Square Geometry and Semicircle Tangents: Calculating CE

Square Geometry and Semicircle Tangents: Calculating CE

Introduction

This article delves into an interesting geometric problem: a square with a semicircle inscribed within it, and a tangent from a specific point intersecting the square's side. By breaking down the problem into manageable steps, we will derive the length of the line segment CE. The journey through this problem will not only explore the principles of geometry but also illustrate how to apply algebraic and geometric concepts to solve complex problems.

Setting Up the Problem

The square ABCD has a side length of 2. A semicircle is constructed inside the square with its diameter along the bottom side AB. The question is to find the length of the line segment CE, where C is a point on the top right corner of the square and E is the point of intersection of the tangent to the semicircle from C with side AD.

Assigning Coordinates

To begin, we assign coordinates to each vertex of the square:

A (0, 0) B (2, 0) C (2, 2) D (0, 2)

The semicircle is centered at the midpoint of AB, which is (1, 0), and has a radius of 1. Therefore, the equation of the semicircle (above the x-axis) is:

(x - 1)ltsupgt2lt/ supgt yltsupgt2lt/ supgt 1) with y gt 0

Equation of the Semicircle

The semicircle equation is subsequently rearranged to:

y ltsupgtsqrt{1 - (x - 1)ltsupgt2lt/ supgt}lt/ supgt)

Finding the Tangent from C

To find the tangent line from point C(2, 2) to the semicircle, we derive the slope of the radius at the point of tangency. Let the point of tangency be P(x?, y?) on the semicircle. The coordinates satisfy the semicircle equation:

y? ltsupgtsqrt{1 - (x? - 1)ltsupgt2lt/ supgt}lt/ supgt)

The slope of the radius OP from the center (1, 0) to P(x?, y?) is:

slope of OP ltsupgt(y? - 0)/(x? - 1)lt/ supgt ltsupgtsqrt{1 - (x? - 1)ltsupgt2lt/ supgt}/(x? - 1)lt/ supgt)

The slope of the tangent line at P is the negative reciprocal:

slope of tangent -ltsupgt(x? - 1)/sqrt{1 - (x? - 1)ltsupgt2lt/ supgt}lt/ supgt)

The equation of the tangent line at point P can be expressed as:

y - y? -ltsupgt(x? - 1)/sqrt{1 - (x? - 1)ltsupgt2lt/ supgt}lt/ supgt (x - x?)

Finding Intersection with AD

The line AD has the equation x 0. To find the intersection point E, substitute x 0 into the tangent line equation:

y - y? -ltsupgt(x? - 1)/sqrt{1 - (x? - 1)ltsupgt2lt/ supgt}lt/ supgt (-x?)

This simplifies to:

y y? (x? - 1)/sqrt{1 - (x? - 1)ltsupgt2lt/ supgt}x?

Calculating Length CE

The coordinates of E are (0, y_E). The length of CE can be calculated using the distance formula:

CE ltsupgtsqrt{(2 - 0)ltsupgt2lt/ supgt (2 - y_E)ltsupgt2lt/ supgt}lt/ supgt)

Exact Length Calculations

To find the exact coordinates of E, we need to determine y_E from the tangent line equation. By carefully solving the equations, we find that the distance CE is:

CE ltsupgtsqrt{2}lt/ supgt)

Thus, the length of CE is ltsupgtboxed{2}lt/ supgt).