Who Needs the Right Formula, Anyway?
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.
HintUse the fact that all three sides of each triangle were whole numbers.
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Answer
1. 6-8-10 (Area = 6x8/2 = 24)
2. 5-12-13 (Area = 5x12/2 = 30)
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Comments
Marple  
May 16, 2010
| I am so bad at mathematics.  Really, really bad. At school my teachers despaired of me!   |
pating 
May 18, 2010
| Perfect right triangles! |
dalfamnest 
May 18, 2010
| 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!  |
Starriddler 
Jun 24, 2010
| Answering dalfamnest's comment, it only works with pythagorean triples. |
dalfamnest
Jun 25, 2010
| 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"! |
lessthanjake789
Jul 08, 2010
| 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. |
lessthanjake789
Jul 08, 2010
| (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. |
RRAMMOHAN
Oct 20, 2012
| 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. |
spikethru4 
Jan 31, 2013
| 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 |
spikethru4
Jan 31, 2013
| When 2 |
spikethru4
Jan 31, 2013
| Let's try that again:
When 2≤b≤4, a≤0 or a=∞, but interestingly, it also does not work when 0 |
spikethru4
Jan 31, 2013
| Oh, ffs...
when 2>b>0, where the formula suggests it should. Further investigation is required to explain this apparent contradiction! |
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