How To Solve For The Radius. Challenging 1970s Math Contest!

There doesn't seem to be enough information to solve this, but the answer works out magically. Thanks to all patrons! Special thanks this month to: Richard Ohnemus, Michael Anvari, Shrihari Puranik, Kyle.

I also credit patrons Pradeep Sekar and Nestor Abad for finding a typo in my original video--thanks!

Maths By Amiya : Perpendicular Chords and Radius

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"The Joy of Game Theory" shows how you can use math to out-think your competition. (rated 4.0/5 stars on 41 reviews)

"The Irrationality Illusion: How To Make Smart Decisions And Overcome Bias" is a handbook that explains the many ways we are biased about decision-making and offers techniques to make smart decisions. (rated 3.5/5 stars on 4 reviews)

"Math Puzzles Volume 1" features classic brain teasers and riddles with complete solutions for problems in counting, geometry, probability, and game theory. Volume 1 is rated 4.4/5 stars on 13 reviews.

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"40 Paradoxes in Logic, Probability, and Game Theory" contains thought-provoking and counter-intuitive results. (rated 4.4/5 stars on 13 reviews)

"The Best Mental Math Tricks" teaches how you can look like a math genius by solving problems in your head (rated 4.8/5 stars on 5 reviews)

"Multiply Numbers By Drawing Lines" This book is a reference guide for my video that has over 1 million views on a geometric method to multiply numbers. (rated 4.3/5 stars on 6 reviews)

Connect with me

[tehArcher]

the secret trick to maths problems, pull some obscure formula from out of nowhere

[AmosdollMusic]

My method to solve it was: 1.) Worked out the line going up to C is a length of 4 (similar triangles in circle geometry)
2.) Taking the bisecting 2 lines as (0, 0), I then used coordinate geometry to find circle center as (2, 1/2)
3.) Then perpendicular bisected the y axis to form a triangle and used Pythagorean theorem to find the radius = root(3.5^2 + 2^2) = root(65) / 2
Maths is awesome! So many ways to solve!

[educatedonealways]

You pick the longest and most obscure way to solve problems

[FrancisACNH]

Nobody:

Problems in the 1970s: *handed out to unborn foetuses in Asia*

[bentalks]

I now know the radius of Super Smash Bros.

Thank you

[jbglaw]

Looking at 4r² = w² + x² + y² + z² I realize that if you draw a cross somewhere into a circle, making each of the cross's quadrants a square, that's the same area as a square drawn around the circle. Geometrical beauty!

[math]

Love the way you challenge the brain with these mind boggling problems. You are doing great work and I look forward to being challenged.

[wallop]

I immediately thought 4. But it’s never that easy

[jvthunder]

You can also use the circumradius theorem of the triangle ABD. The formula is simply R = abc/4L where a, b, c are the side lengths and L is the area. The area is 8*3/2=12. a, b and c is the length of BD, AB and AD in any order. By phytagoras, you get AD = √13 and BD = 3*√5. Thus, plugging in to the formula, we get R = 8*√13*3*√5/(4*12)= √65/2

Don't get it wrong though, your solution is also amazing, I'm just showing my approach when I first saw this problem.

By the way, you have a great channel, thank you for your amazing content and presentation.

Edit: I forgot to say what L stands for.

[shashikalayadav]

Very easy .I am 10 standard student and got right answer using simple chord theorems.

[sofiaatzeni]

I actually solved this in a third way! I used the Pythagorean theorem to solve for BD and AD. Then, I used the cosine theorem with the triangle ABD to find the value of the cosine of the angle ADB, and consequently its sine. At this point, all I needed to do was remember that given a circle and a chord, the chord in equal to 2r sin(Alpha). We obtain AB = 8 = 2r sin(ADB), r = AB/2sin(ADB)

[polarwin]

(x-x0)^2+(y-y0)^2=r^2 : three variables with three conditions (-2, 0), (6, 0), (0, -3)

[monkeseeaction]

I enjoyed the video quite much because when I was in college and doing research on writing unimportant bits of a computational software, we needed an algorithm that takes in the coordinates of three points on a plane and spit out the coordinate of the center of the circle that passes through all the points, and the radius. I remember pulling a neat linear algebra trick where I used the property that the vector that describes the direction of a chord must be orthogonal to the vector of its perpendicular bysector, which is expressed in terms of the unknown circle center coordinate. The two orthogonality conditions directly translate into a linear system and you never even touch r to get the center coordinate, just linear algebra. I pulled out the method, dusted it off and used it to solve this problem. 10 minutes well spent. Thanks for posting these problems; they make quite neat brain exercises.

[abhradeepdey]

This is pretty simple if you know the properties of a circle.
I solved this problem in two steps:
1) Joined the points A and D and used the property "angles of same segment in a circle are equal" to got the relation between the angles ACO and DBO (named O as the intersection point of the chords) and OAC and ODB as ACO=DBO and OAC=ODB. Then applied the concept of similarity to derive the relation: OC/OB = AO/OD after proving triangles ACO and DBO similar by 'AA' similarity criterion. Hence by substituting the known parameters in the above relation got the value of OC=4.

2) applied the property of the the perpendicular bisection of chords by the radius of the circle to get MC=MD=3.5 (named M as the bisection point of chord DC) and AN=NB=4 (named N as the bisection point of chord AB). Then subtracted MC from OC to get OM=0.5 and finally joined XB (named X as the centre of the circle) and applied Pythagoras Theorem : XN^2 + NB^2 = XB^2 (Radius^2). Now from the diagram XN=ON. Therefore substituted it's value in the equation and got the answer as sqrt.(16.25)

[zanissisojevs]

I used formulas for triangle areas and Pythagoras theorem, I combined S=ah/2 with S=abc/4R and found R.
I used formula DO*OC=AO*OB to find CO (O is intersection of AB and CD), drew triangle CBD, used Pythagoras theorem to find sides, calculated area of triangle and calculated R.

[Airhornsman]

You can also do it by:
1) calculate the lengths AD and DB using Pythagoras
2) calculate the angle ADB using trigonometry + the three lengths
3) Lets call the centre point O, the circumflex angle at the centre subtended by the arc AB is twice ADB. So the interior angle AOB is 360 - the angle we just calculated.
4) Because the triangle ADB is isoceles, the angle OAB is (180 - AOB) / 2
5) Use trigonometry to calculate length AO, which is the radius.

[phungpham]

Thank you for your nice explanation.
I have another approach by using basic geometry.
1/Finding the point O, the center of the circle.
Label H as the meeting point of AB and CD, and M and N as the midpoints of AB and CD respectively. Then draw the two bisectors lines from two midpoints which meet at point O. O is the center of the circle. Now we have a rectangle HNOM.
2/Calculating the hypotenuse OH and the radius of the circle:
Using chord theorem: CHxHD=AHxHB---> CH=2x6/3=4---> HN=(7/2)-3=1/2 and HM=AM-2=2
Using Pythegorean theorem: sq OH=sqHN+sq HM= sq (1/2) +sq 2= (1/4)+4=17/4----> OH= (sqrt of17)/2.
EXtend OH to the other sides we have the diameter of the circle (radius=R)
Using chord theorem: (R-OH)x(R+OH)= sqR - sqOH= sq R - (17/4 )=12 ---> sq R= 12+(17/4)= (48+17)/4=65/4.
Thus the answer is R=(sqrt of 65)/2

[mike.]

I had never seen that theorem before, but my mind immediately went to perpendicular bisectors of chords intersecting in the center. I suppose I would have rediscovered it that way!

[DgtalMess]

Challenger Approaching!

Presh Talwalker divides the competition!

[lesadams]

Thank you for this. I also used the co-ordinate method! I was surprised and pleased to learn of the perpendicular chords theorem - that method gives the fastest solution.