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I came across a problem from SASMO(Singapore) Olympiad sample paper. Here is the problem:

enter image description here

I gave this problem to my students and made them solve. Happily, they have solved and told the answer as $22$cm.

Now comes the Tragedy!!!

I asked them to find the length of $OR$. So let us find now.

It is evident that $$RP=OQ=7$$

Let $OR=x \Rightarrow OP=10-x$. Now by Pythagoras theorem, we get $$x^2+(10-x)^2=49 \Rightarrow 2x^2-20x+51=0$$ which has complex roots. Boom!

EDIT: Well then what is the moral? We need to find Maximum perimeter of rectangle that can be inscribed in a quarter circle of radius $7$cm.

Let the Perimeter be $2\lambda$

We have $$OR^2+RP^2=49 \Rightarrow x^2+(\lambda-x)^2=49 \Rightarrow \lambda=x+\sqrt{49-x^2}$$

Take $x=7\sin t$, we get $$\lambda=7(\sin t+\cos t) \leq 7\sqrt{2} \Rightarrow 2\lambda \leq 14\sqrt{2}<20$$

Ekaveera Gouribhatla
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  • Maybe I miss something but why $OR = x \Rightarrow OP = 10-x$? – Qise Mar 30 '23 at 12:32
  • @Qise length+breadth=$10$, since perimeter is $20$ – Ekaveera Gouribhatla Mar 30 '23 at 12:36
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    The hypothesis of the exercise was impossible, that's all. – Anne Bauval Mar 30 '23 at 12:43
  • @AnneBauval this is what I suspected also. Can you point out what is not working exactly? – Qise Mar 30 '23 at 12:47
  • Exactly what you wrote. The perimeter of the rectangle is too large to fit inside the quarter of disc. – Anne Bauval Mar 30 '23 at 12:49
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    The biggest (by perimeter) rectangle that can be inscribed in the way you depict is the square, and the perimeter of that is $<20$. How were they meant to get $22$? – lulu Mar 30 '23 at 12:49
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    @lulu Assume the geometry is possible. The perimeter of the quarter circle is $11+7+7=25$ (with $\pi=22/7$). The diagonal of the rectangle $(7)$ is $3$ less than its half-perimeter $(10)$ giving the answer as $22$. – nickgard Mar 30 '23 at 12:58
  • @nickgard Ah, thanks. I couldn't get past the missing $\pi$. – lulu Mar 30 '23 at 12:59
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    There's a famous (if somewhat insulting) "impossible figure" problem, along these lines, which your students might find entertaining, here. Good lesson in the dangers of mechanically applying formulas, if nothing else. – lulu Mar 30 '23 at 13:34

2 Answers2

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Yes, boom.

The objective may not actually have been to get numbers. It may have been to test critical thinking by checking that the conditions are in fact realistic. In fact a rectangle inscribed in a quarter-circle as shown must have a perimeter less than or equal to that of a square in that position, which we can easily work out as $4(r/\sqrt2)=r\sqrt8$. But your perimeter/radius ratio is $\sqrt{400/49}$ and $8$ is only $392/49$.

See also this answer, and maybe do a little light reading.

enter image description here Illustration source

Oscar Lanzi
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Take the sides of the rectangle to be $a,b.$

Then, $98=2a^2+2b^2\ge (a+b)^2=100.$ Impossible.

Umesh Shankar
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