Solving For The Radius - Math Olympiad Practice

This is good practice for math Olympiad type problems. I thank Mirza for suggesting the problem, and since I couldn't figure it out, I also thank Mirza for sending the solution. Can you figure it out?

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[NestorAbad]

The intersecting chords theorem is very elegant! I did it by means of coordinates: the origin (0, 0) is at the common point of the segments of lengths a and b, so according to the segment lengths there are three known points that lie on the circumference: (0, 0), (0, –a) and (b, c). Let (x, y) be the center of the circle. Then, the square of the distances of this center to that three points must be the same. This gives us the equations

x²+y² = x²+(y+a)² = (x–b)²+(y–c)²

The first one gives us y=–a/2. Replacing this value in the second one gives us x=(b²+ac+c²)/(2b). Finally,

r = √(x²+y²) = √(((b²+ac+c²)/(2b))²+a²/4)

which can be simplified as shown in the video.

[christopherwinqvist]

Drinking game: Drink every time he says "squared".

[benphan-nguyen]

MYD: *comes up with some long complicated answer involving square roots, fractions, quadratics, etc*
Also MYD: Did you figure it out?

[CraftedEasy]

How is the extended part of the already a + c chord assumed to be c ? He said he uses symmetry to know that its c but is he assuming it or is there something I am missing ?

[jeffreysung]

I got the same answer by using circle equation. Knowing three points on circle. (0, 0), (0. a). (b, a+c)

[krishnanshumisra]

we can also do it by taking two triangles just by joining centre with c down and c up, and strt of b and end of d( i did it this way). I got the same result but man the process becomes lenghty. That right angled property just slipped from my mind. Amazing question with amazing solution. Love your videos

[guilhermebasso]

I did it a different way. I used 2 expressions: R² = (a/2)² +(b-X)² and R² = (a/2+c)² +X; X is the distance from C to the horizontal center. It gave me a different expression, although just as big as yours.

[NT-yxos]

Of coarse I figured it out! Too easy. Answer is 3.436mm.

[boredmoroccan]

Thank you I always learn something new with every video.

[GaryTugan]

yep! I totally used algebra, geometry and a coordinate system with (0, 0) set at the intersection of lines a and b. Therefore, I now know two points on the edge of a circle (0, -a) and (b, c). Using this along with the equation of a circle (x-h)^2 + (y-k)^2 = r^2....I plugged both points into my equation, once for each point, and got two equations with 2 unknowns. Solving these I was able to find h and k, the coordinates of my circle's center. Using that info I was able to use a triangle (Using the coordinates) and connecting with radius r, we can use Pythagorean theorem to then solve for r. Got the same thing as you showed. I appreciate seeing how you then simplified it :)). Awesome problem. I was close to solving it geometrically as well, when I suddenly saw how to do it using the coordinate system method outlined above, so I gave that up in favor of.... algebra. :)

[flaviusk]

Solved it at first try, but used coordinates same as
Nestor Abad. I used different origin however, O(0, 0) at the bottom of chord a. Worked nicely with no difficulties. Then I watched the video and I was reminded of the chords theorem, last time used it 20+ years ago, nevertheless a very nice way to solve this. Thank you author for posting this problem, and thank you Mirza for sending it.

[ADARSHKUMAR-lxhn]

That was an elegant approach. We can also proceed by joining the two extremes on the circle, one on the line segment of length c and other on that of length a. Applying cosine rule in the obtuse triangle formed with c as one of its sides will give this newly joined line's length, and using it for sine rule in the opposite triangle will give us the radius on equating to 2R . Anyway, it was a really nice problem and involves some good geometry. Kudos Mr.Presh Talkwarkar, keep posting such Olympiad based problems time to time.

[zuzanazila]

You can use thales theorem. Point between a and c is the right angle of one right angle triangle above the diameter of the circle and the endpoint of c will form right angle of another right angle triangle above the diameter. Then you just setup pythagorean equations and use algebra to solve it.

[bananasrfr]

Can this be solved without the intersecting chords theorem?

[WarmWeatherGuy]

The equation of a circle is R^2 = (x-x0)^2 + (y-y0)^2. There are 3 pairs of (x, y) where the 3 points touch the circle. I used the bottom of the "a" line to be (0, 0) so the 3 (x, y) pairs are (0, 0), (0, a), and (b, a+c). Now you have 3 equations and 3 unknowns. I get that y0 = a/2 and x0 = (b^2 + c^2 + ac)/2b. My value of R^2 is probably what you got but I didn't do the algebra to see if it is the same. R^2 = [(b^2 + c^2 + ac)^2 + a^2b^2]/[4b^2]

[ibrt]

I solved it using coordinates with the center of the circle at (0, 0). Let the lower left point be (x, y). Then the next point working clockwise is at (x, y+a). The third point is at (x+b, y+a+c). Using the equation of the circle there are now three equations involving x, y and r: x^2 + y^2 = r^2, x^2 + (y+a)^2 = r^2, (x+b)^2 + (y+a+c)^2 = r^2. From here it is possible by subtracting one equation from another to solve for y, x and r.

[jakehayes]

great problem. learned much.
I figured it out by constructing a rectangle inside the circle by extending line b. 
I proved to myself that for any rectangle you could construct a circle where the intersection of the diagonals would be the center of the circle and the radius would be the line from the center to the corners. I took a step backwards to prove this.
After that used similar triangles and pythagorean theorem to come up w/ equation for the radius.
I wasn't able to see the beautiful simplification Presh did at the end where it's so much easier to read.
I had the same huge mess divided by so it was right... just not elegant.
It took a while and a couple runs at it but i'm glad I stuck with it. I learned a lot.

[molnez]

I went with a circumcircle approach. I made a coordinate system with Origo at the bottom of the line with length a. Then I drew a line between (0, a) and (b, a+c) making a triangle at the top. Then I found the half way point of that line and made a function that describes a line going through that point and that is also normal to that line. Where this function equals a/2 the center of the circle can be found. Then I find the length of the vector from (0, 0) to the circle center, and that will be the radius. Maybe a bit convoluted, but it works.

[arvindhsekar]

Every time you ask at the end, "Did you figure it out am like, LOL :D

[BokiB]

I got the same solution in much simplier way using the formula P=(xyz)/(4R) where x, y and z are the lenghts of the sides of the triangle and R is the radius of his circumcircle. 🙂