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 rightangled triangles. LottieLu'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!
LottieLu 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.
Hide
Answer
1. 6810 (Area = 6x8/2 = 24)
2. 51213 (Area = 5x12/2 = 30)
Hide
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/(b4). 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 = (4b8 )/(b4). 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 (a4) = 8/(b4). 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/(b4) expands to a=(4(b4)+/(b4), or a = 4 + 8/(b4). The only (+ve) integer solutions for a and b come when b4 is a factor of 8; as Ram says, when b=5,6,8 or 12.
So yes, those are the only two allinteger triangles. However, any rightangled triangle whose short sides satisfy the equation a = 4 + 8/(b4) 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! 
Jimbo
Dec 08, 2014
 Pythagorean triads where two larger sides differ by 1 are: (n^21)/2, n, (n^2+1)/2
So: Area = perimeter gives
0.5 x n x (n^21)/2 = (n^21)/2 + n + (n^2+1)/2
n(n^21) = 2(n^21)+4n+2(n^2+1)
n^21 = 4n + 4
n^24n5=0
(n5)(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^21), 2n, (n^2+1)
Area = perimeter
0.5 x 2n x (n^2)1) = (n^21) + 2n + (n^2+1)
n(n^21)=2n^2+2n
n^21=2n+2
n^22n3=0
(n+1)(n3)=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!

Babe
Dec 17, 2014
 You are all giving me a headache. Geeez! 
jaycr
Dec 17, 2014
 It's easy to tell who is still in school.
I forgot this stuff years ago. 
catmom
Dec 17, 2014
 Pythagorean?? 
auntiesis
Dec 17, 2014
 Huh ? Totally lost. 
LittleGreenMan
Dec 17, 2014
 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) 
Back to Top
 
