## Week 38: 2016 June 4

**Hat Trick**. There are 4! = 24 ways the students can order themselves left to right. We have to multiply that by the number of ways they can wear the hats. Split this latter number up into the possibilities for their wearing four hats, three hats, or two hats. In the first case, there are five possible hats for the first person, four for the second, three for the third, and two for the last person, or 120 options. In the second case, there are four possible subsets of three people wearing hats, and then there are five possible hats for the first person, four for the second, and three for the third, for 240 options. And in the last case, there are six possible pairs of people to wear hats, and then there are five times four choices of hats, for another 120 options. Thus, there are 480 options for hat-wearing in all, times the 24 orders of the students, for a total of 11,520 possible pictures.

**Possible Parabolas**. The interior of a parabola (which can also be seen as the set of all points on the opposite side of the parabola from any tangent to the parabola) is what is known as a *convex* set: if *P* and *Q* are two points in the interior, then the entire line segment *PQ* lies in the interior. (The accompanying diagram illustrates this.) This fact immediately tells us that the entire triangle (and its interior) defined by points *A, B,* and *C* lies in the interior of any parabola through *A, B,* and *C*. On the other hand, if we put different coordinate axes on the plane containing *A, B,* and *C*, as long as the *y*-axis is not parallel to any side of the triangle, we can use quadratic interpolation to find a parabola through all three points with central axis parallel to the chosen *y*-axis. But as that chosen *y*-axis becomes closer and closer to parallel to *AB*, the arc of the parabola from *A* to *B* becomes closer and closer to the segment *AB*. This shows that any point on the opposite side of the segment *AB* from the triangle *ABC* is *not* in the interior of every parabola through *A, B,* and *C*. Hence, the desired region is simply the triangle defined bu *A, B,* and *C*. Since the answer (11,520) from the previous week’s **Hat Trick** problem divided by 1,440 is just 8, the desired area is just ½×(8-1)×4 = **14**.

**Equal Laterals**. From the diagram, we can see that for *any* interior point *P* in equilateral triangle *T* = *ABC* the area of the entire triangle *ABC* is equal to the sum of the areas *PAB* + *PBC* + *PCA* = ½*sPS* + ½*sPQ* + ½*sPR*, where *s* is the side length of the triangle. Since the area of *ABC* is (√3/4)*s*², multiplying by 2/*s* gives us that *PS* + *PQ* + *PR* = (√3/2)*s*. Since in this case *s* is 14 (from the previous week’s **Possible Parabolas**), the answer is **7√3**.

*P*as specified in the problem is

*much*more difficult than solving this problem for an arbitrary point inside the equilateral triangle. So be on the lookout for that kind of monkey wrench in the problems — Coach Newton is kind of mean that way, hoping to sidetrack you.

**Isocircules**. The diagram shows the situation as described in the problem, with the addition of point *M*, the midpoint of segment *BD*. Since by construction arcs *BC* and *CD* are equal (being subtended by the two equal angles at *A*), triangle *BDC* is isosceles, and triangle *BMC* is a right triangle. Hence, the ratio *BM* / *BC* is equal to cos *DBC*, and the desired ratio *BD* / *BC* is equal to 2cos*DBC*. Now, angle *DBC* subtends the same arc as *DAC* which is the same angle as *BAC*, so 2cos*DBC* = 2cos*BAC*. But by the law of cosines, *BC*² = *AB*² + *AC*² + 2(*AB*)(*AC*)cos*BAC*. Using that *AB* = *AC* and dividing out by *AB*², this gives us that *BD* / *BC* = 2cos*DBC* = 2cos*BAC* = 2 – *BC*²/*AB*² = 2 – (*BC*/*AB*)² = 2 – 1/4 = **7/4**.

## Week 39: 2016 June 11

**Phenomenal Packing**. As you can see in the diagram, Ariana’s boxes have sizes (in decimeters) 1, 1, 2, 2, 2, 3, 3, 4, 6, 6, and 7.

## Week 40: 2016 June 18

**Spiral Notebook**. Rather than thinking of them spiraling in the plane, it is easier to think of the two curves as two points on the unit circle, which start at an angle of 0 at time *t* = 1 and return to that same angle at time *t* = 10. (In other words, interpret the distance from the origin as time rather than distance. From this point of view, it is obvious that if each point made 1 lap around the circle, since they are going in opposite directions, that they would intersect once. (No more and no less.) Each lap added by either point increases the number of intersections by 1, so with eight and seven laps respectively, the two curves will intersect exactly **14** times.