Science:Math Exam Resources/Courses/MATH100/December 2019/Question 08

MATH100 December 2019

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[hide] Question 08
Among all rectangles inscribed in a given circle of radius $R$, which one has the largest perimeter? Prove your answer.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

[show] Hint 1

First, draw a picture of a rectangle inscribed in a circle and remember that $R$ is fixed. Also recall (or take on faith) that a rectangle inscribed in a circle has its centre at the same point as the centre of the circle. Second, notice that this is an optimization problem. Try to find a single variable that corresponds to each inscribed rectangle and use it to write down a function for the perimeter.

[show] Hint 2

Use the angle $t$ between two diagonals of a rectangle to parameterize the rectangles inscribed in the circle, as in the first diagram below. Write the perimeter of the rectangle as a function of $R$ and $t$. This is a function of one variable (since $R$ is fixed). Rectangle inscribed in a circle parameterized by angle. Alternatively, use the length of the base to parameterize the rectangle as in the second diagram. Write the perimeter of the rectangle as a function of $R$ and $b$. Rectangle inscribed in a circle parameterized by base

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[show] Solution 1

Refer to the diagram posted in the second hint above. We use the angle measurement $t$, where $0<t<\pi$. The triangle below the angle is isosceles, so the other two angles are equal, and they are equal to ${\frac {1}{2}}(\pi -t)$. Using the cosine of one of these angles, we obtain $b$ the length of the base of the rectangle as $b=2R\cos \left({\frac {1}{2}}(\pi -t)\right).$ The left height of the rectangle forms a right-angle triangle with the base and a diagonal, so we can use Pythagoras' theorem to compute its length, $h$. We computed the length of the base above and the length of the diagonal is $2R$, so the height has length $h={\sqrt {(2R)^{2}-\left(2R\cos \left({\frac {1}{2}}(\pi -t)\right)\right)^{2}}}={\sqrt {4R^{2}\left(1-\cos ^{2}\left({\frac {1}{2}}(\pi -t)\right)\right)}}=2R{\sqrt {\sin ^{2}\left({\frac {1}{2}}(\pi -t)\right)}}=2R\sin \left({\frac {1}{2}}(\pi -t)\right)$ (Note that $\sin((\pi -t)/2)$ is positive if $0<t<\pi$, so we do not need to take the absolute value when we get rid of the square root.) The perimeter of the rectangle is $P(t)=2b+2h=4R\cos \left({\frac {1}{2}}(\pi -t)\right)+4R\sin \left({\frac {1}{2}}(\pi -t)\right).$ To minimize, we compute the first derivative of $P$: ${\begin{aligned}P'(t)&=4R\left(-\sin \left({\frac {1}{2}}(\pi -t)\right)\cdot \left(-{\frac {1}{2}}\right)+\cos \left({\frac {1}{2}}(\pi -t)\right)\cdot \left(-{\frac {1}{2}}\right)\right)\\&=2R\left(\sin \left({\frac {1}{2}}(\pi -t)\right)-\cos \left({\frac {1}{2}}(\pi -t)\right)\right)\end{aligned}}$ So $P'(t)=0\iff \sin \left({\frac {1}{2}}(\pi -t)\right)=\cos \left({\frac {1}{2}}(\pi -t)\right)$, but $\sin(\theta )=\cos(\theta )$ precisely for $\theta ={\frac {\pi }{4}}+k\pi$, with $k=0,\pm 1,\pm 2,\dots$ In our case, the angle $t$ lies between 0 and $\pi$, so $0<{\frac {1}{2}}(\pi -t)<{\frac {\pi }{2}}$ meaning $k=0$. We conclude that ${\frac {1}{2}}(\pi -t)={\frac {\pi }{4}}\Rightarrow t={\frac {\pi }{2}}.$ So, among all rectangles inscribed in a circle with radius $R$, the one with the largest perimeter is the one with angle ${\frac {\pi }{2}}$ between its diagonals. In other words, it is the square.

[show] Solution 2

Let us use the length of the base to parameterize the rectangles instead, as in the second diagram above. In this case $0<b<2R$. The base, diagonal and height of the rectangle form a right-angle triangle, so Pythagoras' theorem applies. Let $h$ denote the length of the height. We have $h={\sqrt {(2R)^{2}-b^{2}}},$ so the perimeter of a rectangle with base $b$ is $P(b)=2\left(b+{\sqrt {(2R)^{2}-b^{2}}}\right)$ To maximize $P(b)$, we compute its derivative with respect to $b$ and set it equal to 0: ${\begin{aligned}P'(b)&=2\left(1+{\frac {-2b}{2{\sqrt {(2R)^{2}-b^{2}}}}}\right)\\&=2\left({\frac {{\sqrt {(2R)^{2}-b^{2}}}-b}{\sqrt {(2R)^{2}-b^{2}}}}\right)\end{aligned}}$ A ratio ${\frac {a}{b}}$ is 0 if and only if the numerator $a$ is 0, so $P'(b)=0\iff {\sqrt {(2R)^{2}-b^{2}}}-b=0\iff b^{2}=4R^{2}-b^{2}\iff b={\sqrt {2}}\cdot R$ Thus, among all rectangles inscribed in a circle of radius $R$, the one with the largest perimeter is the one whose base has length ${\sqrt {2}}R$. We now use this to determine the height ${\begin{aligned}h&={\sqrt {(2R)^{2}-b^{2}}}\\&={\sqrt {(2R)^{2}-({\sqrt {2}}R)^{2}}}\\&={\sqrt {2R^{2}}}\\&={\sqrt {2}}R.\end{aligned}}$ Since both the height and the base are the same, this is again that rectangle which is also a square.