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Hi everyone! This is my first time doing thing, and I am a little frustrated trying to find some information about inradii of a right triangle. :confused: So my question is this... how do you prove that the radius of the inscribed circles is less than half of either leg and less than 1/4 of the hypotenuse. I have the equation for the inradius, but so far, that is not doing me any good. Anyone have any hints? Thanks!
Hello. Welcome to the forums :wave:
If you look at fig. 1 you'll see that the shortest hypotenuse occurs when it forms a 45 deg. angle (actually 135 deg.) with the horizontal axis. Also, the shortest legs occur simultaneously at that angle.
http://www.vbforums.com/attachment.p...5&d=1197020696
So in fig. 2 you have a drawing of this case. You don't need numbers or formulas to prove that r < L/2 where L is the length of either leg. If you move from left to right in fig. 1 you see that as you 'roll down' the hypotenuse, the vertical leg approaches 2r but in the limit when both the horizontal leg and the hypotenuse become infinite, the vertical leg just becomes equal to 2r.
http://www.vbforums.com/attachment.p...6&d=1197020708
As for the second proposition, the triangles OCB and OCA are equal as the line OC forms an angle of 45 deg. with the horizontal axis. Then,
OC = BC = CA = AB / 2
But from the drawing, OC > 2r so that AB / 2 > 2r and finally,
AB > 4r
Of course it could be strictly proved with equations but I don't think they should be used where they aren't necessary.
Thanks for your reply! I hope that works. I will try to construct the proof of that now. It actually seems really easy! :thumb:
If anybody is interested, here is a diagram showing how to work out the radius of a circle inscribed in a right-angled triangle.
In Fig 1, we have the two legs as X and Y, the hypotenuse as Z and the radius as r. We also draw lines from the centre of the circle to the points at which the circle touches each side, and the two ends of Z.
In Fig 2) we just double up what we've got, to form a rectangle of area XY.
In Fig 3), we break it down into the smaller pieces, ignoring the circle.
In Fig 4) we rearrange the pieces to form a rectangle rX, a rectangle rY and a rectangle rZ.
From Figs 2) and 4), we know that therefore XY = rX + rY + rZ, and so:
r = XY / (X + Y + Z)
I can't claim credit for this; it was first written in the Chiu-chang suan-shu, or Nine Chapters on the Mathematical Art, thought to have been written in China sometime between 3BC and 1BC.
zaza
I really appreciate all the replys... one question though, how can I find those ratios using the inradius formula? Is there anyway to do it using sin and cos?? Just a thought! Thanks!
Ok, so I am trying to do this using formulas and am getting stuck. Anyone got any ideas??? Thanks!
-Katie
If you look carefuly at fig.2 in my earlier post you'll realize the coordinates of point C where the tangency occurs between the circle and the hypotenuse are:Quote:
Originally Posted by ktdid_5
xc = r + r*cos(Pi/4) = r + r/Sqr(2) = r[1 + 1/Sqr(2)]
yc = r + r*sin(Pi/4) = r + r/Sqr(2) = r[1 + 1/Sqr(2)]
whereby the distance OC is:
OC2 = xc2 + yc2 = {r[1 + 1/Sqr(2)]}2 + {r[1 + 1/Sqr(2)]}2 = 2*{r[1 + 1/Sqr(2)]}2 = r2[3 + 2*Sqr(2)]
Since the line OC forms an angle of Pi/4 with the x axis, then OC = AC so that
OA = OB = Sqr(OC2 + AC2) = Sqr(2*OC2) = Sqr{2r2[3 + 2*Sqr(2)]} = r*Sqr[6 + 4*Sqr(2)] = r*3.414...
so r is actually less than one third each leg.
I assume you can now work the second question on your own.