Is this Geometry Feasible?

Question

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

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

Answer 1

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

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

Answer 2

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.

Illustration source