+0

# Hard Question

+1
101
8

In certain right triangles$$\sqrt{{h}^{2}-{a}^{2}}=24$$ , where $$h$$ represents the length of the hypotenuse and $$a$$ is the length of one of the legs. Find all possible ordered pairs $$(h,a)$$, where $$h,a\in \mathbb{N}$$ .

Thanks for the help!

Edit: Please don't give me a full answer, just tell me how to go about solving this. I want to figure some of it out by myself.

Edit 2: Fixed minor spelling mistake.

Dec 14, 2018
edited by Guest  Dec 14, 2018
edited by Guest  Dec 14, 2018

#1
+100513
+1

This one is (a little) difficult

Note that...if we square both sides, we get

h^2 - a^2  =  24^2

h ^2 - a^2  = 576

(h + a) ( h  - a)  =  576

We need to find all the divisors of   576.....here they are :

1 | 2 | 3 | 4 | 6 | 8 | 9 | 12 | 16 | 18 | 24 | 32 | 36 | 48 | 64 | 72 | 96 | 144 | 192 | 288 | 576

Now, we need to consider  all the pairs that multiply to 576  ......and we need to be able to split one of the pairs such that the difference of this split  is equal to the other member in the original pair

Note...the larger number in the original pair is the one we want to split....!!!!

Let me get you started....we can work from the "outside-in" to find all the possible pairs that multiply to 576

Pair

576  , 1     no way to split the first number such that the difference of this split = 1

288, 2        split 288 into 145, 143.....their difference = 2

Note that  ( 145 +143) ( 145 - 143)   = 576

192, 3     no  way to split the first number  such that the difference of this split = 3

144, 4        note that 144 spits into  74, 70   .....their difference = 4

Note that (74 + 70)(74 - 70)  =  576

96, 6          51, 45 is the split  we need

(51 + 45) (51 - 45) = 576

You should find  4 more workable pairs....if you get stuck...let me know!!!

Dec 15, 2018
edited by CPhill  Dec 15, 2018
edited by CPhill  Dec 15, 2018
edited by CPhill  Dec 15, 2018
#2
0

Thanks! However, I have a couple of questions. Why do you need to work "outside in" to find the pairs? Why can't you pick any 2 divisors? What if there was an odd number of divisors?

Guest Dec 15, 2018
edited by Guest  Dec 15, 2018
edited by Guest  Dec 15, 2018
#3
0

Hey I just realized that the splits are all on even 2nd numbers

Guest Dec 15, 2018
edited by Guest  Dec 15, 2018
#4
+100513
0

You can't pck just any two divisors.....they may not multiply to 576.....

Also...any perfect suqare will always have an odd number of divisors.....the middle divisor will be the square root....

Sounds like you have a good start.....!!!

CPhill  Dec 15, 2018
#5
0

Thanks for all the help! So the 4 other pairs are (40, 32), (30, 18), (26, 10), (25, 7)?

Guest Dec 15, 2018
#6
+100513
0

YOU GOT IT>>>>>GREAT JOB>>>>!!!!!

This was/is an interesting one.....!!!

CPhill  Dec 15, 2018
#7
0

Thanks a lot for all your help!

Guest Dec 15, 2018
#8
+100513
0

OK.....!!!!

CPhill  Dec 15, 2018