A Circle in a Square, a 5 by 10 Rectangle, and the Root You Throw Away

One radius drawn to the rectangle's far corner turns the whole problem into a right triangle with legs R - 10 and R - 5, and the quadratic that follows has roots 5 and 25. Both satisfy the equation exactly, so rejecting 5 takes geometry rather than arithmetic: at that radius the corner really does touch the circle while the rectangle has already swallowed half the disk. The general a by b rectangle shows the discarded root is a permanent feature of squaring.

A circle is inscribed in a square, so it touches all four sides and its radius is half the side. A 5 by 10 rectangle wedges into one corner with two of its sides lying along the sides of the square, and its far corner just reaches the circle. Find the side of the square.

The side is 50. What makes the problem worth writing about is not the answer but the second root, which satisfies the equation perfectly and describes a picture that does not exist.

Putting the picture in coordinates

The configuration has one degree of freedom, so name it and the rest follows. Put the square’s corner at the origin with its sides along the axes. The square is then $[0, 2R] \times [0, 2R]$ , the inscribed circle has centre $(R, R)$ and radius $R$ , and the rectangle occupies $[0, 10] \times [0, 5]$ , ten units along the bottom and five up the side. Its far corner is the point $(10, 5)$ , and touching the circle means that point sits at distance exactly $R$ from the centre.

\bigl\|(10, 5) - (R, R)\bigr\|^2 = R^2

(1)

Which way round the rectangle lies makes no difference. Swap the 10 and the 5 and equation (1) comes out the same, because the expression is symmetric in the two side lengths, so fixing one orientation costs nothing.

One radius does the whole job

Draw the line the problem is built around: the radius from the centre to the touching corner. With the horizontal and vertical offsets it makes a right triangle whose legs you can read straight off the picture. Horizontally, the centre sits at $x = R$ and the corner at $x = 10$ , so that leg is $R - 10$ . Vertically the centre is at $y = R$ and the corner at $y = 5$ , so that leg is $R - 5$ . The hypotenuse is the radius itself, which is what makes the triangle worth drawing: the unknown appears in all three sides at once.

(R - 10)^2 + (R - 5)^2 = R^2

(2)

Fig. 1 — Drawn at the real proportions. The rectangle is small, which is exactly why the misreading of the problem looks reasonable.

The quadratic, and both of its roots

Expand, cancel one $R^2$ from each side, and the equation collapses to something monic and small.

R^2 - 30R + 125 = 0 \qquad \Longleftrightarrow \qquad (R - 5)(R - 25) = 0

(3)

The discriminant is $900 - 500 = 400$ , a perfect square, so both roots are exact integers and nothing has been rounded away. The factorisation shows there is no third root hiding anywhere. Two candidates, 5 and 25, and the equation holds no opinion about which one you were asking for.

That is the part worth slowing down for. $R = 5$ is not a near miss produced by sloppy arithmetic. Substitute it and the identity is exact:

R = 5: \quad (5 - 10)^2 + (5 - 5)^2 = 25 + 0 = 25 = 5^2

(4)

Why 5 is a root and still impossible

So the rejection cannot be arithmetical. Squaring destroyed a sign. Pythagoras was applied to legs of length $|R - 10|$ and $|R - 5|$ , and at $R = 5$ the first of those is $R - 10 = -5$ . Geometrically that puts the corner on the far side of the centre rather than between the centre and the square’s edge, which is a different picture wearing the same equation.

Extraneous root

An extraneous root satisfies the equation you wrote down and not the problem you meant. Squaring is the usual way to acquire one, because $t^2$ forgets the sign of $t$ . The algebra cannot detect the loss, so the check has to be made against the original configuration rather than against the equation.

Look closely at what equation (1) claims. It says the corner $(10, 5)$ lies on the circle. At $R = 5$ that is true: the disk has centre $(5, 5)$ and radius 5, and the distance from $(5, 5)$ to $(10, 5)$ is exactly 5. The corner really does sit on the circle. What the equation never said is that the rest of the rectangle stays outside the disk.

And at that radius it does not. The square is only 10 wide, so a 5 by 10 rectangle covers its entire lower half, disk included. One point is enough to settle it. Take $(5, 1)$ : its distance to the centre is 4, so it is strictly inside the circle, and it is strictly inside the rectangle too. The rectangle cuts through the circle instead of touching it, and nothing in the arrangement resembles a corner being wedged.

Fig. 2 — The rejected root, drawn honestly. The corner does touch the circle, and the rectangle has already swallowed half of it.

The answer, and a triangle worth recognising

That leaves $R = 25$ and a side of $2R = 50$ . Before moving on, look again at the triangle: legs 15 and 20 against a hypotenuse of 25. That is the 3-4-5 triangle scaled by five, so the quadratic was never strictly necessary. Anybody who recognises the triple checks $15^2 + 20^2 = 25^2$ and stops.

R = 25, \qquad 15^2 + 20^2 = 625 = 25^2, \qquad \text{side} = 2R = 50

(5)

The problem hides that neatly. The legs are $R - 10$ and $R - 5$ , so the triple only becomes visible once you know $R$ , and you only know $R$ once the quadratic is solved. The recognisable structure sits on the far side of the work rather than at the entrance to it, which is a reasonable description of most problems of this kind.

Any rectangle in the corner

Replace 5 and 10 by positive $a$ and $b$ and the derivation runs unchanged.

(R - a)^2 + (R - b)^2 = R^2 \iff R^2 - 2(a + b)R + (a^2 + b^2) = 0

(6)

R = (a + b) \pm \sqrt{2ab}

(7)

The discriminant works out to $(a+b)^2 - (a^2+b^2) = 2ab$ , positive for every genuine rectangle, so there are always two distinct real roots. The ambiguity we just resolved is permanent rather than an accident of these numbers. With $a = 10$ and $b = 5$ the square root is $\sqrt{100} = 10$ , giving $15 \pm 10$ and the two roots from before.

Here is where the general case earns its keep. For the corner $(a, b)$ to be the nearest point of the rectangle to the centre, the centre has to lie beyond the rectangle in both coordinates, which says $R \ge a$ and $R \ge b$ . Test the smaller root against that condition, taking $a \ge b$ without loss of generality.

R_- \ge \max(a,b) \iff \min(a,b) \ge \sqrt{2ab} \iff \min(a,b) \ge 2\max(a,b)

(8)

No pair of positive lengths satisfies the right hand side. The smaller root is therefore impossible for every rectangle, not merely for this one, and the discarded root turns out to be a structural feature of squaring rather than a coincidence of 5 and 10. The larger root always works, since $(a+b) + \sqrt{2ab}$ exceeds $a + b$ and so exceeds both sides. Whatever rectangle you wedge into the corner, the problem has exactly one solution.

Two consequences fall out of the closed form. Written as $R = a + b + \sqrt{2}\sqrt{ab}$ , the radius is the sum of the two sides plus $\sqrt{2}$ times their geometric mean, so a square tile of side $s$ in the corner forces $R = s(2 + \sqrt{2})$ and an irrational side. The tidy 50 we got exists only because $2ab = 100$ happens to be a perfect square. And the rectangle always fits inside the square, since $2R > 2(a+b)$ is larger than either side length, so the configuration can never fail to exist for want of room.

Sources and further reading

The circle touching all four sides, and why its radius is half the side — Inscribed figure
The discriminant, and what it does and does not tell you — Quadratic equation
Roots that solve the equation and not the problem — Extraneous and missing solutions
The 3-4-5 triangle and its relatives — Pythagorean triple

Both roots were solved in exact rationals rather than floating point, and the two configurations were then tested by sweeping a grid of points across the rectangle against the disk. At $R = 25$ not one sampled point lands inside the circle. At $R = 5$ tens of thousands do.