Congratulations to Philip H. for winning the One-Fold Origami Puzzle Raffle! Many thanks to all who participated!

Here is Philip’s solution:

Step 1. Pinch at the halfway point along an edge.

Step 2. Fold and unfold along the indicated diagonal.

Step 3. Fold and unfold along a crease that goes through the lower right corner and the midpoint of the edge that you marked in Step 1.

Step 4. Where the horizontal that passes through the intersection of the two creases you made in Steps 2 and 3 meets the left edge, make a pinch.

Step 5. Make another pinch halfway between the lower left corner and the pinch you made in Step 4.

Step 6. Fold the upper left corner to the bottom edge along a crease that passes through the left edge at the pinch you made in Step 5.

Step 7. Make a pinch where the upper left corner of the square meets the lower edge, then unfold the paper.

Step 8. This is how your origami paper should look.

Step 9. Fold the lower left corner to a point on the diagonal that you creased in Step 2 over a new crease that passes through the mark on the lower edge that you made in Step 7.

Step 10. Finished model. The colored triangle will have the same area as the white margin.

Can you prove that this folding sequence works?

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You could just copy the sqrt(2/3), which is already there in the solution to the horizontal version, to two adjacent sides of the paper, and then the desired configuration is just one step away. The whole sequence takes only 6 folds if you do it in the right way.

Sorry I made a silly careless mistake in my earlier post… But I found a solution with as few folds! Let the square be ABCD. { (1) Fold A to C, making line K. Unfold. (2) Fold A to D, making line L. Unfold. (3) Fold through D such that A goes onto L. The fold cuts AB at X. Unfold. (4) Fold through D such that X goes to Y on K. (5) Fold through Y not along K. Unfold everything. (6) Fold D to Y. } Let the square have side 1. In (3) let A go to M on L. Then L bisects AD and AD = DM, so ADM is equilateral, thus ADX = 30 deg, hence DY = DX = 2 / sqrt(3). Therefore in (6) the “square” created by the fold has side sqrt(2/3), and area 2/3.

Yes, this is also a very nice solution! Thanks for sharing!