top of page

Solution to the Gifted World Challenge #5 posted on Mar 30, 2024

Updated: Jun 19

Hello everyone! Apologies for the delay in posting the solution to this challenge- we had extended this challenge because nobody had provided a proof that only 2 rectangles (4x4 and 6x3) have the same numerical value of area and perimeter. You can see the original question towards the end of this post. Observations on the Responses we Received

This was not an easy question and we appreciate all the students who spent time on this and submitted responses.

  • Many students identified the 2 rectangles that satisfy this condition, through trial and error.

  • Some students went about exploring the problem systematically, starting with one side as having length 1, length 2 etc. That approach allowed them to notice interesting properties- that Perimeter was greater than Area till side length went above a ceerain number, and Area was greater than Perimeter above a certain side length.

  • Some students said that infinite such pairs of rectangles exist and gave some examples beyond 4x4 and 6x3. Unfortunately they were careless and did not check that other rectangles did not satisfy this condition (for example students said that 60x30 and 12x6 satisfied the condition.

Only one student, Aahana Das, gave an answer that is a proper proof.

Aahana Das' Solution

Aahana is in Grade 7 and is probably not introduced to formal algebra yet. She has however worked systematically and methodically like a mathematician on this problem to prove that there are only 2 rectangles that satisfy the criterion. Her agruments are beautifully constructed! We have summarized the key steps of her solution in language that is easier to understand (you can read her original detailed answer below).

  • Aahana calls the shorter length 'breadth' and systematically explores the relationship between area and perimeter for breadth= 1, 2 and so on. She establishes that area and perimeter cannot have equal numerical values if breadth is 1 or 2. In fact, perimeter > area in such cases.

  • Exploring breadth 3, she says Area has to be 3xL (Breadth) and for the Perimeter to be equal to Area, L+L+B+B has to be equal to 3xL which means L has to be twice B (because B+B has to account for 1 L). Thus she arrives at L=6 and 3x6 being one rectangle that satisfies the criterion.

  • She then establishes that for breadths greater than or equal to 5, area is always greater than perimeter (remember that she defines breadth as the shorter side). She does this through the following arguments. The rectangles we consider now have a breadth of at least 5 and a length 5 or larger. Let's call the breadth of such rectangles as 'medium number' and the length as 'larger number'.

    • Now Perimeter (P) is 2(medium number + larger number). Thus, the maximum perimeter of the rectangle is 4x larger number

    • While Area (A) is medium number x larger number; so area A is at least 5 x larger number

    • Thus for breadth 5 or larger, A is always > P!! Because 5x larger number > 4xlarger number (the maximum possible perimeter)

  • Finally she shows how a 4x4 rectangle satisfies the condition.

  • Thus, only 3x6 and 4x4 rectangles have equal numerical value of area and perimeter.

An elegant solution from elsewhere

We took this problem from the fantastic NRICH website. You can see the problem and the approaches of different students here. We really like Bhavik's solution featured on NRICH. We share a modified version of his explanation below.

1.      Divide the rectangle into unit squares and shade all the squares along the borders.

2.      A rectangle always has 4 corner unit squares (except for rectangles where one side is 1 unit)

3.      Each border square contributes 1 unit to the area and 1 unit to the perimeter, except for the 4 corner squares that account for 2 units of the perimeter (because 2 of the sides of the unit square are exposed). Thus, Perimeter= 4 + number of shaded border squares.

4.      Which means that for Perimeter to be equal to Area, there can be only 4 inside squares!

5.      There are only 2 ways to have 4 inside squares- 2x2 and 1 x4


6.      If we now put a ‘unit square border’ around these ‘inside rectangles’, we get either 4x4 or 3x6 rectangles.


Detailed Answer from Aahana Das, Grade 7, GreenWood High, Bengaluru

The Question

Perimeter and area obviously have different units , but is it possible to have two rectangles in which their numerical values are equal ? If so, how many such rectangles are there and what are their dimensions? Assume the side lengths of the rectangles are natural numbers i.e. 1,2,3.... not fractional or decimal values.

You are NOT allowed to use algebra in answering this question.

You can submit your answer here. Even if you don't manage to answer this question completely, share whatever you have done and what you plan to investigate next. How you think is more important than getting the right answer quickly!

(Note that commenting is disabled on this post so that nobody gives away the answer to someone who hasn't solved it yet)

155 views0 comments


bottom of page