So Many Maths Challenge Tricks in One Problem | No calculators allowed!
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- Опубліковано 19 гру 2023
- To solve this problem at all is tricky, but to do it efficiently without a calculator you need to apply all the best tricks of the maths challenges.
I really think you can learn more from this single problem than from a whole series of maths lessons - we'll use Pythagoras, some neat algebra and be clever with indices and surds to avoid any messy calculations!
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Committing a few Pythagorean triples to memory gets you there much faster 😉
Yes it’s amazing how often these come up!
Great video! This is the type of mathematics thinking that you gave me in the JMO math challenge to help me achieve the great result that I got! Thanks again Kevin, and congratulations on nearly 30k subscribers, and for this great solution!
Thanks Tim, great your doing so well and it’s my pleasure to help!
Most calculations are unnecessary.
In the 3-4-5, 5-12-13, 13-85-86 right triangles we can notice a pattern:
The small number squared equals to the sum of the other two.
3²=4+5
5²=12+13
13²=85+86.
This is the case for all right triangles, with the smalles side being an odd number. Examples:
3-4-5
5-12-13
7-24-25
9-40-41
11-60-61
And so on
That’s an interesting observation. I would have to think a bit more but I think the result you’re using is only a unique answer when the smallest side is an odd prime number. For example for 9, you could also have 9,12,15 say. But for 13 it would work, though for 13^2 I think you mean 84 and 85.
Certainly knowing results about Pythagorean Triples is very useful for maths challenges and if you can spot one that works you can save some time, though for a question with working you would also need to give some justification of what you’re doing. One problem here would be that we don’t know for sure in advance that every side length is an integer so you’d have to make it clear why your answer is the only one possible.
It’s a very nice idea though, I like the way you’re thinking about this!
@@MathsaurusIt doesnt have to be a odd number you can also have :
8 - 15 - 17
10 - 24 - 26
so on...
the idea is you write the square of short side and divide by 2 then think two consecutive even numbers that sums to that.
Yes - I think the result I’m thinking of says something like if the shorter side is an odd prime, then it’s the only Pythag triple possible. I think the proof is actually not so bad, we just write eg 13^2=c^2-b^2=(c+b)(c-b) and then deduce c-b must be 1 and c+b is 169 and then solve for b and c. Replacing 13 with any other prime still works but if it’s not prime we may lose the uniqueness.
But yes if we don’t mind the answer is not unique then we can get lots more! An interesting exercise if you haven’t tried it is to show algebraically why that’s possible - it should just be a case of doing some multiplying out of appropriate brackets I think!
@@Mathsaurus @Mathsaurus what I was trying to say was, take any odd number. Let's say 427.
427² = 182,329. Now we're going to find 2 consecutive numbers that sum up to this value. divide that number by 2:
182,329 / 2= 91,164.5(average of 2 consecutive numbers)
So our other numbers are 91,164 and 91,165. So there you go!
427² + 91,164² = 91,165² is indeed a Pythagorean triple! I don't have any idea how this works at all but would like to see an explanation. This is not only for prime numbers like you said though, because 427 is not a prime number.
@@arda9310 I suppose if you start with an odd number 2n+1 and square it you get (2n+1)^2=4n^2+4n+1. So your even and odd number that are roughly half this are 2n^2+2n and 2n^2+2n+1. Then we just need to check that (2n+1)^2+(2n^2+2n)^2=(2n^2+2n+1)^2.
If you multiply all of these out carefully you would get 4n^2+4n+1+4n^4+4n^2+8n^3=4n^4+4n^2+1+8n^3+4n^2+4n and you can check you have the same on the left hand side and the right hand side and so this does always work!
AWESOMEEEE!!
could you use simultaneous equations?
Consider the angle of inclination of the smaller triangle be x then for smaller triangle cosx=4/5 now for middle triangle cosx=12/y equate these and youll get y(hypotaneous)=15 :////
But you can’t do this because the triangles are not similar so the x is different from the first triangle to the second!
@@Mathsaurus i thought the angle between the base of smaller and the perpendicular of the larger triangle is 90 degrees my bad
Man 85 +84+ 13 is not equal to 188
No, it’s the perimeter of the whole shape that’s 188 though, so that’s 3+4+12+84+85