Let ABCD be a square of side of unit length. Let a circle C1 centered at A with unit radius is drawn

Q: Wait, why is the center of $C_2$ at $(r, r)$? Shouldn't it be somewhere else?

That is a fantastic question! Think about the constraints: $C_2$ is tangent to the line $AD$ (which is the y-axis, $x=0$) and the line $AB$ (which is the x-axis, $y=0$). For a circle to be tangent to both the x-axis and the y-axis in the first quadrant, its center must be equidistant from both axes. That distance is exactly the radius $r$. Thus, the center must be at $(r, r)$.

Q: How do we know the distance from $A$ to $O$ is $r\sqrt{2}$?

Imagine a small square formed by the origin $A(0,0)$, the point $(r,0)$, the center $O(r,r)$, and the point $(0,r)$. The distance from the origin to the center $O$ is the diagonal of this square. By the Pythagorean theorem, the diagonal is $\sqrt{r^2 + r^2} = \sqrt{2r^2} = r\sqrt{2}$. It is a beautiful, simple geometric fact!

Q: Why do we reject the smaller slope when solving $m^2 - 4m + 1 = 0$?

Great observation! The quadratic equation $m^2 - 4m + 1 = 0$ gives us two roots: $m = 2 \pm \sqrt{3}$. Both are valid slopes for a line passing through $C(1,1)$ that could be tangent to the circle. However, if you sketch the diagram, you will see that the line $CE$ is quite steep and has a positive slope greater than 1. The root $2 - \sqrt{3}$ is less than 1, while $2 + \sqrt{3}$ is greater than 1. Therefore, based on the geometry of the problem, we must choose the steeper slope.

Question

Let $ABCD$ be a square of side of unit length. Let a circle $C_1$ centered at $A$ with unit radius is drawn. Another circle $C_2$ which touches $C_1$ and the lines $AD$ and $AB$ are tangent to it, is also drawn. Let a tangent line from the point $C$ to the circle $C_2$ meet the side $AB$ at $E$. If the length of $EB$ is $\alpha + \sqrt{3}\beta$, where $\alpha, \beta$ are integers, then $\alpha + \beta$ is equal to _______

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine the square $ABCD$ placed on a Cartesian plane. To simplify the geometry, we anchor vertex $A$ at the origin $(0,0)$. This places $B$ at $(1,0)$, $C$ at $(1,1)$, and $D$ at $(0,1)$.

The first circle, $C_1$, is centered at $A$ with a radius of $1$. It defines the boundary of our space. We then introduce $C_2$, a smaller circle tucked into the corner at $A$.

Because $C_2$ is tangent to both the x-axis ($AB$) and the y-axis ($AD$), its center $O$ must be located at $(r, r)$, where $r$ is its radius. This symmetry is the first key to unlocking the problem.

### The Trapped Circle

To find $r$, we consider the line segment from the origin $A(0,0)$ to the center $O(r,r)$. The length of this segment is $r\sqrt{2}$.

If we extend this line until it hits the edge of the large circle $C_1$, the total distance from $A$ to the edge of $C_1$ is the radius of $C_1$, which is $1$. The segment from $O$ to the edge of $C_1$ is simply the radius of the small circle, $r$.

Thus, we establish the following relationship:
$$r\sqrt{2} + r = 1$$

Factoring out $r$, we get $r(\sqrt{2} + 1) = 1$. Rationalizing this expression, we find the radius of the small circle:
$$r = \sqrt{2} - 1$$

### The Tangent Line

We draw a line from $C(1,1)$ that is tangent to the small circle $C_2$. Let the slope of this line be $m$.

Using the point-slope form, the equation of the line $CE$ is $y - 1 = m(x - 1)$. This rearranges to the standard form:
$$mx - y + 1 - m = 0$$

### The Algebra of Tangency

For this line to be tangent to $C_2$, the perpendicular distance from the center $O(r,r)$ to the line $mx - y + 1 - m = 0$ must equal the radius $r$. Using the distance formula, we write:

$$\frac{|mr - r + 1 - m|}{\sqrt{m^2+1}} = r$$

Simplifying the numerator, we obtain $|(m-1)(r-1)|$. Squaring both sides, we note that since $r = \sqrt{2}-1$, then $r-1 = \sqrt{2}-2$. Squaring this yields:
$$(r-1)^2 = 2 - 4\sqrt{2} + 4 = 6 - 4\sqrt{2} = 2(3 - 2\sqrt{2}) = 2r^2$$

Substituting this back into our squared distance equation:
$$\frac{2r^2(m-1)^2}{m^2+1} = r^2$$

The $r^2$ terms cancel, leaving us with $2(m-1)^2 = m^2 + 1$. Expanding this, we arrive at the quadratic equation:
$$m^2 - 4m + 1 = 0$$

### Final Calculation

Solving $m^2 - 4m + 1 = 0$ using the quadratic formula gives $m = 2 \pm \sqrt{3}$. Geometry dictates we choose the steeper slope, $m = 2 + \sqrt{3}$.

To find the intersection $E$ on the x-axis ($y=0$), we substitute $y=0$ into our line equation:
$$-1 = (2 + \sqrt{3})(x - 1)$$

Solving for $x$:
$$x - 1 = -\frac{1}{2+\sqrt{3}} = -(2-\sqrt{3}) = \sqrt{3}-2$$
$$x = \sqrt{3}-1$$

The length $EB$ is the distance from $B(1,0)$ to $E(\sqrt{3}-1, 0)$, which is $1 - (\sqrt{3}-1) = 2 - \sqrt{3}$. Comparing this to the form $\alpha + \sqrt{3}\beta$, we identify $\alpha = 2$ and $\beta = -1$.

The final result, $\alpha + \beta$, is **1**.

Enter Numerical Value:

Visualized Solution

Coordinate System Setup

Geometry of Circle $C_{2}$

Distance from $A$ to $O$

Calculating Radius $r$

Equation of Tangent $C E$

Applying Tangency Condition

Simplifying the Distance Equation

Substituting $r$ and Squaring

Solving for Slope $m$

Selecting the Correct Slope

Finding Intersection $E$

Calculating Length $E B$

Final Result

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Trapped Circle

The Tangent Line

The Algebra of Tangency

Final Calculation

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