Right $\triangle ABC$ has hypotenuse $AB$. Square $BCDE$ has $BC$ as one of its sides. Suppose that the area of $BCDE$ is a prime number. If $AB$ and $AC$ are each integers less than $20$, how many possibilities are there for their lengths?

LeoIsTheBest Apr 17, 2019

#1**+1 **

I think there's only one possibility. If the area of that square is a prime number, then either its width or its height has to be 1. Let's say its width. Any value for the width other than 1, when multiplied by the height will give a product that isn't prime.

.

Guest Apr 17, 2019

edited by
Guest
Apr 17, 2019

edited by Guest Apr 17, 2019

edited by Guest Apr 17, 2019

#4**+3 **

I believe this is correct

AB is the hypotenuse, so AC and AB are legs

So

BC^2 = AB^2 - AC^2 = area of BCDE

BC^2 = (AB + AC) (AB - AC)

If we let (AB - AC) = 1......then BC^2 will have only one prime factor, (AB + AC)

So we have the following possibilites

AB AC BC BC^2

19 18 √37 37

16 15 √31 31

15 14 √29 29

12 11 √23 23

10 9 √19 19

9 8 √17 17

7 6 √13 13

6 5 √11 11

4 3 √7 7

3 2 √5 5

2 1 √3 3

CPhill Apr 17, 2019