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### Small Sudoku

Ariel is solving Sudoku puzzles. Inspired by the evening’s theme, Ariel thinks about 4-by-4 puzzles, in which the solution must contain each of the numbers 1 through 4 in each row, in each column, and in each of the 2-by-2 squares in the corners of the grid. In the example diagram, the space labeled with a “?” must be filled with a three, because the lower left 2-by-2 already has a 1, a 2, and a 4.

*What is the smallest number of clues (filled-in squares) that can be given for a 4-by-4 sudoku so that it will have a unique solution? In other words, so that there will only be one way to fill in the rest of the squares while obeying all the rules?*

### Puny Pegs

Raven is playing peg solitaire. Different puzzles have different starting configurations, but all of the pegs are on the grid points of a square grid. On each turn, you can jump any peg over any other adjacent peg horizontally or vertically, removing the peg jumped over.

The diagram shows a 6-by-6 game in progress, where black circles represent pegs and white circles represent empty spaces. Peg A could jump over either peg B or peg C, removing that peg and landing on space D or space E, respectively. It can’t jump over peg F because it isn’t adjacent, and it can’t jump over peg G because the space where it would land is occupied.

*What is the smallest square board for peg solitaire in which you can start with every grid point except one occupied and end up with all but one empty?*

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## Solutions to week 72

**The Third Circle**. The accompanying diagram shows the givens as described in the problem, with the point of intersection *G* of lines *s* and *t* and some auxiliary line segments added in gray and in dashed lines. Note that since the radii of circles *a* and *b* are both *AB*, triangle *ABC* is equilateral with all angles equal to 60°. Let *x* be the measure of angle *EBC*. Since *EC* and *CF* are both radii of the same circle, and segments *AF, BE, AC,* and *BC* are all equal, triangles *AFC* and *BEC* are congruent. Hence, angle *FAC*=*GAC* is also *x*. Angle *ABG* is 60-*x*, and since triangle *ADB* is isosceles, angle *ADG* is also equal to 60-*x*. But now angles *ADG, GBA, BAC,* and *CAG* add up to 60-*x* + 60-*x* + 60 + *x* = 180-*x*, so the only bit of the angles of triangle *DAB* left, namely angle *GAD*, must also have measure *x* to reach the total of 180°. Since triangle *DAF* is isosceles, angles *ADF* and *DFG* must both be 90-*x*/2. But since *ADG* is 60-*x*, this means that angle *GDF* is 30+*x*/2. Subtracting the measures of the other two angles of triangle *DGF* from 180°, we get that the desired angle *DGF* is 180 – (30+*x*/2) – (90-*x*/2) = **60 degrees**.

**The Second Bisection**. Again, we start with a diagram that shows the givens as in the problem, as well as the intersection point *H* of lines *s* and *v* and some auxiliary segments. Since *u* is the perpendicular bisector of segment *AB*, *GA* = *GB*. Since they are both radii of circle *a*, *GA* = *AB*. Hence, triangle *GAB* is equilateral, and all of its angles are 60°. Let the measure of angle *AGD* be *y* degrees. Then since triangle *AGD* is isosceles, angle *ADG* is also *y*, and angle *DAG* is 180-2*y*. Since *v* is the angle bisector of *DAG*, angle *HAG* is 90-*y*, and subtracting angle *BAG*=60°, angle *HAB* is 30-*y*. On the other hand, angle *DBG* subtends the long arc *DG*, which has measure 180+2*y*, so the measure of angle *DBG* is 90+*y*. Subtracting angle *GBA*, we have angle *ABH*=30+*y*. Subtracting both *HAB* and *ABH* from 180°, we obtain angle *BHA* = 120°. Since this is the obtuse angle between lines *s* and *v*, we conclude that the acute angle between those lines is also **60 degrees**.

## Recent Weeks

**Week 72**: The Second Bisection & The Third Circle, solutions to Eye Like It & Pythagorean Poster

**Week 71**: Eye Like It & Pythagorean Poster, solutions to In The Groove & On The Path

**Week 70**: In The Groove & On The Path, solutions to Not Dozens & Festive Factorial

**Week 69**: Not Dozens & Festive Factorial, solutions to Snug Circle & Congruence Time

**Week 68**: Snug Circle & Congruence Time, solutions to Card 61 & Card T

Links to all of the puzzles and solutions are on the Complete Varsity Math page.

**Come back next week** for answers and more puzzles.