# Problem of the Week (POTW)

Over the summer, a problem will be posted each week for students to work on and submit their solutions. These problems are curated by the HMMT Problems staff — the same people who write the HMMT November and February tests — and are intended to increase in difficulty every week. After the submission deadline closes, the problem writers will review submissions and post one of their favorites alongside the next POTW. Discussion of the POTW through any form of media is closed until the submission deadline passes.

### This Week's Problem (July 26 to July 30)

The squares of the following figure are filled with the integers $1, 2,\ldots , 64$, such that for each $1 \leq i < 64$, the numbers $i$ and $i + 1$ are in adjacent squares. What is the minimum possible sum of the $28$ numbers in the gray squares?
Note: This problem is challenging! Feel free to submit any progress you find towards the solution.
This week's problem will be released on Monday, July 26 at 12 PM EDT (UTC-0400).

### Schedule

The current POTW and past week’s solution will be posted on Monday, July 26 at 12 PM EDT (UTC-0400).
The deadline for submission is Friday, July 30 at 06 PM EDT (UTC-0400).

### Last Week's Problem

The answer to last week's problem of the week was $\boxed{\frac{5\sqrt{26}-7}{14}}$.
One method, used by several submissions, was to convert the entire problem into algebra. If we introduce a coordinate system on the diagram so that the lower left corner is $(0,0)$ and $(4,4)$, we need to solve $(x-0.5)^2+(y-0.5)^2 = (x-1.5)^2+(y-3.5)^2=(x-3.5)^2+(y-2.5)^2.$ While not particularly elegant, it is possible to deduce $(x, y) = (13/7, 12/7)$ and solve the problem.
Viraj S. from Kent School, Connecticut submitted a slick solution, which observes that if we let $\Omega$ be the circle that we are interested in, $\Omega_1, \Omega_2,$ and $\Omega_3$ be the three smaller circles, and $\Omega '$ be the circle that passes through the centers of the small circles, then
Since the radius of $\Omega_1$, $\Omega_2$, and $\Omega_3$ are all $0.5$, the radius of $\Omega$ is simply $R' = R + 0.5$ and both $\Omega$ and $\Omega '$ share the same center (this is true since $\Omega$ is tangent to $\Omega_1$, $\Omega_2$, and $\Omega_3$).
Here's a handy diagram to visualize this:
From there, one can finish the problem using the property $[ABC] = \frac{AB \cdot BC \cdot AC}{4R'}$
Using Pythagoras theorem we get that, $BC = \sqrt 5$, $AC = \sqrt{13}$, and $AB = \sqrt{10}$... Now, we will find the area of triangle ABC... The easiest method [is] to subtract [from] the area of the square DEFA the areas of the triangles DBA, BEC, and CFA. $[ABC] = 3 \cdot 3 - \frac{3 \cdot 1}2 - \frac{2 \cdot 1}{2} - \frac{3 \cdot 2}2 = \frac7 2$
Here's another diagram to help understand how we found the area of this triangle:
Rearranging the triangle-area expression and inserting these values gives $R = R' - \frac{1}{2} = \frac{\sqrt{5} \sqrt{13} \sqrt{10}}{4 \cdot \frac{7}{2}} - \frac{1}{2} = \frac{5\sqrt{26}}{14} - \frac{1}{2}.$

### Archive

Past POTWs and their solutions can be found in the archive.