Day 21: One and a Half Circles in a Square

Form a triangle with vertex $A$ being the top left corner of the square, vertex $B$ being the center of the radius $r$ circle, and vertex $C$ being the point where the horizontal through $B$ intersects the left side of the square.

Then $\Delta ABC$ is a right triangle with hypotenuse $AB = 27 + r$ and legs $AC = 54 - r$ and $BC = 27$ . Thus, using the Pythagorean Theorem, we have that

$(27 + r)^{2} = (54 - r)^{2} + 27^{2} \Longrightarrow 54r = 54^{2} - 108r \Longrightarrow r = \dfrac{54^{2}}{162} = \boxed{18}$ .

We first scale the diagram down so that the given radius is $1$ for easier calculation, then we will multiply back; it does not affect the result. Now draw in this triangle: Using Pythagoras' Theorem we can find the height of the square in terms of $r$ : $2 = \sqrt{r^2+2r}+r$

And solving: $r^2-4r+4=r^2+2r \\ 6r=4 \\ r = \frac{2}{3}$

Hence the required answer is $\frac{2}{3} \times 27 = \boxed{18}$ .

Let the square be $ABCD$ . $A$ and $B$ the two top vertices. Let the midpoints of $AB$ and $CD$ be $M$ and $N$ respectively and the center of the circle be $O$ . Due to symmetry, we find that $MON$ is a straight line and that:

$MN = MO + ON = \sqrt{AO^2 - AM^2} + ON = \sqrt{(27+r)^2 - 27^2} + r = 54$

$\Rightarrow \sqrt{(27+r)^2 - 27^2} = 54 - r$

$\Rightarrow (27+r)^2 - 27^2 = (54 - r)^2$

$\Rightarrow 27^2+54r+ r^2 - 27^2 = 54^2 - 108r + r^2$

$\Rightarrow 162r = 54^2 \quad \Rightarrow r = \boxed {18}$

Suppose that this square is named ABCD such that A and B is the center of the circles having a midpoint of 27. And O is the center of the circle touching these circles and the bottom of the square having a radius of r. Draw a line from the midpoint of AB, M to O and from B to O. From the circle theorem, we have that BMO is a right triangle, so, MB = 27 and BO = 27+r Because MO is a perpendicular on AB and M is the midpoint of AB, MN = AB = 54, where N is the midpoint of CD. So MO = 54-r. By applying the Pythagorean theorem, we have, r = 18.

by phythagorth : (27+r)^2=(2r)^2 + (27)*2 in the end of the equation we will get r=18

Firstly, denote the square as ABCD clockwise from top left to bottom left. Now make the centre of the circle (of radius r) point P and let the midpoint of AB be Q. Draw AP, BP and QP. Now QP is perpendicular to AB (tangent to a radius). Using Pythagoras's Theorem: $(AQ)^{2}$ + $(QP)^{2}$ = $(AP)^{2}$ . For simplicity, let ${AQ}$ = ${QB}$ = ${x}$ (27). Now $x^{2}$ + $(2x-r)^{2}$ = $(x+r)^{2}$ . Expanding and rearranging gives ${r}$ = $\frac{2x}{3}$ = $\frac{54}{3}$ = 18.