Below is an interesting problem from the 2026 ICTM State Competition. It is a geometry question from the individual contest.
Math competitions are often won in the margins, but rarely is a single question so deceptive that it fools nearly every student in the room—and even the contest organizers themselves.
At the 2026 ICTM State Competition, one geometry problem from the individual contest did exactly that. It was a masterclass in “hidden assumptions.” By the time the dust settled, only one person had walked away with the correct answer. Even the original official answer key was wrong.
To solve the problem, one must first identify the volume of the original rectangular solid. The volume, V, = 5*12*4=240. Thus, the transformed solid must have twice the volume of the original, so the transformed solid has volume 480. In order to find potential values of k and w, it is intuitive to set up the following equation:
12*(5-k)*(4+w)= 480
Simplifying,
(5-k)*(4+w)=40
Now, it seems like all one must due to solve the question is find the ordered pairs (k,w) that satisfy this equation. Since k and w are positive, there are not that many possible values of k and w, and simple “plug and chug” logic reveals the following three solutions for k and w: (4, 36), (3, 16), (1,6).
Adding up the sums of the pairs yields: 4+36+3+16+1+6=66
This is the answer that most students who solved the question, and also ICTM obtained. However, a few hours after the individual contest, ICTM issued a correction and changed the accepted correct answer to the problem to 174. This was quite surprising to most people because the answer of 66 seems logically correct. I was quite confused when I first saw this, wondering where 174 came from.
Looking back at the work above, there is a critical flaw. The equation (5-k)*(4+w)=40 is correct, but so is (5-k)(4+w)= -40. This is because volume can never be negative. Although the ordered pair solutions for k and w would result the rectangular solid extending into the negative x or negative z directions, the problem never stated this wasn’t allowed, and the volume of the solid is still 480. The problem only stated that k and w had to be positive integers.
Solving for the values of k and w that satisfy the equation reveals the following ordered pairs for k and w:
(6, 36), (7, 16), (9, 6), (10, 4), (13, 1)
Adding up the sums of these ordered pairs yields: 6+36+7+16+9+6+10+4+13+1=108.
Adding 108 to our original answer of 66 (because those solutions are still valid as well) yields: 108+66=174.
Overall, these “hidden” solutions that were overlooked by so many people, including the creators of the contest, demonstrate the true beauty of math. The solutions were always there, hidden behind a simple absolute value sign, that most people were unable to see.
During the contest, when I solved the problem, the thought of these additional solutions never crossed my mind. Although I ended up getting second place in the individual geometry contest due to this question and the question cost me 1st place, I am still very happy with my results and am excited for next year’s ICTM math team state competition.