This is adapted from the 1994 Putnam, A2. Thanks to Nirman for the suggestion!

Let R be the region in the first quadrant bounded by the x-axis, the line y = x/2, and the ellipse x2/9 + y2 = 1. Let R‘ be the region in the first quadrant bounded by the y-axis, the line y = mx and the ellipse. Find the value of m such that R and R‘ have the same area.

As usual, watch the video for a solution.

Putnam Ellipse Areas Question

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“All will be well if you use your mind for your decisions, and mind only your decisions.” Since 2007, I have devoted my life to sharing the joy of game theory and mathematics. MindYourDecisions now has over 1,000 free articles with no ads thanks to community support! Help out and get early access to posts with a pledge on Patreon.

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Answer To Putnam Ellipse Area Question

(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).

Thanks to Tony, Puggan and Nestor for alerting me of a mistake in an earlier version of the post!

There’s an ingenious trick to solve this problem. Consider the change of coordinates:

x1 = x/3
y1 = y

The change of coordinates transforms the ellipse into a circle with radius 1, and the regions R and R‘ into circular sectors S and S‘.

Since the change of coordinates is a linear transformation, the areas of R and R‘ are multiplied by the same non-zero factor, so they will have same area if and only if S and S‘ have the same area. By symmetry of the circle, this happens if and only if S and S‘ are reflections about the line y1 = x1.

The bounding line of y1 = 3x1/2 in region S is reflected to give the bounding line:

x1 = 3y1/2
y1 = 2x1/3

We now use the reverse change of coordinates from S‘ to R‘:

x = 3x1
y = y1

This results in the line y = 2x/9, so m = 2/9.

Reference

Putnam 1994, A2 solution, John Scholes
https://prase.cz/kalva/putnam/psoln/psol942.html

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