Perimeter puzzles: composite curved boundaries reduce to arcs you already know

0:00A boundary made of curves looks like it needs something new. It does not. Every one of these shapes is a list of arcs, and you already know what an arc is worth: its share of a circle, times that circle's way round. So the job is writing the list down honestly - which arc, what radius, what share - then adding. The arithmetic is never the hard part. The reading is.

0:28And two puzzles show the two ways that reading pays off. Here is the first. Two circles of the same size, and each one passes through the other's centre. They cross at two points, above and below. The question asks for the length of the outer boundary - the part of each circle that is not inside the other one. The inner arcs are dotted, and they are not counted.

0:57But keep an eye on them - they are about to do the work. And notice what you have not been given. Nobody has told you a single angle. Draw the line from one centre to the other. Then the line from each centre to the crossing point above. Look at what those three lines are. The first joins two centres, and each circle passes through the other's centre, so it is a radius.

1:25The other two run from a centre to a point on its own circle, so they are radii too. Three lines, all the same length. That is an equilateral triangle, and every angle in it is sixty degrees. Nobody gave you an angle. Three equal lengths handed you one. A length condition delivering an angle is what this example exists to teach. The same triangle appears below the line, so the arc of each circle lying inside the other subtends a hundred and twenty degrees.

2:04A hundred and twenty out of three hundred and sixty is a third. So the dotted arc is a third of its circle, and the arc we actually want is the other two thirds. Two circles, two thirds each: the outer boundary is eight thirds of pi times the radius. And the sixty degrees came from the distance between the centres, not from the shapes being circles. Slide them apart until they are two radii apart and the circles merely touch: the angle collapses to nothing and the whole of both circles is outer boundary.

2:45Second puzzle, and this one you should guess before you calculate. Two points, and two routes between them. The first route is one semicircle arching over the whole distance. The second route is three semicircles standing on three pieces of the same line - one dipping below, one arching above, one below again - and the three pieces are deliberately unequal. Which route is longer? The single arch, the three bumps, or neither?

3:17Commit to an answer. Most people say the bumpy one. The eye is counting wiggle. It sees the second route going up and down, and reads the movement as extra distance. But a semicircle's length does not know about wiggle. It is fixed entirely by the width it stands on - the arc is pi times half the diameter, and nothing else enters. Flip one of those bumps to the other side of the line and the shape changes completely while the length does not change at all.

3:53So the question is not how much the route moves. It is how much line the arcs are standing on. Now the arithmetic, and it is one line. Take the whole distance as twelve centimetres. The single arch stands on twelve, so its arc is six pi. Split the twelve as five, four and three: the arcs are two and a half pi, two pi, and one and a half pi.

4:21Six pi. Split it as eight, three and one instead: four pi, one and a half pi, half a pi. Six pi again. Pi comes out of the sum, and what is left is the pieces adding back to twelve - which they must, being pieces of it. Every way of cutting twelve into up to six whole pieces was tried - a thousand and twenty-four of them. One answer, every time.

4:50And it is not that nothing can depend on a split: the same sweep on the area those semicircles enclose gives ten different answers across the fifty-five three-way cuts. The boundary is the thing that does not care. So here is a page of them. A rectangle with a semicircular cap on each end: two straight sides of eighty, two caps on a width of sixty, so a hundred and sixty plus sixty pi.

5:22An arch - a big semicircle with a smaller one cut out and two short feet closing it: ten pi plus four. A square of side ten with a semicircle bulging out of every side: the straight edges vanish and only the four arcs are left, twenty pi. A triangle of side twelve with a semicircle on each side: eighteen pi. Notice what is happening. Not one of these needed a new idea.

5:52Each one needed a list. And three of them are the same statement wearing different clothes. A right-angled triangle with sides six, eight and ten, carrying a semicircle on each: twelve pi. A base of twelve with one big semicircle over it and three little ones of four underneath: also twelve pi. That second one is the race again, with numbers in it. And a base of twenty with a big arc over it, one small arc above the left half and one below the right: ten pi on top, five pi and five pi from the halves, twenty pi.

6:36Equal again - and that is the flip, made concrete. One arc moved to the other side of the line, the shape completely different, the length identical. Now a warning, because these exercises tell you to use twenty-two sevenths for pi. Of those nine boundaries, only two come out a whole number of centimetres. And which two is not luck. It has a seven underneath, so a boundary lands whole exactly when its multiple of pi is a multiple of seven.

7:12Take the grid square, whose eight arcs come to fifty-six pi. Fifty-six is eight sevens, so it reads a hundred and seventy-six centimetres exactly. But that number is a fact about twenty-two sevenths, not about the figure: with pi itself the boundary is a little under a hundred and seventy-six, because twenty-two sevenths is a little over pi. Say which one you used. One more thing about that figure: all eight of its arcs are equal, four semicircles on a width of fourteen and four quarter circles of radius fourteen.

7:52That is no coincidence - a semicircle on a diameter and a quarter circle on that same number as radius are both pi times half of it, whatever pi you use. Last, two flowers, and here the reading is the whole question. A square with four petals inside it, and the only thing you are told is where the arcs are centred: at the midpoints of the sides. Take the midpoint of one side and draw the circle of radius half the side.

8:26It passes through both corners of that side and through the centre of the square - all three the same distance away. That is why the petals meet where they meet. And the piece of it that bounds a petal runs from a corner to the centre, which is a right angle at the midpoint. A quarter circle - not a semicircle, which is the mistake this question is built to catch.

8:52Four petals, two arcs each, eight quarter circles of radius seven: twenty-eight pi. The second flower is a hexagon, and now the arcs are centred at the corners. That works because a regular hexagon's distance from centre to corner is exactly its own side. So from any corner, both neighbouring corners and the centre are all one side away. The arc bounding a petal runs from a neighbouring corner to the centre, and the angle at the corner is sixty degrees.

9:28A sixth of a circle. Twelve of them, on a radius of forty-two: a hundred and sixty-eight pi. And the petal count is not an extra fact to remember. Seen from the centre, neighbouring arc-centres are ninety degrees apart in the square and sixty in the hexagon - four to fill the turn, and six. Each flower has as many petals as its frame has sides. None of it was a new formula - all of it was reading the picture and writing the list down.