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

[@MindYourDecisions]

I just noticed the video length is 6:28 on mobile--what an accidentally perfect time. 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! (You can get early access to videos by supporting on Patreon--such support makes a huge difference.) www.patreon.com/mindyourdecisions

[@yassineloulou4533]

Lmao, good job though

[@blue_link_3461]

I think there's a typo at 4:56, it's supposed to be (-2+p)²+q²=r² not (-2-p)²+q²=r². That or I didn't understand what you did.

[@chunfengmugu]

5:40 how can you get the value of q with r's value unknown?

[@v6790]

@@chunfengmugu he didn't show but he said something about eliminating the r^2 term by subtracting

[@quantumcity6679]

I have best method for finding the radius... it's by using ptolemy's theorem.. 😅

[@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!

[@lecinquiemeroimage]

Second question: How to solve the value of the vertical segment, which is not shown in the figure? Greetings from Paris. professeur essef, in mathematics (active for over a year, on YT & Wiki, mainly in astronomy and astrophysics).

[@Jerryfan271]

@@lecinquiemeroimage angles ADC and CBA are equal.

[@lecinquiemeroimage]

@@Jerryfan271 Not at all! Neither for the angles ADC and ACB (at least, you make two confusions) ... Answer (to justify): let us write I the intersection of the two lines (AB) and (CD); then IC = 4 (different from ID = 3) P.S: angles ADC and ACB would be equal [to 90°] if [A;B] were a diameter, and then IC = ID.

[@Jerryfan271]

@@lecinquiemeroimage um, you conclude IC=4 but that in itself implies SAS similarity of ADI and CBI meaning ADC and ABC must be the same angle. I used same-arc subtending to get that result. Since arc AC subtends both ADC and ABC, and D and B are on the major arc, the angles are equal. So IC=4 iff ADC=ABC. if I am somehow wrong and ADC=ABC is false then why not give a counterexample?

[@prithujsarkar2010]

wait whaaaat u doin maths now a days ? niceeee

[@tehArcher]

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

[@RodelIturalde]

The formula isn't obscure nor pulled out of nowhere. It isn't that hard to find either. But you have to work a bit to find it and prove it. If you can't prove the formula you use, you shouldn't use the formula at all, since you don't know what you are doing.

[@Marcel-vz7vp]

RodelIturalde shut up nerd

[@RodelIturalde]

@@Marcel-vz7vp lol. Sorry for knowing stuff and actually trying to learn something. I should probably go ignorant like you and call everyone who knows more then basic addition and multiplication nerds.

[@r3mxd]

RodelIturalde *than. Damn can’t even use then and than properly.

[@spartankongcountry6799]

@@RodelIturalde Well said.

[@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!

[@ultimatedeatrix9149]

that's actually 4 times the area of the squares drawn from the quadrants but nice way to see it :)

[@gvc76]

@@ultimatedeatrix9149 jbglaw stated it correctly.

[@brettstafford9665]

I think this is wrong??

[@barttemolder3405]

It is correct. a² is the size of the surface of a square with sides a, b² same for b, and r² for a square with sides r. A square around the circle would have sides of the length of the diameter of the circle, that's 2r. Its surface is (2r)² = 4r².

[@mrreg]

thankyou so much for that,that is much easier to visualize!

[@sofiaatzeni570]

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)

[@lettsstartcooking]

Hmm good idea. Clean and simple

[@plamenpenchev262]

Just solved in the same way. Never knew "the power of a point", etc. Using phone calculator and twice proving -- 30 min. If using sin() = sqrt(1-cos()^2) instead of sin(acos()) the same formula is obtained as that of the presenter.

[@lucastellez2558]

This was my approach too

[@-Chicken_]

I only understood half of this. Please someone explain 🥺

[@ninoinoz4437]

Or you could find the sine of angle ADB by O/H i.e. 3/BD as you have just calculated BD.

[@jvthunder6548]

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.

[@zoesan4517]

This is also what I did. I was very happy to have the same result

[@devanshusharma9768]

It is also a secret and hidden formula.by the way can u pls derive it for me❤❤❤❤

[@primus4cameron]

@@devanshusharma9768 If you draw the line (new chord) AD and bisect it, the extended bisection will go through the circle's centre. Likewise the bisection of AB will go through the centre, parallel to CD but 2 units distant. You end up with lots of CONGRUENT triangles and with PYTHAGORAS it's easy to solve without any formulas, just the most rudimentary algebra.

[@devanshusharma9768]

@Ayush Ojhau did very intresting approach to the problem

[@Аня-л3э8г]

I did the same

[@polarwin]

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

[@ilPescetto]

Yeah, too easy

[@thesteaksaignant]

@@ilPescetto Yeah I used that method too. I was just worried about mistakes during solving but it turned out alright for me ^^"

[@なぎさ-t9o]

I solved it by using this method too.

[@Phendoxia]

@@ilPescetto o

[@outandabout259]

Just learned that couple weeks ago in school.

[@bentalks7775]

I now know the radius of Super Smash Bros. Thank you

[@harshghuman2411]

That’s what I thought also lol

[@advaitpai1615]

Lmao

[@onetwentyeight128]

Ben Talks i clicked this video because of that

[@tmsmeister7603]

at least I'm not the only one who noticed

[@santiagoarosam430]

By calculating the power of the intersection point of the proposed chords with respect to the circumference, we obtain the length of the upper section of the vertical chord: 2x6=3c ⇒ c=4. Taking this chord as height, we construct a rectangle inscribed in the resulting circumference with height H=3+4=7 and base B=6-2=4. The center of the constructed rectangle coincides with that of the circumference and its radius R is half the diagonal; its value is obtained by Pythagoras: R²=(B/2)²+(H/2)² = 2²+3.5² = 16.25 ⇒ R=√16.25 ⇒ R=4.0311

[@242math]

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

[@wallop6740]

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

[@janus9148]

6+2=8 (diameter) then 8÷2=4 (radius) That's what I thought

[@dennisdegouveia5439]

@@janus9148 he said chord AB, so 6+2 is not the diameter. It just happens in this case that they're similar, which means this chord is very close to the centre of the circle.

[@janus9148]

@@dennisdegouveia5439yeah, i thought the chord AB was in the center

[@gabym.9836]

Janus no not unless the center point is on the line which would make it the diameter

[@bradenvester4308]

It is never that easy. No matter how sure you are.

[@AuraBOI]

Super Math Bros Ultimate

[@angusclark6170]

Kyledude252 I wanted to make that joke

[@tejashnayak6090]

Bruh

[@Megacooltommydee]

Don't you make fun of people with lisps. /s

[@anotheraggieburneraccount]

Numbers weave into a tower of math

[@elijahshadbolt7334]

I visited this video just to make sure someone commented that

[@zanissisojevs3388]

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.

[@monkeseeaction21987]

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.

[@mx.fuzzypants1911]

Thanks, now I know the radius of the Super Smash Bros logo

[@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.

[@shashikalayadav884]

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

[@educatedonealways6281]

You pick the longest and most obscure way to solve problems

[@abhradeepdey9054]

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)

[@SoumojitPan]

It could be done by properties of triangle[ABD] Ex-radius,R=abc/4Δ Here a,b,c are three side lengths and Δ is area of the triangle. So, a=6+2=8 b=√(6²+3²)=√45=3√5 c=√(3²+2²)=√13 Δ=(1/2)(8)(3)=12 Therefore, R =[8×(√13)×(3√5)]/[4×12] =[24√65]/[48] =[√65]/2 =4.031

[@pratikparikh8027]

The co-ordinate geometry solution is beautiful

[@adarshkaranwal2400]

I m a Indian and too much interested in mathematics and your videos today 5 may I sat for one of the thoughest exam of india to pursue BSC in MATHEMATICS and this question appeared in exam thank you Mind your decision .....U guys r doing a fantastic job..

[@Coronavirus-dk7mq]

Which exam was it?

[@D1g1talMess]

Challenger Approaching! Presh Talwalker divides the competition!

[@tx2016]

Pomm popopoooomm pOpopOpoooOmm!!! 😂👌

[@phungpham1725]

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)= AHxHB=2x6=12-----> sqR - sqOH= 12-----> sq R - (17/4 )=12 ---> sq R= 12+(17/4)= (48+17)/4=65/4. Thus the answer is R=(sqrt of 65)/2

[@FrancisBSCR]

Nobody: Problems in the 1970s: *handed out to unborn foetuses in Asia*

[@arnavambekar5665]

Peppermint Cookie 😂😂😂🤣🤣🤣👍👍✌🏼👍👍👍👍👍👍👍👍😂😂😂

[@kirigayakazuto7432]

foetuses lmao

[@Mathematician6124]

Math should be esteemed.

[@CatnipMasterRace]

I worked out the angle ABD using trigonometry. I worked out length AD using Pythagoras' theorem. The angle AOD (O being the centre of the circle) is 2 times angle ABD. I then used the cosine rule to work out the radius (2 of the side lengths of triangle AOD). Gave me the same answer.

[@puiyuyeung4243]

We use the same way to solve the problem🙌🏻

[@k_rvs6771]

(6+2):2

[@josh4563]

and here i am solving math problems by gut feeling

[@plumberdash9729]

Then congrats.....you are a human

[@EaglePicking]

My gut feeling immediately said "4" and I was only off by .031 so that worked out well enough.

[@MothaLuva]

I noticed that the horizontal line is not going exactly through the center. So, my solution was not 4, but „4 and a little bit“. So, I was exactly on spot. „A little bit“ might not exactly be a mathematical term but in real life it works pretty well in solving real life problems (as opposed to the brain masturbation presented here).

[@matthewberling7934]

@@MothaLuva good enough for the girls I go with

[@AJCham]

Not being familiar with the power of a point, I had a different approach to this. I duplicated both lines, rotated 180°. The line A(A`) is a diameter of the circle, and also the hypotenuse of the triangle AB(A`). AB has length 8, and B(A`) is unknown. Representing B(A`) as x, and using Pythagoras, we get a diameter of √(8² + x²). CD(C`) has lengths 4 and (6+x), so C(C`) gives the diameter a value of √(4² + (x + 6)²). Then we can solve (x + 6)² + 16 = x² + 64 Which gives x = 1. Substituting this into the aforementioned formula for the diameter A(A`), we get √(8² + 1²) = √65 So r = (√65)/2

[@sriniramiah6445]

Hi Presh, Thanks again for the great videos. i think you can solve this problem by using simple triangle properties without using coordinate systems or having to know other formulae. This is what I did: - I labelled the 4 vertices A, B, C and D clockwise starting from top. P is the point of intersection of the chords. - GIVEN: DP=2, CP=3, BP=6. SOLVE radius R= ? - Mark the center of the circle as O. Connect O to A,B,C & D - OA=OB=OC=OD = R. Drop a perpendicular line from O to chords AC and BD. Label the lengths of these as x & y. - Now, AP*3 = 2*6 , AP= 4. - Triangles AOC and BOD are isosceles. so it follows: 4-y/R = 3+y/R ; y = 0.5 2+x/R = 6-x/R; x = 2. Now applying Pythagoras theorem , we get R^2 = (4-y)^2 + x^2 R^2 = 3.5^2 + 2^2 = 16.25 ==> R = 4.031.

[@rinkeborg]

This is what I did. Easy.

[@mike1024.]

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!

[@fuckinepicful8329]

First time I finally solved a challenge from talwalker. Now i can finally go to sleep.

[@power-l5z]

I did it another way; note that you have a triangle with base 2+6=8 and height 3 and thus area 12 at the bottom of the diagram, with sides 2+6=8, sqrt(13), and 3sqrt(5). then the circumradius of that triangle is 1/2 abc/A = 1/2 * 24sqrt(65)/12 = sqrt(65) as desired.

[@solarfluxman8810]

I like what you did, but you ended up with the diameter.

[@lesadams647]

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.

[@NurulIslam-dw8yp]

I did it with circumcircle and circumradius. The answer matched!

[@XJWill1]

Call the intersection point P. Then triangles APD and BPC are similar and allow us to solve for CP=4. Then inscribe a rectangle in the circle with the bottom horizontal line as AB, and with a height of 1 (the height is chosen as CD - 2*DP = 7 - 2*3 = 1 since an inscribed rectangle must be centered vertically and horizontally in the circle). The diagonals of an inscribed rectangle are diameters of the circle. Since the rectangle is (8,1) in size, we know the diameter of the circle is (1^2 + 8^2)^0.5 by the PYTHAGOREAN THEOREM and thus the radius is sqrt(65)/2

[@firstlast5528]

We can use the sine rule in the ABD triangle: 2R=AD/sinB...etc

[@seanmatthew8232]

ye thats what i did. got the answer pretty quick that way.

[@jitendramohan7500]

That's trigonometry method but we need a pure geometric sol

[@dlevi67]

@@jitendramohan7500 1) trigonometry IS geometry. 2) Who says we _need_ a purely geometric solution?

[@moretzzz9500]

I click this on my recommendations. Now i feel smart.

[@3bydacreekside]

Yep

[@godson200]

Nobody Literally nobody: Le me after watching this video: mothemoticon

[@g-rex5440]

I solved it without knowing that special formular and that wx=yz draw the center (approximately) of the circle and mark it with O AB=8 so the bisector is 4 OB=r distance from chord AB to O labeled with x this is our first triangle: r²=x²+4² secound triangle: OD=r distance chord CD to O labled with y = AB/2 - 2 = 2 third side is x+3 so: r²=(x+3)² + 2² equaling both formulars x²+4²=(x+3)² + 2² x²+4²=x²+6x+3²+2² 4²-3²-2²=6x x=(16-9-4)/6 x=1/2 r²=x²+4² r²=(1/2)² + 4² r²=1/4 + 16 r²=(1+16*4)/4 r=root(65)/2

[@ABaumstumpf]

Same thing. Takes about a minute cause it is rather easy and no special knowledge needed.

[@bificommander7472]

Yep, that's the one I went with too.

[@alwaon3820]

Thank you! I think this way is more intuitive and easy to follows without knowing any special formula, etc.

[@marcobertini8768]

Same 😁

[@vishesh_soni]

Exactly , I too did it on my first attempt. 😀😀

[@mdatik5517]

This is one of the most liked channels by me.

[@059echo]

I never knew id get to knw a new formula !! i wanted it to be solved by geometry n was aghast when the formula was mentioned thinking the geometrical deduction wld be skipped ... Thanks for actually showing how the formula is derived !

[@jdnoflegend9719]

i used Trigonometry, arc tan twice to find angle ADB, then angle at center = twice angle at circumference. then split angle intto 2 (isosceles triangle) & cosine to find the Radius for those puzzled at Intersecting Chords Theorem wx=yz , u can prove it using Similar Triangles ACP & DBP , where P is the intersection point

[@StretchyDeath]

Ditto. Inscribed Angle Theorem + Law of Cosines FTW

[@safi3968]

"These math videos avaliable for free on youtube, builds confidence for students"... wish I could say the same....

[@federicogiustini9907]

I would've used trigonometry, but your method is way cooler and more elegant.

[@61rmd1]

Anch'io ho usato la trigonometria! Chiamando R l'incrocio delle corde, trovi facilmente RC, poi BC, quindi usi il teorema della corda col triangolo BCD, in cui l'angolo alfa = atan(2); da lì trovi r=sqrt(13)/sin(atan2) Su questo profilo c sono una bella serie di problemi, molto più tosti di questo...saluti dall'Italia!

[@federicogiustini9907]

@@61rmd1 grazie per le informazioni! Sono uno studente di ingegneria matematica, questi video sono per me il pane 👍👍

[@geo8046]

I solved it with a system of equations. The circumference's equation is x² + y² + ax + by + c = 0, where a,b,c are the parameters that can define every circumference on the plane. The problem gives you the coordinates of three points on the circumference so I solved for the equation of the problem's circumference by solving for a, b, and c. The way to do it is to make a system of three equations of the form x² + y² + ax + by + c = 0, but substituting for x and y the values of every given point. The final equation was x² + y² - 4x + -y + 12 = 0, and the radius of any circle is given by the formula 0.5sqrt( a² + b² -4c ) which leads to 0.5 root 65.

[@ravenheartFF]

I used power of points first to get the other part of the vertical, then found the total length of each chord. I set the intersection of the chords as my origin, then bisected each to find the coordinates of the center of the circle. Using that, I calculated the distance to the point (-2,0), which is the radius 4.031.

[@JohnDixon]

I did it with inverse trig functions and used the fact that arc ACB is equal to 2(arctan(2/3)+arctan(2))

[@mcandelaria8769]

Normal people:math Me, and intellectual: waluigi for smash

[@SateliteRaider1]

Intellectual mispelling *an. Heh

[@jackrabbit8255]

Okay KZbin recommendations. I guess it’s big brain time.

[@MichaelRothwell1]

Great problem. I solved it using geometry, knowing that wx=yz, but not knowing the formula for r^2. Then I did the general case, and derived the formula that you used and proved. Very neat & satisfying!

[@murdock5537]

Great! Many thanks! Another way solving the problem: w = 2 x = 6 x = 4 (via Chords Theorem) z = 3 N = point of intersection AB and CD = N(0; 0) y - z = 4 - 3 = 1 = 2(1/2) x - w = 6 - 2 = 4 → (1/2)4 = 2 → M = center of the circle = M(2;1/2) D = N(0;0) - (0;3) = D(0;-3) → r^2 = (2 - 0)^2 + (7/2)^2 = 65/4 → r = 2√65 🙂

[@sayedabdullah95]

For this Just Construct perpendicular on both the chords and you will get .5 a and 4 unit of the side of triangle in which the hypotenuse is the radius.

[@benv6875]

Exactly: I did it in my head not knowing the formula; .5 sq + 4 sq = 16.25. Take the square root = 4.031. Why make things so complicated. You can easily see the half way point of the line 7 is 3.5, which is .5 from the centre and the base is 4.

[@ZackJRich]

I solved it like this: Area of the triangle ABD is |AB|•|BD|•|AD| ÷ (4R) where R is radius of the circle |AB| = 8 |BD| = 3 * sqrt(5) |AD| = sqrt(13) And area is of course 8*3/2 = 12 Solving for R is easy

[@JamalAhmadMalik]

So did I!💕

[@skybuildergaming7618]

Same

[@weirdoctopus]

@@mevnesldau8408 I thought you were dead due to pneumonia , welcome back from realm of dead , btw big fan here

[@milzanmurtadha1926]

Where the 4R come from?

[@CthulhuYar]

IMHO- It's much more easier and more obvious then the example in video

[@dominicclwong]

A more “human” approach is: 1. Let r be the radius 2. Add a chord EF parallel and symmetric to CD, and another chord GH parallel and symmetric to AB; 3. CF and BG should pass through the centre of the circle, i.e. they are diameters, so equals to 2r 4. |> CDF and |> ABG should be right angle triangles 5. For |>ABG, let AG = a; then for |>CDF, CD = 3 + a + 3 = a + 6 6. For |>ABG, AG^2 + AB^2 = BG^2 for |>CDF, CD^2 + DF^2 = CF^2 7. For |>ABG, a^2 + 8^2 = (2r)^2 for |>CDF, (a+6)^2 + 4^2 = (2r)^2 8. a^2 + 8^2 = (2r)^2. ..........i (a+6)^2 + 4^2 = (2r)^2 ......ii 9. ii - i, 12a + 36 + 16 - 64 = 0, => a = 1 10. Using i, 1+ 64 = 4r^2, Answer: r = sqrt(65) / 2

[@JulianPlaysPiano]

Yes this is much simpler - no advanced theorems needed, just symmetry and Pythagoras! I expected Presh to solve it this way and was surprised he didn't. I can even imagine him describing it: "Let's make a second copy of the chords, rotated by 180 degrees, now draw a red triangle here, and a blue triangle here". Much more elegant.

[@garavelustagaravelusta9717]

1) Calculate remaining length as 4 2) Divide the horizontal line to two by drawing a perpendicular line towards it from the radius 3) Form a right angled triangle by merging the radius with point B. The equation is as follows: r^2 = 4^2 + height^2 4) To find the height, form a rectangle by drawing a perpendicular line towards vertical line from the radius. The vertical line is divided to two. Since its total length is 7, the distribution would be like 3, 0.5 (vertical side of the rectangle, hence the height), 3.5 when you look at the schematic carefully. 5) r^2 = 16 + 0.5^2 = 16.25 6) r = sqrt(16.25)

[@julianvickers]

Solved using a mix of trigonometry and circle geometry. Used circle geometry (angle at center = 2x angle at radius), then used cosine rule to calculate the radius.

[@alexchou1984]

Just subscribed to your channel! Proposing another solution: Connecting CB and AD (or AC and BD), we can also see that we get two similar triangles through inscribed angles based on the same pair of points. This allows us to get the top length to be 4. From there, we can use Pythagorean theorem on the 3.5 and 2 to get the final answer as well.

[@4zdr456]

the feeling when you solved it with the easiest way possible, but you can't explain it due to language barrier

[@mokiplay8515]

Alternate title: "finding the radius of Australian super smash Bros logo!"

[@joshuakurian4827]

Someone probably already mentioned this with different wording, but you can set the the three points A, B, and D on a coordinate system. The radius of the circle through A,B, and D is the circumradius of the points. To find the circumcenter, one can draw any pair of two lines among the three points, and find their intersection. Then find the distance between the intersection and a point.

[@WillSmith-wv6he]

Solved it with only the Pythagorean theorem. Draw center of Circle (C), Draw segment perpendicular intersection to horizontal chord. Call length of the segment "a". This bisects the chord, each half of the chord being 4. Draw a radius from C to intersection of chord with edge of circle. This gives a right triangle with sides, a and 4 and hypotenuse r. 4^2 + a^2 = r^2. Now draw perpendicular bisector from C to vertical chord. Notice that bisected chord length is 3 + a, and bisector is 2. From here you have another triangle with sides, 2 and 3+a, and hypotenuse r. 2^2 + (3+a)^2 = r^2. Solve for a, 1/2. Then plug in for r.

[@ArkajyotiBanerjee]

Applying the Intersecting Chords Theorem, the length of the remaining segment of the chord is (2 x 6) / 3 = 4. Join AC. Join BC. Let's say that the angle ABC is x. Then the angle subtended by the chord AC at the center of the circle is 2x. (Recall that the angle subtended by a chord at the center of a circle is twice the angle subtended by the chord at any point of the circle) Let the intersection point of the two chords be P, then from triangle CPB, we have sin x = 2 / sqrt(13). Mark the center of the triangle as O, then from triangle AOC we have 2r sin(2x / 2) = sqrt(2^2 + 4^2); or, 2r * (2 / sqrt(13)) = sqrt(20); or, r = sqrt(65) / 2. PS: Nice question, by the way. It feels good to have solved a geometry question using mainstream geometric theorems and not brute-forcing a system of equations to get the value of r.

[@udith]

I have another easier solution. Mark the intersection as origin, AB as x-axis CD as y-axis. Then we will get A(-2,0), B(6,0) and D(0,-3). According to some theorem in 9th/10th class the perpendicular bisector of cord passes through centre. Join BD. Find equations for perpendicular bisectors of cord BD and AB. Solve for x and y we will get the centre as (2,1/2) now find the distance between centre and B

[@adarshchaurasiya1219]

When you draw the perpendicular bisector of two non parallel chords their intersection point always be the centre of circle, it's mentioned in class 10 NCERT

[@godson200]

Yeah bhai... Mein abhi 12 mein hu aur mujhe sharam aa rhi h ki ye mere se hua nahi kese

[@godson200]

12 mein aate aate sirf calculus calculus hota h

[@furret.mp4]

The new Smash DLC Fighter looks amazing!

[@dddodu0195]

This is an easy task indeed. How do you construct the circle? Locus lines for the center of the circle are the perpendiculars of AD and DB. The coordinates of the 3 points can be read off directly, the straight lines AD and DB can be easily specified, the slopes of the vertical can be determined immediately. The midpoints of the routes remain to be determined (arithmetic mean of the respective coordinates). The point of intersection of the vertical lines results in the center of the circle. The radius is then e.g. the length of the line MB. In detail: AD: y = -1,5x-3 */* Midpoint (-1/-1,5) */* Vertical line y=(2/3)x-(5/6) */* DB: y=0,5x-3 */* Midpoint (3/-1,5) */* Vertical line y=-2x+4,5 */* point of intersection of the vertical lines M(2/0,5) */* radius r^2=MB^2=16,25 */* r=sqrt(16,25)=4,03 */*

[@cassandrachurch7318]

Much more simple with circle geometry. One chord is 7 units, the other is 8. The perpendicular to the midpoints of chords passes through the centre. Make a triangle, the radius is the hypotenuse, go 4 left from (6,0), then 0.5 up (3.5 - 3). Pythagoras’ theorem.

[@InvaderMixo]

Do you have a video on "Power of a Point"? I've never encountered that before! Great math exercise. I love circle geometry. Thanks for presenting it, Presh.

[@ObsidianParis]

He'll surely have a Powerpoint presentation about it :)

[@Tfin]

And he just takes it as a given. You might not know the other equation, but surely you know this! Um, no?

[@Shiraori999]

I think the w*x=y*z formula comes from the fact that the triangle with sides w and y is a similar triangle to the triangle with sides z and x. Two sides of 2 similar triangles are proportional to each other which gives the formula y/w=x/z. From which you can get the w*x=y*z formula.

[@rjheisenberg839]

00:36 OA×OB=DO×OC * O is the point of intersection

[@AAAAAA-gj2di]

Clearly, the other part of the vertical cord = 4. (3×4 = 2×6). Now if we take that point of intersection as our Origin of the cartesian plane with vertical chord as Y-axis and horizontal one as X-axis, this problem becomes way to easy. The center lies on (0.5, 2) [centre perpendicularly bisects every chord] and one of the circumference points is (0,4). We can also use (6,0) or another two of those 4 points. EDIT: The geometry method is similarly derived too. Both the methods use the same concept

[@Marcel-vz7vp]

Shadow Ninja idc nerd

[@JLvatron]

That radius formula's proof is impressive. Nice!

[@znhait]

The thing about math is that there are almost always multiple ways to solve a problem. I can typically understand the algebra involve in both proving and using a formula. My question is always the same: what is the motivation. That’s what math teachers need to be better at: explaining the why’s. Sometimes it’s good enough to say that you “guess” based on previous knowledge.

[@maximashin8526]

You can just find some angles (using tan (a)) and use the law of sines on one of the triangles that lays on the circle... Easy.

[@4zdr456]

I did this way too. i didn't even understand why was it hard

[@btdpro752]

I don't understand

[@maximashin8526]

@@btdpro752 read about the law of sines, and I'm sure you'll see it

[@В.Гоподинов]

that feeling when I wrong on ≈0.031

[@Clockpie5436]

6 + 2 /4?

[@one_exp99]

Now I can solve the radius of the Super Smash Bros logo

[@jeanf6295]

I have drawn the picture on a grid, eyeballed the center, checked that it worked by computing the distance to the three given points.

[@PawStreetProwlers]

A practical application for finding the center and edges of circles is CNC machining. There are radius probes that you can use, but I found they are clumsy and less accurate than just tapping off any 4 points on the edge and calculating the center and radius.

[@Susp7]

3:48 why is it w²+x² ?? Shouldn't there be a -2wx

[@matthijsgeerlings]

The intersection of the two lines I called point E. Then I shove the line CD to the middle so that AE= 4 & BE=4. There thene is ane equal distance (x) berween point C and the top of the circle and point D and the bottem of the circle. So (3 + x)•(4+x)=16 and (3+x+4+x)/2=r. Then I solved for x

[@yurenchu]

Initially, I was looking for a method like this (shifting the chords and reapplying intersecting chords theorem), but I couldn't figure it out (I didn't think of labeling x). Cool method! I'm glad to see that it can indeed be done that way.

[@1ciricola]

I did look over a hundred or so of the 2.3 thousand comments. My solution was similar to yours. I posted it already. So I thought I would pass it along to you personally: There have been over two thousand comments to this problem over five years. So, the approach I took has probably already been submitted. Regardless, this is my solution to the problem: The two parts of the vertical line segment CD have lengths three and four. So the midpoint of the vertical cord is 3.5 units; a horizontal line drawn through that midpoint is the diameter of the circle. If the horizontal cord AB is "pushed up"so that it coincides with the above diameter line, then its new length will be (2+x) + (6+x). So, by the Intersecting Chords Theorem: (2+x)(6+x) = (3.5)(3.5) The solutions to the resulting quadratic equation, x² + 8x -.25 = 0, are .03 and -8.03. x = .03 is the useable root. The diameter length is (2+x) + (6+x) = 8 + 2x = 8 + .06 = 8.06 Therefore the circle’s radius = 4.03

[@bryandcs4356]

I was able to do it, there is theorem that says a^2+b^2+c^2+d^2=4R^2 Where a, b, c and d are chords

[@ObsidianParis]

Yep, this is called the Presh Talwakar Theorem :)

[@w.jordan1736]

Nice, how can I proof this?

[@bryandcs4356]

@@w.jordan1736 I have a book, I wish I could send a picture

[@turtlelink6845]

Switch two adjacent chords to prove instantly.

[@yurenchu]

@Willian Honório, - "Nice, how can I proof this?" Consider the figure at 4:19. Let a+b be the horizontal chord (AB), and c+d be the vertical chord (CD). Copy the vertical chord and shift it horizontally to the right, until it touches the circle circumference at F (at top) and E (at bottom). Then CDEF is a rectangle with diameter 2r, vertical side (c+d) and horizontal side (b-a). Using Pythagoras: (2r)² = (c+d)² + (b-a)² 4r² = c²+d² + 2cd + b² + a² - 2ab However, due to the intersecting chords theorem (which we also used to compute c = ab/d = 2*6/3 = 4), we know that ab = cd, so the terms 2cd and -2ab cancel eachother; and therefore we're left with 4r² = a² + b² + c² + d²

[@andrewmander-jones8204]

I've just seen this problem. My solution: (a) compute the length of the upper part of CD is 4 by similar triangles (or the power law as mentioned). (b) Imagine the vertical diameter through the centre. The circle is symmetrical about this line. Draw the vertical line EF parallel to CD reflected in the vertical diameter. By symmetry it will intersect AB at a distance of 2 units from B, and thus will be 4 units along AB from CD. (c) Join CE, length 4, and consider the right-angled triangle CDE. DE will pass through the centre of the circle and thus be a diameter. Its length will be sqrt(CD^2+CE^2) = sqrt(65). Thus the radius is sqrt(65)/2.

[@Stelios2711]

sin(

[@alexsere3061]

I used an extended version of The sine theorem a/sinA=abc/area=2R

[@georgiteodosiev9037]

I solved it with the theorem of intersecting chords, then I drew a chord, parallel to CD, starting from point A, whitch by symmetry is equal to 1. Then I connect this new drawn 's vertex to B to get a right triangle. By the theorem that 2 chords, intersecting on a point of the perimeter of the circle , making a right angle, form a triangle with a hypotenuse that is equal to the diameter, I get that d=sqrt (1^1+8^2), whitch is d=sqrt (65). Then the radius is d/2=sqrt(65)/2, and that's my answer. Did YOU figure it out?! Thanks for seeing my solution, sub and ya...😂😀

[@MetaDriver33]

I also solved this way. In fact, after finding y (= 4), you can flip the chords around either vertical or horizontal. The result is the same: r^2 = 8^2 + 1^2 = 7^2 + 4^2 = 65.

[@MintyBlaziken]

Presh Talwalker smash reveal?

[@DergaZuul]

It can be directly solved in one step using radius of circle around triangle formula, and Pythagorean theorem of course. More work to simplify the answer but very straight forward.

[@johnmichaelleslie4835]

I ran across this problem on Twitter and did it this way, what do you think? Inscribed angles: Angle of a chord from a point on the circumference is half the angle of that chord from the centre So angle of AD from B is half the angle of AD from X (= centre of circle) Length AD (Pythagoras) = sqrt(3^2+2^2) Tan = Opposite / Adjacent Angle AD from B = atan(3/6) Angle AD from X = 2 * atan(3/6) Say Y is the centre of AD, length AY = sqrt(3^2+2^2) / 2 Make a right-angle triangle AX, AY, XY (As going at 90 degrees from the centre of a chord is a diameter, so goes through the centre) Angle AY to centre of circle = 2 * atan(3/6) / 2 = atan(3/6) AX = R = radius Sin = Opposite / Hypotenuse R = (sqrt(3^2+2^2)/2) / sin(atan(3/6)) = 4.0311

[@jhawk2402]

First see the missing length is 4 (because 2*6 = 3*x), then use the formula that the diameter is equal to the square root of all of the lengths squared and added, so √(3^2 + 4^2 + 2^2 + 6^2) = √(9 + 16 + 4 + 36) = √65. This means the radius is √65/2

[@pipertripp]

slick. What's the name of that theorem?

[@Gandarf_]

pipertripp check the video

[@jhawk2402]

@@pipertripp idk, sorry. The place I learned it from never gave the name

[@pipertripp]

@@Gandarf_ thx mate. Great little proof of it too.

[@nguthang821]

2R =BD/ ( sin(DAB) Sin DAB = 3/( ROOT 13) => R = (root 65 )/2

[@vrela_kosilica9285]

excuse me, how

[@chenchangpo]

FratarGaspetra Law of sines.

[@keatonrix2512]

You can also use cosine rule and the fact that the angle at the circumference is half the angle at the center.

[@joaoolivio8378]

ngu thang I solved it this way as well

[@stablexit]

me: no, this can't be this easy my brain: but, it is me: then, how you solve that? my brain: 6+2=8 8/2= 4, the radius is 4

[@ayeitsjoe8221]

But the line doesn’t cut directly through the center of the circle, so that won’t be accurate

[@udith]

@@ayeitsjoe8221 r/whoooooooshhhhh

[@thomasgoetz8723]

@@ayeitsjoe8221 Are you sure? I can remember an article where a school teacher sent a student to detention because the student insisted the teacher was wrong. And why did the student insisted? Because the teacher came up with the idea/appraoch pi = 3 so calculation would be easier. So in the US everything is possible.

[@Carnezz]

Lol I did the same. 4 was basically correct though

[@Apollorion]

That would only be true if AB was a diameter though. But if AB was a diameter, then, due to the circle theorem of Thales, AB would be the hypotenuse of ADB and then AD^2+BD^2 would be equal to AB^2 the latter of which is 64. But that it isn't, because, also calculated via the theorem of Pythagoras: AD^2=3^2+2^2=9+4=13 and BD^2=3^2+6^2=9+36=45 and hence the sum of these two squares is 58 which clearly isn't 64. iow: the center of the circle lies not on the chord AB.

[@ghhdcdvv5069]

تمرين جميل جيد . رسم واضح مرتب. شرح واصح جيد . شكرا جزيلا لكم والله يحفظكم ويحميكم ويجميعا . تحياتنا لكم من غزة غلسطين .

[@Utars]

My method is way more simple: ABD is a triangle inscribed in a circle. The radius of the circumscribed circle is (abc)/(4S), where a,b and c are the sides of the triangle, and S is a square of the triangle. The sides AD and DB can be calculated using the Pythagorean theorem, AB is already known to be 8. Let DH be the height of triangle ABD. The square of the triangle is (1/2)*AB*DH Therefore the (abc)/(4S) formula turns into (√13*√45*8)/48, which is equal to √65/2

[@dekiller78]

Solved this in second with sin theorem (A/Sin(a) = 2R )

[@neriaseeman198]

Me too

[@argyrisgiannisismanes8600]

Same here

[@clumsycookie3659]

Me as well

[@igxniisan6996]

“The perpendicular bisectors of all the chords inside a circle always passes through its centre.”

[@ronniecoles7335]

4.031 is rounded to 4 so I’m right and I done it all in my head, yayy!

[@stanleystanleystanley7246]

Here's a simpler geometry proof. By intersecting chords, the top portion of CD is 4. Next, take AB and draw a mirror image chord above (call it A'B'). Likewise, take CD and draw a mirror image chord on the right (call it C'D'). Use Pythagorean theorem on any of 4 diameters: AB', A'B, CD', C'D.

[@mathisnotforthefaintofheart]

From the moment you have A,B and D, you can use perpendicular bisectors to establish the Center. Then it's easy

[@Rainier214]

I clicked on this because the circle in the thumbnail looked like the Smash Ball.

[@bizboy9508]

Literally just the smash logo

[@hlobet_]

CUT OUT THE CIRCLE, FOLD IT IN HALF TWO TIMES, THAT IS THE RADIUS

[@avinashbalakrishna1927]

Thank You, doing the same process every time but not thought of a theorem🙏.

[@gvigary1]

Nice. I used the coordinate method but set my origin at the midpoint of the horizontal chord, so I knew the centre was (0,c) and it had to pass through (4,0) and (-2,-3). From there it fell out. Very pleasing!