As shown in the figure, circle $C_1$ with center $O_1$ and radius $r_1$ touches the square $VWXY$ at points $P$ and $Q$, while circle $C_2$ with center $O_2$ and radius $r_2$ touches the square $VWXY$ at points $R$ and $S$. The two circles touch each other at $T$.
Given $r_1 = 1$ cm and $\overline{VY} = \overline{VW} = 4$ cm, $r_2 =$ cm.

Question

A rectangle $ABCD$ has sides $AB = 45 \, \text{cm}$ and $BC = 26 \, \text{cm}$. Point $E$ is the midpoint of side $CD$. Find the radius of the incircle of the triangle $\triangle AED$.

Answer 1

Both circles sit snugly in opposite corners of the square, each tangent to the two sides meeting at that corner, and they touch each other. A quick way to solve this is to work along the diagonal of the square that passes through both corners, $W$ and $Y$.

The center of a circle tucked into a right-angle corner, tangent to both sides, always lies on the diagonal bisecting that corner angle, at a distance $r\sqrt{2}$ from the corner, since the perpendicular distance to each side is $r$, and the diagonal makes a $45^\circ$ angle with each side.
The diagonal $WY$ of the square has length $\sqrt{4^2 + 4^2} = 4\sqrt{2}$ cm.
Center $O_1$ (near corner $W$) lies at distance $r_1\sqrt{2} = 1 \cdot \sqrt{2} = \sqrt{2}$ cm from $W$, measured along $WY$.
Center $O_2$ (near corner $Y$) lies at distance $r_2\sqrt{2}$ cm from $Y$, so it is $4\sqrt{2} - r_2\sqrt{2}$ cm from $W$ along the same diagonal.
Since the circles touch, the gap between $O_1$ and $O_2$ along the diagonal equals $r_1 + r_2$: $$\left(4\sqrt{2} - r_2\sqrt{2}\right) - \sqrt{2} = r_1 + r_2 = 1 + r_2$$

Simplify the left side: $$4\sqrt{2} - \sqrt{2} - r_2\sqrt{2} = 3\sqrt{2} - r_2\sqrt{2}$$ So the equation becomes: $$3\sqrt{2} - r_2\sqrt{2} = 1 + r_2$$ $$3\sqrt{2} - 1 = r_2(1 + \sqrt{2})$$ $$r_2 = \frac{3\sqrt{2} - 1}{\sqrt{2} + 1}$$

Multiply numerator and denominator by $(\sqrt{2} - 1)$ to rationalize: $$r_2 = \frac{(3\sqrt{2} - 1)(\sqrt{2} - 1)}{(\sqrt{2} + 1)(\sqrt{2} - 1)} = \frac{6 - 3\sqrt{2} - \sqrt{2} + 1}{2 - 1} = 7 - 4\sqrt{2}$$

This matches the coordinate-geometry method exactly, confirming $r_2 = 7 - 4\sqrt{2}$ cm, about 1.34 cm.

\[ \boxed{r_2 = 7 - 4\sqrt{2} \text{ cm}} \]

Answer 2

Both circles sit snugly in opposite corners of the square, each tangent to the two sides meeting at that corner, and they touch each other. A quick way to solve this is to work along the diagonal of the square that passes through both corners, $W$ and $Y$.

The center of a circle tucked into a right-angle corner, tangent to both sides, always lies on the diagonal bisecting that corner angle, at a distance $r\sqrt{2}$ from the corner, since the perpendicular distance to each side is $r$, and the diagonal makes a $45^\circ$ angle with each side.
The diagonal $WY$ of the square has length $\sqrt{4^2 + 4^2} = 4\sqrt{2}$ cm.
Center $O_1$ (near corner $W$) lies at distance $r_1\sqrt{2} = 1 \cdot \sqrt{2} = \sqrt{2}$ cm from $W$, measured along $WY$.
Center $O_2$ (near corner $Y$) lies at distance $r_2\sqrt{2}$ cm from $Y$, so it is $4\sqrt{2} - r_2\sqrt{2}$ cm from $W$ along the same diagonal.
Since the circles touch, the gap between $O_1$ and $O_2$ along the diagonal equals $r_1 + r_2$: $$\left(4\sqrt{2} - r_2\sqrt{2}\right) - \sqrt{2} = r_1 + r_2 = 1 + r_2$$

Simplify the left side: $$4\sqrt{2} - \sqrt{2} - r_2\sqrt{2} = 3\sqrt{2} - r_2\sqrt{2}$$ So the equation becomes: $$3\sqrt{2} - r_2\sqrt{2} = 1 + r_2$$ $$3\sqrt{2} - 1 = r_2(1 + \sqrt{2})$$ $$r_2 = \frac{3\sqrt{2} - 1}{\sqrt{2} + 1}$$

Multiply numerator and denominator by $(\sqrt{2} - 1)$ to rationalize: $$r_2 = \frac{(3\sqrt{2} - 1)(\sqrt{2} - 1)}{(\sqrt{2} + 1)(\sqrt{2} - 1)} = \frac{6 - 3\sqrt{2} - \sqrt{2} + 1}{2 - 1} = 7 - 4\sqrt{2}$$

This matches the coordinate-geometry method exactly, confirming $r_2 = 7 - 4\sqrt{2}$ cm, about 1.34 cm.

\[ \boxed{r_2 = 7 - 4\sqrt{2} \text{ cm}} \]

Show Hint

The Correct Option is C

Solution and Explanation

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