Half Angles, Area and the Radii

Transcript

0:00Introduction

Hello everyone! Take a triangle with sides thirteen, fourteen and fifteen, and no angle given. Can you find its area, the radius of the circle inside it, and the radius of the circle around it? Sounds like a lot of work, right? But the answers come out beautifully clean, and by the end of this video you will get them all in a few lines. This comes up often in JEE, so let's begin!

0:26Half angles from the sides

So, a quick recall. On the figure, small a, b, c are the sides opposite the angles capital A, B, C, and s is half the perimeter, a plus b plus c, over two. The area delta, the little triangle, is the square root of s, times s minus a, times s minus b, times s minus c. Okay? Now we want the half angle, A over two, from the sides alone. First, tell me: the angle A lies between zero and one hundred and eighty degrees, so is sine of A over two positive or negative? It is positive, yes! The half angle lies between zero and ninety degrees, so we take the positive root. Sine of A over two is the square root of s minus b, times s minus c, over b times c: the two brackets without small a, over the two sides at the angle A, right? Cos of A over two is the square root of s, times s minus a, over b times c. Now the bracket with small a goes on top, with s. Clear? Now you try. Tan is sine over cos, so what is tan of A over two? Divide them, and b times c cancels: under the root, s minus b times s minus c, over s times s minus a. Multiply inside, top and bottom, by s times s minus a. The top becomes delta squared, so this is also delta, over s times s minus a. Got it?

1:48Three kinds of radii, all from the area

Now the radii. Capital R, the circumradius, is the radius of the grey circle through the three corners, with centre O, okay? It is small a, over two times sine A, which is also a times b times c, over four times delta. Right? Small r, the inradius, is the radius of the green circle inside, with centre I, touching all three sides. Small r is delta over s, and by our tan formula, also s minus a, times tan of A over two. Clear? Now a new one. r with a small subscript one, read r one, is the exradius opposite A, the radius of the excircle: a circle outside the triangle, which you can picture below B C, touching side small a and the other two sides extended. r one is delta over s minus a, or s times tan of A over two. With me? So compare. Small r is delta over s, and r one is delta over s minus a. Which is bigger? It is r one, every time, because s minus a is smaller than s, and dividing by a smaller number gives more. Likewise, r two is delta over s minus b, and r three is delta over s minus c. Got it?

2:58How the radii are linked

Now, these radii are linked. Let me ask you this: what is one over r one, plus one over r two, plus one over r three? Flip each formula and add. Flip each one: one over r one is s minus a, over delta, and so on. So the top is s minus a, plus s minus b, plus s minus c, and since a plus b plus c is two s, that is three s minus two s, just s. And s over delta is one over small r. Beautiful, isn't it? Two more from the course: r one plus r two plus r three, minus small r, equals four times capital R, and r one r two, plus r two r three, plus r three r one, equals s squared; the small two on top means squared. Right? And O I squared equals capital R squared, minus two times capital R times small r, where O I is the distance between the centres O and I, okay? Also, tan of A over two, times tan of B over two, is s minus c, over s, and so on. Clear?

4:01Example 1: area, R, r and exradii of the 13, 14, 15 triangle

Now, remember our challenge? Let's solve it. Small a is thirteen, b is fourteen and c is fifteen; find delta, capital R, small r and the three exradii. Pause the video and try it first. Done? Okay, let's check. The perimeter is forty two, so s is twenty one. Then s minus a is eight, s minus b is seven, and s minus c is six. Okay? Delta is the square root of twenty one times eight times seven times six; the dots mean times. Don't multiply blindly. Twenty one is three times seven, eight times six is three times four squared, and there is one more seven. So under the root we have three squared, times seven squared, times four squared, and the root is three times seven times four, which is eighty four. Neat, right? Capital R is a b c over four delta: thirteen times fourteen times fifteen, over four times eighty four. Fourteen times fifteen is two hundred and ten, and four times eighty four is three hundred and thirty six. Two hundred and ten over three hundred and thirty six simplifies to five over eight, and thirteen times five over eight gives sixty five over eight. So capital R is sixty five over eight. Clear? Small r is delta over s, eighty four over twenty one, which is exactly four. Now, before you divide again, guess: which exradius will be the largest, r one, r two or r three? It is r three, because s minus c, which is six, is the smallest bracket. r one is eighty four over eight, which is twenty one over two; r two is eighty four over seven, which is twelve; and r three is eighty four over six, which is fourteen. And that solves our challenge!

5:44Example 2: find a:b:c when r1 = 2r2 = 3r3

Now a harder one. If r one equals two times r two, which equals three times r three, find the ratio a to b to c, written with colons. Pause the video and try it first. Done? Okay, let's check. Many students see one, two, three and write r one to r two to r three as one to two to three. Is that right? No, no, careful, it is the other way round! r one is the biggest, since r two is only half of r one. It is the reciprocals, one over r one, one over r two and one over r three, that go one to two to three. Okay? Now flip r one equals delta over s minus a: s minus a equals delta over r one. Delta is the same for all three, so s minus a, s minus b and s minus c also go one to two to three. Right? So let s minus a be k, s minus b be two k, and s minus c be three k, for some positive k. Add all three: three s minus two s is s, and the right side is six k. So, as the arrow says, s equals six k. With me? Then small a is six k minus k, five k; b is four k; and c is three k. So a to b to c is five to four to three. Well done!

6:56Quick revision: tan A/2, R, r, r1 and how the radii link

So, quick revision. Half-angle ratios are positive, and tan of A over two is delta, over s times s minus a. Capital R is a b c over four delta, small r is delta over s, and r one is delta over s minus a. The reciprocals of the exradii add up to one over r, and four times capital R is r one plus r two plus r three, minus small r. Now try the practice problems on this page. All the best!