Too big to count
A diagonal of a polygon is a straight line joining two corners that are not next to each other. How many diagonals does a polygon with 20 sides have?
Drawing all of them and counting is hopeless: pairs of them cross thousands of times, and it is easy to miss one or count one twice. So shrink the problem until it can be done by hand, and look for the method there.
Count the small cases
A triangle has no diagonals, because every corner is next to both of the others. A square has 2, the two lines corner to corner.
A pentagon has 5: from each of its corners two diagonals leave, to the two corners not beside it. A hexagon has 9: three long ones through the center and six shorter ones.
The counts so far are 0, 2, 5 and 9 for 3, 4, 5 and 6 sides. They grow by 2, then 3, then 4, which suggests 14 for 7 sides. A pattern in a list is a guess. The small cases are worth more for what they show about why.
A pentagon with its 5 diagonals in gold. Each corner sends out two, to the two corners not next to it.
A hexagon with its 9 diagonals in gold: AD, BE and CF through the center, and six shorter ones.
The reason in the small cases
Stand at one corner of a polygon with n corners. A diagonal can go to any corner except three: the corner itself and its two neighbors, which are joined to it by sides. So each corner sends out n − 3 diagonals. In the hexagon, n − 3 = 3.
There are n corners, so that makes n(n − 3) diagonal ends. Each diagonal has two ends and is counted once from each, so the number of diagonals is .
This is a reason, not a pattern. Nothing in it used n = 5 or n = 6, so it holds for every polygon.
From one corner of a hexagon, diagonals go to the 3 corners that are neither the corner itself nor its two neighbors: 6 − 3 = 3.
A second count agrees
There is another way to count. Any two of the n corners are joined by a line, and there are pairs of corners. Of those lines, n are sides, and the rest are diagonals: .
The two answers are the same, because . Two different arguments reaching one formula is strong evidence that neither has slipped.
Check, then use
Before trusting the rule on the case you cannot draw, test it on cases you counted. For n = 6 it gives 6 × 3 ÷ 2 = 9, and the hexagon has 9. For n = 4 it gives 4 × 1 ÷ 2 = 2, and for n = 3 it gives 0.
Now use it on the case you could not count: n = 20 gives 20 × 17 ÷ 2 = 170 diagonals. The second count agrees: 20 × 19 ÷ 2 = 190 lines between corners, less 20 sides, is 170.
The diagonals counted from drawings, beside the rule. They agree for every polygon from 3 to 7 sides.
The same move on handshakes
How many handshakes are there when 30 people each shake hands once with everyone else? Shrink it to 4 people, A, B, C and D: the handshakes are AB, AC, AD, BC, BD and CD, which is 6.
The small case shows the reason. Each of the 4 people shakes 3 hands, which gives 4 × 3 = 12, and each handshake is counted by both people in it, so there are 12 ÷ 2 = 6. With n people the count is , and 30 people make 30 × 29 ÷ 2 = 435 handshakes.
The usual mistakes
Answering with the small case. The hexagon has 9 diagonals, which says nothing directly about 20 sides; what carries over is the method.
Trusting the list without the reason. The gaps 2, 3, 4 suggest the next count, but a list can stop following its pattern, and only the corner count says why every polygon follows the rule.
Forgetting to halve. n(n − 3) counts every diagonal from both of its ends, so for 20 sides it gives 340, twice the true 170.
Keeping a rule that a counted case contradicts. If a rule gave 8 for the hexagon, the 9 drawn diagonals would end it, however well it fitted other cases.
Cakes and garden paths
In the first application, 1, 2, 3 and 4 straight cuts across a cake are counted by hand, and the reason each new cut adds one more piece than the cut before gives a formula for any number of cuts. In the second, paths 1 to 4 half-meters long are paved by hand, and the way a pattern begins gives every longer count.
Worked example: A Round Cake Cut into Tasting Pieces with Straight Cuts, and the Cuts Needed for 50 Pieces
Question A baker cuts a large round cake into tasting pieces for a food fair. Each cut is straight and runs right across the cake. The pieces are sold by weight, so they need not be the same size, and she places the cuts so that no two are parallel, no three pass through one point, and every crossing is inside the cake. (a) How many pieces do n cuts made in this way give? Find a formula in n. (b) What is the smallest number of cuts that gives at least 50 pieces?
1.Solve the smallest cases. One cut makes 2 pieces. Two crossing cuts make 4. Three cuts, each crossing the other two at different points, make 7, and four cuts make 11.
One cut makes 2 pieces, two cuts make 4, and three cuts make 7. 2.The counts 2, 4, 7, 11 go up by 2, 3 and 4: each new cut seems to add one more piece than the cut before it.
Four cuts make 11. The counts go up by 2, 3 and 4. 3.See why. The nth cut crosses each of the n − 1 earlier cuts once, at n − 1 different points inside the cake, and those points split it into n segments. Each segment divides one piece into two, so the nth cut adds exactly n pieces.
The newest cut crosses the 3 earlier cuts at 3 points, so it is split into 4 segments and adds 4 pieces. 4.(a) The whole cake is 1 piece before any cut, so n cuts give 1 + (1 + 2 + ⋯ + n) = 1 + n(n + 1)2 pieces. Check: n = 4 gives 1 + 10 = 11.
(a) n cuts give 1 + n(n + 1)2 pieces. 5.(b) Nine cuts give 1 + 9 × 102 = 46 pieces, fewer than 50, and ten cuts give 1 + 10 × 112 = 56 pieces. The smallest number of cuts is 10.
(b) Nine cuts give 46 pieces and ten give 56, so the smallest number is 10 cuts.
Answer: (a) 1 + n(n + 1)2 pieces; (b) 10 cuts, since nine cuts give only 46 pieces and ten give 56
Common mistakes
- Assuming each cut doubles the pieces, since 1 cut gives 2 and 2 cuts give 4. Three cuts give 7, not 8: the third cut crosses only two earlier cuts, so it passes through only three pieces.
- Taking 2n pieces, as for cuts that all pass through the center. Cuts through one point break the rule that no three meet, and they add only 2 pieces each, so 10 such cuts give 20 pieces, not 56.
Worked example: A Garden Path Paved with Rectangular Slabs, and the Patterns That Look the Same from Both Ends
Question A garden path from the back door to the gate is 1 m wide and 5 m long. It is paved with identical slabs 1 m long and 0.5 m wide, with no gaps and no cutting. A slab can lie across the path or along it. The path has a door end and a gate end, so a pattern and its reverse count as two patterns when they differ. (a) In how many different patterns can the path be paved? (b) How many of those patterns look the same from both ends of the path?
1.Measure the path in half-meters: 5 m is 10 half-meters. Let an be the number of patterns for a path n half-meters long. A slab across the path fills 1 half-meter of its length. A slab along the path fills only half the width, so the other half beside it must also be a slab along the path: slabs along the path come in side-by-side pairs, and each pair fills 2 half-meters.
Measure in half-meters. A slab across fills one half-meter; slabs along the path come in pairs that fill two. 2.Solve the shortest paths. a1 = 1 (one slab across), a2 = 2 (two across, or one pair along), a3 = 3 and a4 = 5.
The shortest paths: 1, 2, 3 and 5 patterns for 1, 2, 3 and 4 half-meters. 3.Every pattern begins either with a slab across, leaving a path n − 1 long, or with a pair along, leaving a path n − 2 long. So an = an − 1 + an − 2, and the counts run 1, 2, 3, 5, 8, 13, 21, 34, 55, 89.
Every pattern begins with a slab across or a pair along, so an = an − 1 + an − 2. 4.(a) The path is 10 half-meters long, so it can be paved in a10 = 89 patterns.
(a) A path of 10 half-meters has a10 = 89 patterns. 5.A pattern looks the same from both ends when it is its own mirror image about the middle, 5 half-meters from each end. A slab across fills a single half-meter, so it cannot lie over the middle, which falls between the 5th and 6th half-meters. Either a join between slabs lies on the middle, and the door half can be any of a5 = 8 patterns, with the gate half its mirror image; or a pair along covers the middle, and the 4 half-meters before it can be any of a4 = 5 patterns.
At the middle there is either a join, with a5 = 8 choices for the door half, or a pair along, with a4 = 5 choices before it. 6.(b) 8 + 5 = 13 of the 89 patterns look the same from both ends. Check the method on a path 4 half-meters long: it predicts a2 + a1 = 3, and of its 5 patterns exactly three read the same both ways: four slabs across, a pair along between two slabs across, and two pairs along.
(b) 8 + 5 = 13 patterns look the same from both ends.
Answer: (a) 89 patterns; (b) 13 patterns
Common mistakes
- Letting a single slab lie along the path next to a slab across it. A slab along the path covers only half the width, and a slab across cannot fill the other half, so slabs along the path always come in side-by-side pairs.
- Halving the 89 patterns to count those that look the same from both ends. Halving counts pairs of mirror images, which is a different question, and 89 is odd; the symmetric patterns are counted by what lies over the middle.