that can fit inside it . Draw an isosceles triangle with two legs of length one, and split the triangle symmetrically so you think of your triangle as two back-to-back right triangles. Then label the height $x$ and width $\sqrt{1-x^2}$. Choose one angle to be $\theta$. Then you should see that $$A = \frac{bh}{2}$$ or equivalently we can define $A$ as a function in terms of $\theta$, since we already have $b,h$ defined in terms of $\theta$. Let $$\\ A(\theta) =\ \frac{\cos(\theta)\sin(\theta)}{2} = \frac{\sin(2\theta)}{4}$$ Since $A(\theta)$ is only the area of our right triangle, we need to double it to account for the area of the entire isosceles triangle. Let $$A_I(\theta) = 2A(\theta) = \frac{\sin(2\theta)}{2}$$ where $A_I(\theta)$ is the area of the isosceles. Similarly, we can define $P$ as a function of $\theta$. The perimeter of the isosceles is $P =$ leg $1$ + leg $2$ + leg $3= 2+b_I$, where we have $b_I = 2b = 2\cos(\theta)$. Hence, $$P(\theta) = 2+2\cos(\theta)$$

Now we can define $r$ as a function of $\theta$ via the relation $$r(\theta) = \frac{A_I(\theta)}{P(\theta)} = \frac{\sin(2\theta)}{4(1+\cos(\theta))}$$ Now you can find when $r'(\theta) = 0 $ and optimize $r(\theta)$

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Let the unknown triangle's base be $2l$.
Draw a diagram and use Pythagoras' Theorem to obtain the height of the triangle as $\sqrt{1-l^2}$.
Now use the triangle's area formula to obtain the area $$\sqrt{1-l^2}\times\frac {2l}2=l\sqrt{1-l^2}$$ The perimeter of the triangle is $2+2l$.
Thus the radius of the circle is $$\frac{2l\sqrt{1-l^2}}{2+2l}=\frac{l\sqrt{1-l^2}}{1+l}$$ And the area of the circle is $$\pi\left(\frac{l\sqrt{1-l^2}}{1+l}\right)^2=\frac{\pi l^2(1-l^2)}{(1+l)^2}=\frac{\pi l^2(1-l)}{1+l}$$

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    $\begingroup$ since base is 2l, shouldn't the area of triangle be lsqrt(1-l^2) $\endgroup$ – Lying Cat Nov 3 '14 at 15:19
  • $\begingroup$ @user3290793 Why is this accepted? I don't see how it answers the maximum case :O $\endgroup$ – Sawarnik Nov 3 '14 at 15:30
  • $\begingroup$ check: $(2+2l)^2=4(1+l+l^2)$? $\endgroup$ – robjohn Nov 3 '14 at 19:01
  • $\begingroup$ @robjohn No, it isn't :/ $\endgroup$ – Sawarnik Nov 3 '14 at 19:02
  • $\begingroup$ @Turtwig Ah yes thanks my mistake $\endgroup$ – Abraham Zhang Nov 4 '14 at 2:30
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Hint: Let the other side be $L$. You should be able to find an equation for the radius of a circle inscribed in a $1-1-L$ isosceles triangle. If not, the center has to be on the bisector of the vertex angle. A little geometry and you can derive it. Then take the derivative, set to zero ....

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  • $\begingroup$ what do i take the derivative of? $\endgroup$ – Lying Cat Nov 3 '14 at 5:54
  • $\begingroup$ You equation should be $r(L)=$ something. You want $\frac {dr}{dL}=0$ $\endgroup$ – Ross Millikan Nov 3 '14 at 5:55
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First assume that $\triangle ABC$ has $AB = AC = 1$, and denote $S$ = area of $\triangle ABC$, and $P$ = perimeter of the triangle, then:

$r = \dfrac{2S}{P} = \dfrac{2\cdot \dfrac{1}{2}\cdot 1\cdot 1\cdot \sin A}{1+1+2\sin A} = \dfrac{1}{2}\cdot \dfrac{\sin A}{1+\sin A} = f(A)$.

Taking derivative of $f$ w.r.t $A$ we have: $f'(A) = \dfrac{\cos A}{2(1+\sin A)^2} = 0 \iff \cos A = 0 \iff A = \dfrac{\pi}{2}$. This shows that $A = \dfrac{\pi}{2}$ is a maxima, and thus $r_{max} = f(\dfrac{\pi}{2}) = \dfrac{1}{4}$, and the circle with maximum area equal to: $\pi\cdot \left(\dfrac{1}{4}\right)^2 = \dfrac{\pi}{16}$.

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  • $\begingroup$ i don't get it. $\endgroup$ – Lying Cat Nov 3 '14 at 6:12
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    $\begingroup$ Shouldn't it be this? :/ $$ \dfrac{1}{2}\cdot \dfrac{\sin A}{1+\sin \frac{A}2} =f(A)$$ $\endgroup$ – Sawarnik Nov 3 '14 at 9:05
  • $\begingroup$ true. I will fix it. $\endgroup$ – DeepSea Nov 3 '14 at 9:19
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    $\begingroup$ good :) but when will you fix it? :O $\endgroup$ – Sawarnik Nov 3 '14 at 9:52

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