Olympiad question - math tricks to find the median

A list of numbers has the number n appear n times, for n ranging from 1 to 200. What is the median of the list? Thanks to Debapriya for the suggestion! This question comes from the 2021 AMC 12A, Problem 16.

2021 AMC 12A Problems/Problem 16
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[chanle]

I solve it using a geometric way. It is hard to explain without any graph or animation but I would try my best.
Rather than listing the number in a long sequence, I stack them like a right-angled triangle like so:
1
2 2
3 3 3
4 4 4 4
…
To find the median of the sequence, I would create an inscribed similar triangle, which share the top vertex, and its area is half of the big triangle. By simply comparing the area, we can find that height of the small triangle is 1/√2 of the big one. So if we put 200 into the calculation, we should get 141.42. Since 141.42 pass the level of 141, the next question would be whether the median would be 141.5 or 142. Next, I divide each level in n part, where n is the number of level. If the height is less than 142+1/142, that means the median lay between 141 and 142, so the median would be 141.5. But in this case, 141.42>141.01, so the median would be 142.

[mike.]

It would have been better to focus on the 10050 as soon as you estimated 141 as a close guess than to keep using that inequality. Once you discovered that (141)(142)/2 was too small, you should have added 142 to it and noticed that was big enough to solve the problem. That's what you mentioned as a second thought, but the first way you did it with all that number manipulation was too much work.

[wehpudicabok]

I got most of this right, but then I made a really subtle mistake at the end. When I realized that "141 is too small and 142 is too big, " I remembered that we were looking for a median, and decided the answer must be 141.5. I had forgotten that we were looking for the median not of all the values of numbers in the set, but the values of all members of the list. In this context "too small" meant "doesn't get you to the midpoint of the list, " and so both of the numbers around the midpoint would therefore be 142. This is the sort of logical leap it's very easy to stumble on in higher areas of mathematics, in my experience.

[koroslav]

Alternate solution. It is the first integer for which the sum of all the nubers above is smaller or equal than sum below. So solve
(200+n)(200-n)/2-n(n+1)/2=0
Multiply by 2 and expand
40000-n^2 -n^2-n=0
2n^2+n-40000=0
Which is simple quadratic formula and the positive solution is approx 141.17 which means that answer to original question is 142

[nedmerrill]

Got this by associating the sequence with a right triangle, from 1 at the apex down to 200 at the base, as the numbers increase linearly. The median would occur at the half area point of the triangle, with is at height/sqrt(2), or 200/sqrt(2), or about 141+. After that it's trial and error as Presh shows.

[beng]

Once you get to the inequality: n(n+1)>20100, we can note that 20100 = 100 × 201 = 150 × 134 = (142+8)(142-8) = 142² - 64. Therefore, n=142 is enough to past the median but only by 64. Therefore, the median is 142.

[Kshitij.with.nature-channel]

This is the best problem i have seen, amazing, thank you❤😃

[tryingtomakeanamebelike]

My method for solving:
Based on the problem shown, there is one 1, 2 twos, 3 threes, etc. The number of numbers can be represented as the arithmetic series
1+2+3…+200
Use summation formula
=20100
There are 20100 numbers, meaning the median is the sum of the 10050th number and the 10051st number
Use trial and error with summation formula
70(141) = 9870
9870 numbers up to 140
9870+141=10011
After adding 142 we see the 10050th and 10051st number are both 142
Answer is 142

[MathFromAlphaToOmega]

Here's a harder version: For each permutation of those numbers, keep track of the middle number. What is the average of all those middle numbers?

[gauravjaiswal]

142
Using the formula of n(n+1) /2 we get the middle location then again by using this formula we can try to get the value of no at this location

[daakudaddy]

Cant believe that I actually solced it, and sopvwd it indeoendently using the same thinking as you

[mike.]

3:38 I started going "what the heck are you doing?" for why you were doing that much work. I think every person who would have gotten this far in the question can use long multiplication to multiply three-digit numbers much faster than all you did.

[zuFuchs]

additional info √2/2: the 2 represents half the aera of the rectangle being the triangels arrea. √2 origins from the diagonal of a 1 x 1 square. Math has such an inner beauty...

[iAmOrenWhoRU]

in general, the median is about the ceiling of n/√2
more specific:
c = count of elements = n*(n+1)/2
mp = middle position = mp=(c+1)/2
cmv = ceiling of n/√2
if mp-(cmv*(cmv-1)/2)==0.5, there are two values for the median: cmv-1 and cmv (example: n=3: (2 values), 2, 3, (2 values))
if (cmv*(cmv-1)/2)==mp, the median is cmv-1 (example: n=10: (27 values), 7, (27 values))

[r]

A heuristic approach. More often used in computer programming.

[ultrametric]

You can do this with straight algebra without having to estimate anything. Adding up we have (1 + 2 + .. + 200) + (2 + 3 + .. + 200) + (3 + 4 + ... + 200) + ... + 200 and it's fairly easy to write down an explicit formula for the sum using SUM(1..n) = n(n+1)/2 and SUM(k..n) = SUM(1..n) - SUM(1..k-1). Also easy to count the terms and so take the average.

[shakyasajesh]

Found 141.5>n>140.5 by solving quadratic of n^2+n-20, 100. Which implies n is greater than 140. So i found the 142th trianglar number which was 10, 153. This means the last 141 was at 10, 001 (cool palindrome btw) which is before the midpoint 10 500 so answer is 142

[jacobgoldman]

Follow up question is value of mean other than writing (1^2+2^2+…+200^2)/20100.

[strikerstone]

hard but unique and fun, i tried this before but today i did it

[bryananthony]

number of terms = 200/2(2 + (200-1)1 = 20100 Hence the terms giving the median are terms 10050 and 10051,
n/2(2 + 1(n - 1)) = i0050 giving n^2 +n -20100 = 0. n=141.28, since n is a whole number and n is greater than 141, n = 142.Similarly term 10051 gives
n = 142 The median is (142 + 142) / 2 = 284/2 = 142