Using distances alone to classify a triangle or a quadrilateral

0:00You can now measure the gap between any two points from their coordinates alone. So here is the obvious next question. If shape is made of lengths, and lengths are now arithmetic, is shape now arithmetic too? Partly. And the interesting part of this topic is exactly where the word partly bites. Three lengths will tell you whether three points make a triangle, and they will do it with total certainty.

0:29Four lengths will not tell you whether four points make a square, and no amount of care with the arithmetic will fix that. The lengths are not wrong. There are simply not enough of them, and this whole video is about learning to notice when. Start with three points, and the one thing three separations always settle. Take the two shorter gaps and add them. If the total beats the longest gap, the three points make a triangle.

1:01If the total exactly equals the longest gap, they do not. They lie on one straight line, and the shortest way from the first to the last runs through the middle one. That is not a very flat triangle. It is not a triangle at all. The equality is the collapse. I ran that test over a field of forty-eight points, which gives seventeen thousand two hundred and ninety-six triples.

1:30Six hundred and twenty-eight of them are straight lines and sixteen thousand six hundred and sixty-eight are triangles. And I checked every one of those verdicts a second time by a completely different route, one that never computes a distance at all: the area enclosed. Three points on a line enclose nothing. The two routes agreed on all seventeen thousand two hundred and ninety-six. Here are three of them worked out. P at three across, two up. Q at two left, three down. R at two across, three up.

2:09P to Q: the differences are five and five, so the gap squared is fifty. Q to R: the differences are four and six, so the gap squared is fifty-two. P to R: the differences are one and one, so the gap squared is two. Notice what I have not done. I have not taken a single square root. The lengths are the roots of fifty, fifty-two and two, which is about seven point zero seven, seven point two one and one point four one.

2:42Seven point zero seven and one point four one add to eight point four eight, which comfortably beats seven point two one. A triangle. Now look again at those three squares, and do not take roots for this either. Fifty plus two is fifty-two. Exactly. Two of the squares add to the third, and that is the converse of Pythagoras: a right angle. The one thing worth being careful about is where the right angle sits.

3:15It is not at the corner where the biggest number turned up. It is at the corner the longest side does not touch. The longest side here is Q to R, and the corner it misses is P. So the right angle is at P. Over the whole field, two thousand one hundred and sixty-four triples carry a right angle, and on every single one of them the angle sat at the vertex the longest side avoided.

3:45Putting it at a corner the longest side touches got the right answer zero times out of two thousand one hundred and sixty-four. Now the other verdict, with three students sitting at desks on a grid. The first is at three across, one up. The second at six across, four up. The third at eight across, six up. First to second: differences of three and three, so eighteen. That is three roots of two.

4:16Second to third: differences of two and two, so eight. That is two roots of two. First to third: differences of five and five, so fifty. That is five roots of two. Three roots of two plus two roots of two is five roots of two. Not approximately. Exactly. The three of them are on one line, and the area their three desks enclose is nothing at all. The middle student is sitting directly on the shortest path between the other two.

4:50Now the trap, and it is a good one, because the picture will lie to you and so will a calculator. Take one point at one across five up, another at two across three up, and a third at two left and eleven down. The three gaps squared are five, two hundred and twelve and two hundred and sixty-five. As decimals that is two point two four, fourteen point five six and sixteen point two eight.

5:20Two point two four plus fourteen point five six is sixteen point eight, which overshoots sixteen point two eight. Not collinear. But now watch what two decimal places can hide. Take a point at the origin, one a thousand across and one up, and one two thousand across. Both short gaps come out as one thousand point zero zero and the long one as two thousand point zero zero. The decimals say straight line.

5:53The area says otherwise. That triangle encloses a thousand square units. It is not close to being a line; it just prints like one. I built six such figures and four of the six fooled a two-decimal reading. The test is the exact comparison, and the decimals are only ever a sanity check. So three lengths close the triangle case completely. Now four points, and watch it fail. Measure the four sides of a quadrilateral, and suppose they all come out the same.

6:32It is very tempting to write down square, and on most of the examples you will ever meet, square will be the right answer. It is still not what you proved. Four equal sides is a rhombus. Push a square over and every side keeps its length while the corners stop being right angles. To measure how blind that rule is, I built forty squares and a hundred and twenty rhombi that are not squares, and asked the four-sides rule about all one hundred and sixty.

7:05It said square to all forty of the squares. It also said square to all one hundred and twenty of the rhombi. It did not refuse a single figure. Not one. That is worse than a rule that is sometimes wrong: it is a rule with no power to separate anything. So take a second, independent measurement. Here is the standard way, on four points. One at one across seven up, one at four across two up, one at one left one down, and one at four left four up.

7:42The four sides squared: thirty-four, thirty-four, thirty-four, thirty-four. All four sides are the root of thirty-four. Now the two diagonals, which is the measurement the sides could not give you. Both come out sixty-eight. Equal sides and equal diagonals, and that pair of facts does force a square. Add the diagonals to the same test and it now says square to all forty squares and refuses all one hundred and twenty rhombi.

8:15Two more numbers, and a rule that separated nothing became a rule that separates everything. There is a shorter route to the same verdict, and comparing the two is where this topic gets interesting. Take the four sides, then just one diagonal, and finish with the converse of Pythagoras. Two adjacent sides squared, thirty-four and thirty-four, add to sixty-eight, which is the diagonal squared. So that corner is a right angle.

8:48Four equal sides plus one right angle is a square. Six measurements have become five. And now look at what made that work. Sixty-eight is exactly twice thirty-four. That is not a coincidence in this example. In every one of those hundred and sixty four-equal-sided figures, the two diagonals squared add to four times a side squared. So equal diagonals, a right angle at a corner, and a diagonal squared being twice a side squared are three ways of saying one thing.

9:26All three agreed on all one hundred and sixty figures, and all three held on exactly the forty squares. The two routes are one fact read twice. It is worth checking the other half of that pair too, because equal diagonals on their own are even weaker than equal sides. An isosceles trapezium has equal diagonals. It is nothing like a rectangle. I built twenty-five rectangles and thirty-six such trapezia and asked the diagonals-only rule about every one of them.

10:05It said square to every single one of them and refused none. Same failure, from the other direction. So neither measurement is doing the work by itself. It is the pair that closes the case. That is the actual lesson here, and it is not really about squares. A classification needs enough independent measurements to force the definition shut. Which measurements you take is your choice. How many you need is not.

10:37Here is that put to work. Four friends sit on a grid of desks, and two of them disagree about whether their seats make a square. The seats are at three across four up, six across seven up, nine across four up, and six across one up. All four sides squared come out eighteen. Four equal sides. So far the disagreement is not settled. The diagonals: thirty-six and thirty-six. Both equal, and both exactly six, because one diagonal runs flat and the other runs straight up.

11:14Equal sides and equal diagonals. It is a square, and the friend who said so is right. Notice that the whole disagreement was settled by two numbers neither of them had. Not by looking harder at the picture. The same six measurements will also tell you when a figure is something else, or nothing at all. Take four points whose sides all come out eight when squared, and whose two diagonals both come out sixteen. Equal sides, equal diagonals: another square.

11:51Here is a stranger one. Four points where the second is reached from the first by a step of three across and two down, and the third by exactly the same step again. Three of the four are on one line. There is no quadrilateral to name. The question had no answer to give. And here is a third set: opposite sides squared matching in pairs, ten and ten, eighteen and eighteen. So it is a parallelogram.

12:22But the diagonals squared are four and fifty-two. Wildly unequal. So it is a parallelogram and nothing more. Four equal sides would have made it a rhombus. Equal diagonals as well would have made it a square. It has neither. One more idea, and it turns the whole thing around. Instead of asking what a set of points is, ask where a point has to be. Which points are equally far from seven across one up, and from three across five up?

12:57Write the two separations squared and set them equal. The x squared terms cancel, the y squared terms cancel, and what is left is x minus y equals two. That is a line, not a point. Being equally far from two places does not pin you down; it puts you on a road. And it is a very particular road. It passes through the midpoint of the two, and it crosses the segment between them at a right angle. The perpendicular bisector.

13:31I tested that over a grid of two hundred and eighty-nine points and four different pairs. Every point that was equally far from the pair was on the line, and every point on the line was equally far. Both mismatch counts were zero. A line through the same midpoint but running along the pair instead of across it fails immediately: fifteen points on it that are not equally far, and twenty-three equally far points it misses.

14:05So if one condition gives you a whole line, getting back to a single point takes a second condition. Find the point on the vertical axis equally far from six across five up and from four left three up. The words on the vertical axis have already done half the work: such a point looks like nought comma something, so one unknown is gone before any algebra starts. Expand, cancel, and you get y equals nine. The point is nought comma nine, and its distance from each of them, squared, is fifty-two.

14:43It is exactly where the vertical axis crosses the perpendicular bisector, which is the same answer arriving from the other side. The same move on the horizontal axis, for two other points, gives seven to the left of the origin. That is the shape of this whole topic. A length is a measurement. A verdict is a measurement plus enough other measurements to close the definition. Counting how many you need is the skill, and the rhombus is the figure that teaches it.