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PuMAC Combinatorics A 2022 Problem 2

Ten evenly spaced vertical lines in the plane are labeled ℓ1, ℓ2, . . . , ℓ10 from left to right. A set {a, b, c, d} of four distinct integers a, b, c, d ∈ {1, 2, . . . , 10} is squarish if some square has one vertex on each of the lines ℓa, ℓb, ℓc, and ℓd. Find the number of squarish sets.

I tried solving this problem using some ideas from co-ordinate geometry but none of it worked as there were too many equations to solve.

The official solution used a very decent approach, it made the question really simple but I'm unable to understand the highlighted part. This is the offical solution:

Without loss of generality, assume that a < b < c < d. Then, it is easy to see that {a, b, c, d} is squarish if and only if the distance between ℓa and ℓb equals the distance between ℓc and ℓd. In other words, we must count the number of subsets {a, b, c, d} of {1, 2, . . . , 10} with d−c = b−a. To do this, we proceed by casework on k = d − c. • If k = 1, we find that for each value of d, the maximum possible value for a is d − 3, so there are d − 3 possible combinations of the four numbers. Then, d ranges from 4 to 10 inclusive, for a total of 7 + 6 + 5 + 4 + 3 + 2 + 1 = 28 combinations. • If k = 2, each value of d gives d − 5 possible combinations. Then, d ranges from 6 to 10 inclusive, for a total of 5 + 4 + 3 + 2 + 1 = 15 combinations. • If k = 3, each value of d gives d − 7 possible combinations. Then, d ranges from 8 to 10 inclusive, for a total of 3 + 2 + 1 = 6 combinations. • If k = 4, there is only 1 combination a = 1, b = 5, c = 6, d = 10. Summing yields a total of 28 + 15 + 6 + 1 = 50 sets {a, b, c, d}.

The problem is that I am failing to proof that why the distance between ℓa and ℓb has to be equal to the distance between ℓc and ℓd.

Can anyone help ?

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    $\begingroup$ You've got a vertex on each line. Call the vertices $A,B,C,D$ with the obvious correspondence to the lines. Opposite edges of a square are equal and parallel, so the line segments $AB$ and $CD$ are of equal length and parallel. Now drop a perpendicular from point $A$ to line $b$, and from point $C$ to line $d$. You should see some congruent triangles that let you finish off the proof. $\endgroup$
    – Gerry Myerson
    Commented Jun 30 at 9:15
  • $\begingroup$ Making any progress, Aashita? $\endgroup$
    – Gerry Myerson
    Commented Jul 3 at 12:28
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    $\begingroup$ I’m voting to close this question because OP refuses to engage. $\endgroup$
    – Gerry Myerson
    Commented Jul 5 at 4:35
  • $\begingroup$ I'm sorry, there might have been a glitch due to which I wasn't notified about the comment made 2 days ago. The solution was very helpful and I could solve the question with the help of that, but it wasn't given in the form of an answer so I couldn't mark it as the correct one and writing thanks in the comments was not permitted so I upvoted the answer to show my acknowledgement. $\endgroup$
    – Aashita
    Commented Jul 5 at 12:49
  • $\begingroup$ I encourage you to write up your solution and to post it as an answer. That may sound odd, but it's actually encouraged on this site. Then, you can accept your own answer – that's also quite an acceptable thing to do. $\endgroup$
    – Gerry Myerson
    Commented Jul 5 at 12:54

1 Answer 1

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Assume the points forming the square ABCD are - A,B,C,D on la, lb, lc and ld respectively.

Now, Since opposite sides of a square are Equal and Parallel.

⇒ AB = CD

Then drop a perpendicular from AE to lb and from DH to ld.

Now we can proof Δ ABE ≅ Δ CDH by AAS

AB = CD

∠ ABE = ∠ CDH

∠ AEB = ∠CHD

Thus, AE = CH by CPCT

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