Centres of a Triangle

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. Draw any triangle and mark three points in it: where its medians meet, where its altitudes meet, and the centre of the circle through its corners. Three different constructions, so you would expect three unrelated points. But here is the surprise: they always lie on one straight line. Today we find the four centres of a triangle, which come up often in JEE, and by the end you will check this surprise on a real triangle. So, let's begin.

0:33Centroid: where the medians meet

So, here is our triangle. Vertex A is x one, y one, in round brackets; the small one below the x is a subscript, labelling the first vertex, okay? Then B is x two, y two, and C is x three, y three. Each median joins a vertex to the mid-point of the opposite side, and the dashed medians in the figure meet at G, the centroid. Now make a guess: what is the simplest formula for G that treats all three vertices equally? Yes, exactly, just the average of the vertices! G equals x one plus x two plus x three, over three, and then, y one plus y two plus y three, over three. Clear?

1:14Incentre: where the angle bisectors meet

Next, the incentre, where the angle bisectors meet. We need the sides: small a is B C, opposite vertex A; small b is C A, opposite B; and small c is A B, opposite C, okay? The incentre is a weighted average of the vertices, with the side lengths as weights. Now here is where many students slip. Side B C touches vertices B and C, so you might think its length, small a, multiplies one of those two. Does it? No, no, be careful, because small a goes with vertex A, the vertex facing side B C; each vertex gets the side it does not touch. So capital I, the incentre, equals small a times capital A, plus small b times capital B, plus small c times capital C, all over small a plus small b plus small c. Coordinatewise means we do this once for x, and once for y. Got it? So the x coordinate of I is small a times x one, plus small b times x two, plus small c times x three, over the sum of the sides, and the same for y. Clear?

2:17Circumcentre and orthocentre

Now the circumcentre, written capital O, where the perpendicular bisectors meet. The perpendicular bisector of a side passes through its mid-point at right angles to it, and the three of them meet at the point O marked in the figure. Right? And what is special about O? It is equidistant from the vertices: O A, O B and O C are all equal. So the circle with centre O passes through all three corners. Fourth, the orthocentre, written capital H, where the three altitudes meet. An altitude is the line from a vertex perpendicular to the opposite side, like the dotted line from C in the figure. With me? Now a quick question for you: in a right triangle, where do the three altitudes meet? At the right-angled vertex, yes! The two legs, the sides forming the right angle, are perpendicular, so each leg is an altitude, and they meet at the right-angled vertex, where H sits. And the circumcentre of a right triangle is the mid-point of the hypotenuse, the side opposite the right angle. Remember both facts; they save time. Okay?

3:24The Euler line

Now, back to our puzzle. In the figure, H, G and O sit on one straight line, the red line labelled Euler line. Collinear means lying on one line, got it? And there is more: H G is to G O as two is to one, and the colon on screen is read, is to. So, is G nearer to H, or nearer to O? Nearer to O, correct! G to O is one part, and H to G is two parts; students often flip this. So by the section formula for internal division, G equals H plus two times O, all over three; the double arrow means, gives. Rearranging, H equals three times G minus two times O, right? So any two of O, G and H give you the third. And there is one special case: in an equilateral triangle, all four centres coincide. So there, find the centroid, and you have found all four. With me?

4:18Example 1: incentre and inradius of a 3-4-5 right triangle

Let us try example one. Find the incentre and the inradius of the triangle zero comma zero, four comma zero, and zero comma three. The inradius is the radius of the circle inside the triangle that touches all three sides. Pause the video and try it first. Done? Okay, let's check. Call the vertices A, B and C, in that order. By the distance formula, B C is the square root of sixteen plus nine, which is five; C A is three and A B is four. So small a, b and c are five, three and four. Now the weights: which vertex gets the weight five? Vertex A, correct, as B C is the side opposite A, right? So I equals five times zero comma zero, plus three times four comma zero, plus four times zero comma three, all over five plus three plus four. That is twelve comma twelve over twelve, so the incentre is one comma one. Now the inradius, small r, with the right angle at A, the origin. For a right triangle, r equals leg plus leg, minus the hypotenuse, all over two. So three plus four minus five, over two, gives one. And it fits: one comma one is one unit from both axes, which are two sides of this triangle.

5:32Example 2: find O, G and H, then check H = 3G - 2O

Now a harder one, on the very triangle in our figure, with vertices A at zero comma zero, B at six comma zero, and C at two comma four. Find O, G and H, and check that H equals three times G minus two times O, okay? Pause the video and try it first. Done? Okay, let's check. Both A and B lie on the x axis, with mid-point three comma zero, so the perpendicular bisector of A B is x equals three, and O is three comma k. Clear? Now tell me, which condition will give us the value of k? Yes, O is equally far from the vertices, so O A equals O C, okay? Squaring, nine plus k squared equals one plus k minus four, the whole squared; the small two up top means squared. The k squared cancels, so nine equals seventeen minus eight times k, and k is one. So O is three comma one, and G, the average of the vertices, is eight over three, comma, four over three. Got it? From the Euler line, H equals three times G minus two times O, right? Three times G is eight comma four, and two times O is six comma two, so H is two comma two. With me? Let us check with two altitudes. Side A B lies on the x axis, so the altitude from C is vertical: x equals two, the dotted line. Careful, a vertical line has no slope, so we read it off directly, okay? Next, the altitude from A is perpendicular to B C, whose slope is four over minus four, which is minus one. When both slopes exist, perpendicular slopes multiply to minus one, so this altitude has slope one: y equals x. It meets x equals two at two comma two, just as the Euler line said, and our puzzle is settled!

7:25Quick revision: four centres, right triangle, Euler line

So, quick revision. The centroid averages the vertices; the incentre weights each vertex by its opposite side. In a right triangle, the orthocentre is the right-angled vertex, and the circumcentre is the mid-point of the hypotenuse. On the Euler line, H G to G O is two to one, so H equals three times G minus two times O, okay? Now try the practice problems on this page. All the best!