Walk From My House

Method 01 : Apply Pythagorean theorem :

Let the vertical distance be denoted as $x$ and the hypotheniuse be denoted as $y$ , then by Pythagorean theorem , we have $15^2 + x^2= y^2$ , where $x$ and $y$ are positive integers.

So we want to minimize the value of $x$ such that $y$ is still an integer.

We can apply trial and error for $x=1,2,3,\ldots$ to get the smallest solution of $x=\boxed8$ and that will be the answer.

---

Method 02 : Apply the parametric equations for Pythagorean theorem :

If all the sides of a right triangle are integers, they will satisfy the parametric triplets: $(m^2-n^2,2mn, m^2 + n^2)$ .

and if one of the sides (that is not a hypothenuse) is an odd number, in this case, it's 15, then $m^2 -n^2= 15 =16-1 = 4^2- 1^2 \Rightarrow m=4,n=1$ , thus the other (non-hypothenuse) side is equal to $2mn = 2\times4\times1 =\boxed8$ .

Using the Pythagorean Theorem, if we denote the vertical distance in miles as $a$ and the hypotenuse in miles as $b$ , then $15^2+a^2=b^2$ .

Rearranging yields $a^2-b^2=-225 \Rightarrow (a+b)(a-b)=-225$

Now we can factorise $-225$ as $-(5^2\times3^2)$ , and $a+b$ and $a-b$ will have to use all these factors between them. Since $a$ and $b$ are both positive, the second bracket $(a-b)$ in our equation must be negative. Also, $|a+b| > |a-b|$ , so now there are just two cases to try:

1. $a+b=25$ , $a-b=-9$
2. $a+b=75$ , $a-b=-3$

The first case gives the desired answer $a=\boxed{8}$ . The second gives another solution of $a=36$ , but this is not the minimum distance so is not what we need.