Question

Let \(A B C D\) be a square with \(A(0,0), C(2,2)\). If \(\mathrm{M}\) is mid- point of \(\mathrm{AB}\) and \(\mathrm{P}\) is a variable point on \(\mathrm{BC}\). The smallest value of \(\mathrm{DP}+\mathrm{PM}\) is (1) \(\sqrt{13}\) (2) \(\sqrt{2}+\sqrt{5}\) (3) \(2 \sqrt{3}\) (4) \(1+2 \sqrt{2}\)

Short answer

The smallest value of DP + PM is \( \sqrt{2} + \sqrt{5} \).

Step-by-step solution

Determine the Coordinates of Points

Identify the coordinates of all the points in the square. We know: \(A = (0,0)\), \(C = (2,2)\). Since it's a square, B and D can be derived as follows: \(B = (2,0)\), \(D = (0,2)\).

Calculate the Midpoint M of AB

The midpoint formula states: \( M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \). Applying this to points A and B, we get: \( M = \left( \frac{0 + 2}{2}, \frac{0 + 0}{2} \right) = (1,0) \).

Express the Coordinates of Point P on BC

Point P lies on BC, where B is (2,0) and C is (2,2). Therefore, P can be represented as \( (2, y) \) where y is between 0 and 2.

Write the Distance Formulas for DP and PM

Using the distance formula, calculate DP and PM. Distance DP: \( DP = \sqrt{(2-0)^2 + (y-2)^2} = \sqrt{4 + (y-2)^2} \). Distance PM: \( PM = \sqrt{(2-1)^2 + (y-0)^2} = \sqrt{1 + y^2} \).

Minimize DP + PM

Combine the expressions and minimize: \( DP + PM = \sqrt{4 + (y-2)^2} + \sqrt{1 + y^2} \). To minimize, consider the symmetry and geometric placement in the square. After analyzing and differentiating, it is found that setting y = 2 form the minimum distance.

Calculate the Minimal Value

Calculate the specific distances when y = 2: \( DP = \sqrt{4 + (2-2)^2} = 2 \) and, \( PM = \sqrt{1 + 2^2} = \sqrt{5} \). Therefore, \( DP + PM = 2 + \sqrt{5} \).

Key concepts

coordinate geometry

Coordinate geometry is the study of geometric figures using a coordinate system. In this exercise, we use the Cartesian coordinate system where every point is identified by a pair of coordinates, \(x\) and \(y\). For example, point \(A(0,0)\) and point \(C(2,2)\) in the problem.
The primary tools we use in coordinate geometry include the distance formula, the midpoint formula, and various methods to find equations of geometric figures like lines and circles. Coordinate geometry helps us solve problems by translating geometric figures into algebraic equations, which are easier to handle analytically.

midpoint formula

The midpoint formula helps us find the middle point between two given points in the Cartesian plane. For two points \(A(x_1, y_1)\) and \(B(x_2, y_2)\), the midpoint \((M)\) is given by:
\[M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)\]
In the given problem, our points are \(A(0, 0)\) and \(B(2, 0)\). Substituting these into the formula, we get:
\[M = \left( \frac{0 + 2}{2}, \frac{0 + 0}{2} \right) = (1, 0)\]
This tells us that the midpoint M of segment AB is at the coordinates (1, 0).

distance formula

The distance formula allows us to calculate the distance between two points in the Cartesian plane. For points \(A(x_1, y_1)\) and \(B(x_2, y_2)\), the distance \(d\) between them is given by:
\[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]
In our exercise, we used the distance formula to calculate DP and PM. For DP, where D(0,2) and P(2,y), the formula becomes:
\[DP = \sqrt{(2-0)^2 + (y-2)^2} = \sqrt{4 + (y-2)^2}\]
For PM, where P is (2,y) and M is (1,0), we get:
\[PM = \sqrt{(2-1)^2 + (y-0)^2} = \sqrt{1 + y^2}\]
Thus, using these formulas, we can compute the distances accurately.

minimizing distance

Minimizing distance involves finding the shortest path or minimum value where distances are considered. In the given problem, we aim to minimize the expression \(DP + PM\).
By setting up our equations for DP and PM, we combine them as:
\[DP + PM = \sqrt{4 + (y-2)^2} + \sqrt{1 + y^2}\].
We try different values of \(y\) to find when this sum is minimized. Analyzing the function, differentiating, or using symmetry about the square helps identify that the smallest value for y is 2, simplifying the process. When \y = 2\, the distances become:
\[DP = 2\] and \[PM = \sqrt{5}\].
Thus, the minimum value is the sum of these two distances:
\[DP + PM = 2 + \sqrt{5}\].
This principle is useful not only in solving this problem but also in finding shortest paths and optimizing routes in various geometric configurations.