### Brain Teasers

# Who Needs the Right Formula, Anyway?

Math
Math brain teasers require computations to solve.

This morning I asked my math class to find the areas of two right-angled triangles. Lottie-Lu's hand shot up immediately, so I asked her the first one.

"24" she announced.

She was right (for a change!) and I made the mistake of asking her how she did it.

"Easy. Just add all the sides together."

I was about to correct her and say it was just a coincidence, when I noticed that her method worked on my second triangle as well!

Lottie-Lu must have caught my frustrated thoughts. "Well, it works, doesn't it?" she chirped.

What were the side lengths of the two triangles I had drawn on the board this morning? They were all whole numbers.

"24" she announced.

She was right (for a change!) and I made the mistake of asking her how she did it.

"Easy. Just add all the sides together."

I was about to correct her and say it was just a coincidence, when I noticed that her method worked on my second triangle as well!

Lottie-Lu must have caught my frustrated thoughts. "Well, it works, doesn't it?" she chirped.

What were the side lengths of the two triangles I had drawn on the board this morning? They were all whole numbers.

### Hint

Use the fact that all three sides of each triangle were whole numbers.### Answer

1. 6-8-10 (Area = 6x8/2 = 24)2. 5-12-13 (Area = 5x12/2 = 30)

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## Comments

I am so bad at mathematics. Really, really bad. At school my teachers despaired of me!

Perfect right triangles!

Poor Marple! It's good for you!!

Now, for those who, at the other end of the "mathspectrum", find it too easy - how about a proof that there are only two solutions?

Have fun!

Now, for those who, at the other end of the "mathspectrum", find it too easy - how about a proof that there are only two solutions?

Have fun!

Answering dalfamnest's comment, it only works with pythagorean triples.

That's true, starriddler, but only because I stated that the sides are all whole numbers. The challenge that I'm adding to my teaser is "prove that there are only these two solutions"!

well, what we want is something that shows, if (as pythagorean triples are generally set up):

(ab)/2 = a + b + c.

substitute "square root of (a sqaured plus b sqaured)" for c so we can deal only with two variables.

after some fun simplfying, you can end up with

a = (4b-/(b-4). from here, you can see that plugging in any of the 4 values (5, 12 or 6, for b will yield the apprropriate a value.

i'm a little stuck, but i'd imagine something around here shows that only those 4 values for b will yield (positive) whole number answers for a. i'll think on it.

(ab)/2 = a + b + c.

substitute "square root of (a sqaured plus b sqaured)" for c so we can deal only with two variables.

after some fun simplfying, you can end up with

a = (4b-/(b-4). from here, you can see that plugging in any of the 4 values (5, 12 or 6, for b will yield the apprropriate a value.

i'm a little stuck, but i'd imagine something around here shows that only those 4 values for b will yield (positive) whole number answers for a. i'll think on it.

(stupid smiley faces) well, what we want is something that shows, if (as pythagorean triples are generally set up):

(ab)/2 = a + b + c.

substitute "square root of (a sqaured plus b sqaured)" for c so we can deal only with two variables.

after some fun simplfying, you can end up with

a = (4b-8 )/(b-4). from here, you can see that plugging in any of the 4 values (5, 12 or 6, or 8 ) for b will yield the apprropriate a value.

i'm a little stuck, but i'd imagine something around here shows that only those 4 values for b will yield (positive) whole number answers for a. i'll think on it.

(ab)/2 = a + b + c.

substitute "square root of (a sqaured plus b sqaured)" for c so we can deal only with two variables.

after some fun simplfying, you can end up with

a = (4b-8 )/(b-4). from here, you can see that plugging in any of the 4 values (5, 12 or 6, or 8 ) for b will yield the apprropriate a value.

i'm a little stuck, but i'd imagine something around here shows that only those 4 values for b will yield (positive) whole number answers for a. i'll think on it.

Taking forward the formula of lessthanjake789, the equation can be written as (a-4) = 8/(b-4). As a, b and c are +ve integers, b has to be 5 or more but not more than 12. This means that b can be 5,6, 8 or 12. These give only two unique combinations 6, 8, 10 and 5, 12, 13.

I enjoyed solving this. Thanks, dalfamnest, for a wonderful puzzle.

I enjoyed solving this. Thanks, dalfamnest, for a wonderful puzzle.

a=(4b-/(b-4) expands to a=(4(b-4)+/(b-4), or a = 4 + 8/(b-4). The only (+ve) integer solutions for a and b come when b-4 is a factor of 8; as Ram says, when b=5,6,8 or 12.

So yes, those are the only two all-integer triangles. However, any right-angled triangle whose short sides satisfy the equation a = 4 + 8/(b-4) and b>4 will have its area equal to its perimeter (e.g. b=7, a=20/3, c=29/3).

When 2

So yes, those are the only two all-integer triangles. However, any right-angled triangle whose short sides satisfy the equation a = 4 + 8/(b-4) and b>4 will have its area equal to its perimeter (e.g. b=7, a=20/3, c=29/3).

When 2

When 2

Let's try that again:

When 2â‰¤bâ‰¤4, aâ‰¤0 or a=âˆž, but interestingly, it also does not work when 0

When 2â‰¤bâ‰¤4, aâ‰¤0 or a=âˆž, but interestingly, it also does not work when 0

Oh, ffs...

when 2>b>0, where the formula suggests it should. Further investigation is required to explain this apparent contradiction!

when 2>b>0, where the formula suggests it should. Further investigation is required to explain this apparent contradiction!

Pythagorean triads where two larger sides differ by 1 are: (n^2-1)/2, n, (n^2+1)/2

So: Area = perimeter gives

0.5 x n x (n^2-1)/2 = (n^2-1)/2 + n + (n^2+1)/2

n(n^2-1) = 2(n^2-1)+4n+2(n^2+1)

n^2-1 = 4n + 4

n^2-4n-5=0

(n-5)(n+1)=0 gives n = 5 or -1 (impossible)

n=5 is only answer for triads of this form

i.e. 5, 12, 13

--------------------

triads where larger sides differ by 2 are

(n^2-1), 2n, (n^2+1)

Area = perimeter

0.5 x 2n x (n^2)-1) = (n^2-1) + 2n + (n^2+1)

n(n^2-1)=2n^2+2n

n^2-1=2n+2

n^2-2n-3=0

(n+1)(n-3)=0 n = 3 or -1 (impossible)

n=3 is only value where 2 larger sides differ by 2. ie 6, 8, 10

Although this does not prove they are the only solutions for all triads, it does prove that out of an infinite number of triads whose larger terms differ by either 1 or 2, ther are only 2 solutions!

So: Area = perimeter gives

0.5 x n x (n^2-1)/2 = (n^2-1)/2 + n + (n^2+1)/2

n(n^2-1) = 2(n^2-1)+4n+2(n^2+1)

n^2-1 = 4n + 4

n^2-4n-5=0

(n-5)(n+1)=0 gives n = 5 or -1 (impossible)

n=5 is only answer for triads of this form

i.e. 5, 12, 13

--------------------

triads where larger sides differ by 2 are

(n^2-1), 2n, (n^2+1)

Area = perimeter

0.5 x 2n x (n^2)-1) = (n^2-1) + 2n + (n^2+1)

n(n^2-1)=2n^2+2n

n^2-1=2n+2

n^2-2n-3=0

(n+1)(n-3)=0 n = 3 or -1 (impossible)

n=3 is only value where 2 larger sides differ by 2. ie 6, 8, 10

Although this does not prove they are the only solutions for all triads, it does prove that out of an infinite number of triads whose larger terms differ by either 1 or 2, ther are only 2 solutions!

You are all giving me a headache. Geeez!

It's easy to tell who is still in school.

I forgot this stuff years ago.

I forgot this stuff years ago.

Pythagorean??

Huh ? Totally lost.

Thank you all for your comments and votes (if you did so) ...

Also your prayers (also if you did so).

Every night I say mine. Peace to Dad (BadBunnee) who died in 2008. Peace with the Lord for my Big Bro (BadBunnee02) who is terminally ill with the same disease (PCA).

And to thanks to God that I was able to get through another day without algebra.

LGM

(still laughing but not for the reason(s) many of you think)

Also your prayers (also if you did so).

Every night I say mine. Peace to Dad (BadBunnee) who died in 2008. Peace with the Lord for my Big Bro (BadBunnee02) who is terminally ill with the same disease (PCA).

And to thanks to God that I was able to get through another day without algebra.

LGM

(still laughing but not for the reason(s) many of you think)

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