________________

On a road trip to the West, the teammates use their down time to hone their geometric and visualization skills.

________________

Divide by Three

Team member Carolyn spots a mesa with a silhouette that she models on paper as half of a regular hexagon, as shown below. She challenges the others to cut it into three pieces of the same shape, where no two are of the same size. (The pieces do not have to be the same shape as the original.) Carolyn states that she has found two solutions. Pieces may be turned over.

How does Carolyn do it?

Cutting the Domino

Team member Brad has some pieces of paper in the shape of 1-by-2 rectangles, and cuts each into three pieces. The three pieces have the same shape but may be different sizes. Pieces may be turned over.

1. Find the two ways he did this where the pieces are triangular.
2. Christy asks what is the smallest length possible for the cuts that accomplish Brad’s task. She knows it’s one of several solutions where the pieces have an L-shape.

Solutions to week 115

In Thanksgiving Split, Bob can force Joe to take 3/4 of the cake. In Easy as Pie, Alice cuts the pie into pieces of sizes 1/5 and 4/5. Beth then cuts to leave pieces of sizes 1/5, 2/5, and 2/5. Alice then leaves pieces of sizes 1/5, 1/5, 1/5, and 2/5, and gets 3/5 of the pie.

Thanksgiving Split answer explained:
Joe cuts the cake into pieces of size s and 1 – s, where s ≤ 1/2. Bob should always cut the smaller piece in half, forcing Joe to take s + (1 – s)/2 = (1 + s)/2 of the cake. Thus, Joe should cut the cake into equal halves and Bob cuts one of those in half, forcing Joe to take 3/4 of the cake.

Easy as Pie answer explained:
Alice can cut the pie into pieces of size s and 1 – s, where s is no larger than 1/2. There are four different situations to consider. Of these four, the most Alice can get is 3/5 of the pie.

1. For s ≤ 1/5, Beth cuts to leave pieces of sizes s, (1 – s)/2, and (1 – s)/2. Alice then leaves pieces of sizes s, (1 – s)/4, (1 – s)/4, and (1 – s)/2 and gets (1 + s)/2 of the pie. For s = 1/5, Alice gets 3/5 of the pie.

2. For 1/5 ≤ s ≤ 1/3, Beth cuts to leave pieces of sizes s, (1 – s)/2, and (1 – s)/2. Alice then leaves pieces of sizes (1 – s)/4, (1 – s)/4, s, and (1 – s)/2 and gets 3(1 – s)/4 of the pie. For s = 1/5, Alice gets 3/5 of the pie.

3. For 1/3 ≤ s ≤ 3/7, Beth cuts to leave pieces of sizes s, (1 – s)/2, and (1 – s)/2. Alice then leaves pieces of sizes (1 – s)/4, (1 – s)/4, (1 – s)/2, and s and gets (1 + 3s)/4 of the pie. For s = 3/7, Alice gets 4/7 of the pie.

4. For 3/7 ≤ s ≤ 1/2, Beth cuts to leave pieces of sizes ε, s, and 1 – s – ε, where ε is arbitrarily small. Alice then cuts the middle piece and gets 1 – s of the pie. For s = 3/7, Alice gets 4/7 of the pie.

Recent Weeks

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

Come back next week for answers and more puzzles.