Detour of an Ant

We call the first and the second points on the x-axis that the ant walks on M and N respectively. If M is (a,0) then N is (a+7, 0).

The distance the ant walks is |AM| + |MN| + |NB| = |AM| + |NB| + 7. To minimize the distance, therefore, is to minimize the sum |AM| + |NB|.

If we draw a parallelogram MNBP, then |PB| = |MN| = 7 and therefore P = (2, 13). Also, |AM| + |NB| = |AM| + |MP|.

Let P' be the reflection of P across the x-axis. Then

|AM| + |MP| = |AM| +|MP'| ≥ |AP'| (triangle's inequality)

As P' = (2, -13), |AP'| = $\sqrt{(2-(-18))^2+((-13)-8)^2}$ = 29

The shortest distance that the ant could have walked is 29 + 7 = 36.

Let (x, 0) and (x+7, 0) be the points where the ant begins and ends its path on the x-axis. Use the distance formula on the points (-18, 8) and (x,0), and on the points (x+7,0) and (9, 13), to find radical expressions for the distances of the straight-line paths the ant takes off of the x-axis; add 7 to the sum of those expressions to find an expression for the total distance traveled.

Differentiate that expression with respect to x, and set the derivative equal to zero to find that x = -218/21; substitute that x value into your total distance expression to find a total distance of 36 units.

The shortest distance that the ant could have walked can be considered to be the distance travelled along the x-axis (7 units) plus the length of the hypotenuse of a right angle triangle with height equal to the total units travelled in the y direction (8 + 13 units), and base equal to the total units travelled in the x direction minus the distance travelled along the x-axis (27 - 7 units).

This allows the distance to be calculated using the Pythagorean theorem:

Shortest distance = \sqrt{(8 + 13)^2 + (27 - 7)^2} + 7 = 36 units

C is point which is symmetry of B over \­( x \­)-axis, so with each point Q chosen uniformly on the \­( x \­)-axis, we have QB=QC translate it them directly to the left for 7 units, we have Q turn into P, and C into D we have CD = PQ = 7 With each point P chosen uniformly on the \­( x \­)-axis, we have \­(AP+PD \geq AD) with AD is constant because A,D are fix points. We have the distance the ant could walk is: \­(AP + PQ + QB \geq AD +CD ) with AD+CD is a constant. We have B(9;13) $\Rightarrow$ C(9;-13) $\Rightarrow$ D(2;-13) We have AD= $\sqrt{ (x_D-x_A)^2+(y_D-y_A)^2}$ = $sqrt{ (2-(-18))^2+(-13-8)^2}$ =29 The shortest distance that the ant could have walked is AD+CD=7+29=36

First, create B', which is made by moving B back 7 spaces. B'= (2,13) There are 20 units of difference between the x coordinates of A and B' Since B' is 13 units up and A is 8 units up, we partition the 20 units into two segments of ratio 13 to 8. The place where the segments come together is also the coordinate on the x axis which creates the least distance for the ant. There are then two triangles created by the path of the ant. the legs are 8 by 160/21 and 13 by 260/21 using the Pythagorean theorem and adding them up, the distance is 29 To account for the shift of 7 spaces from B to B', we add 7 to get 36.

We will instead assume that the final point for the ant is $C=(9,-13)$ , since for any walk of length $k$ from $A$ to $B$ that walks along the $x$ -axis we can find a symmetric walks from $A$ to $C$ where the last portion of the walk is reflected in the $x$ -axis. If we let $K(k,0)$ be the point on the $x$ -axis where the ant walks to, then if we cut out the portion of the plane between $(k,0)$ and $(k+7,0)$ , then we have an equivalent problem, except the ant just steps on the $x$ -axis once. By the triangle inequality, the length of $AC$ is less than or equal to the sum of the lengths of $AK$ and $KC$ . So the minimum walking the ant can do is when $K$ is on $AC$ and the ant just walks along $AC$ . The length of $AC$ (in the shrunken plane) is $\sqrt{(8+13)^2 + (9 + 18 - 7)^2} = \sqrt{20^2 + 21^2} = 29$ . Thus, the minimum distance the ant could have travelled is $29 + 7 = 36$ .