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A quicker alternative:

Total distance travelled D= (8 + x + x/2) miles
=> (3/2)x = D-8

Given, x is a positive integer. When we plug in the answer choices in place of D, only answer choice (D) yields a positive integer value of x.

Answer D
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Given the info we know that the horizontal leg is 8, vertical leg is Y and hypotenuse X, so:

X^2-Y^2 = 64 or:

(X+Y)(X-Y) = 64

Since the travel south of X/2 is still more than X/3 away from the beginning:

Y-X/2>X/3 or Y>5X/6

From difference of squares above and that X and Y are integers, possibilities are:

1 64
2 32
4 16
8 8

So the approach is:

X+Y=larger
X-Y=smaller

64 and 1 would mean X is not an integer (2X=65)

Try 32 and 2:

X+Y=32
X-Y=2

2X=34; X=17, Y=15

But do these satisfy the constraint above ?:

Y>5X/6 15>85/6 or

90>85 ? YES

So total distance is:

8+X+X/2 = 8+17+8.5 = 33.5

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A bicyclist travels 8 miles due west at a constant speed. Next, she rides x miles in a straight line in a direction somewhere between north and east, traveling at half the speed. She stops when she is due north of her starting point, at which time she is y miles from her original location. She then rides, at 1/3 of her original speed, due south for x/2 miles, at which point she ends her trip, more than x/3 miles from her starting point.

If x and y are integers, how many total miles did she cycle?

Attachment:
Screenshot 2025-03-27 at 10.31.04 AM.png
Screenshot 2025-03-27 at 10.31.04 AM.png [ 30.99 KiB | Viewed 1011 times ]

8^2 + y^2 = x^2
8^2 = (x^2 - y^2) = (x+y)(x-y)

Case 1: y > x/2 + x/3 = 5x/6
Case 2: x/2 > y + x/3 ; y < x/2 - x/3 = x/6

1: x + y = 16; x - y = 4; x = 10; y = 6; x/6 = 10/6 = 5/3; 5x/6 = 25/3 = 8 1/3; Not feasible since no cases are possible
2. x+ y = 32; x- y=2; x = 17; y = 15; x/6 = 17/6 = 2 5/6; 5x/6= 85/6 = 14 1/6; y=15 > 5x/6=14 1/6; Feasible

Total distance travelled = 8 + x +x/2 = 8 + 17 + 17/2 = 25 + 8.5 = 33.5 miles

IMO D
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