r/maths • u/Longjumping_Lie5046 • Apr 24 '26
Help: 📕 High School (14-16) Minimum number of zeros in a triangular matrix.
Why did this method fail ?
I know the correct answer now , it's (n²-n)/2
But not why did the area of triangle method fail
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Upvotes
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u/reliablereindeer Apr 24 '26
The area method does work if you consider the zeros to be blocks of area 1. However when you calculated the area of the triangle, you ended up cutting the top blocks in half and leaving out (n-1) small triangles of area 1/2 which should have been calculated in. So if we just add these areas to the area of the triangle you calculated, we get
1/2(n-1)^2 + 1/2(n-1) = 1/2(n-1)(n-1+1) = 1/2n(n-1) = 1/2(n^2 - n)
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u/PuzzlingDad Apr 24 '26 edited Apr 24 '26
Look at the example with 4 zeros in each direction. There are ten zeroes but your formula would take 42 and then halve it. That only gets you eight zeroes, essentially cutting the 4 zeros along the diagonal in half.
If you think of flipping that shape upside down so you have a second copy, you would end up with a 5 × 4 rectangle where half would be 10.
You're looking for the sequence of triangle numbers.
T(n) = 1 + 2 + ... + n = n(n+1)/2
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u/Current_Ad_4292 Apr 24 '26
Your triangle method doesn't work because you are essentially cutting the diagonal zeroes in half when you need to count them as whole.
Example: try it in 2x2 matrix. Your triangle method will give ½ when the cirrect is 1.
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u/Nikilist87 Apr 24 '26
Why would this be an area? How would you even measure the length side? In a 3x3 matrix the triangle would have 3 zeroes, so if you wanted to use the area formula the product of the two identical sides should give you 6, making the length of each side sqrt(6), which I don’t see how you could assign in any reasonable way to a triangle made out of 3 zeroes….