Math Olympiad | Can you find Radius of the Yellow circle? | (Step-by-step explanation)

[@MrPaulc222]

Same here, but I seem to prefer the quadratic formula for the final calculation. It's just a preference. Your videos are the clearest I've seen of all the maths channels on youtube and I thank you for that.

[@PreMath]

Glad you like them! You are very welcome! Thanks for your feedback! Cheers! 😀 You are awesome. Keep it up 👍

[@kennethstevenson976]

Very good step by step solution to this problem

[@PreMath]

Thanks for your feedback! Cheers! 😀 You are awesome. Keep it up 👍

[@MuriloMBarquette]

Glad to see that I still can solve these tasks, even though I'm 45 years out of school. Thanks for your videos! Greetings from Rio de Janeiro - Brazil

[@Ramkabharosa]

Extend the line AC to meet the circle again at G. Then |AG| = 2r - 6. By tangent-secant theorem, |AE|² = |AC|.|AG|. So (27- r)² = 6.(2r - 6). ∴ 729 - 54r + r² = 12r - 36. So r² - 66r + 765 = 0. ∴ (r - 15)(r - 51) = 0. Since r < 27, r cannot be 51, so r =15.

[@hasanthaeranga3027]

Im from sri lanka.i gather large amount of mathematical knoweldge from your channel.i wish you be a proud channel in youtube.thank you very much ❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤

[@SladeMacGregor]

Awesome problem Professor!

[@KAvi_YA666]

Thanks for video.Good luck sir!!!!!!!!!

[@dirklutz2818]

Beautiful!

[@MultiYesindeed]

Great thanks again

[@PreMath]

Excellent! You are very welcome! You are awesome. Keep it up 👍

[@niconickolasgomez]

👌

[@じーちゃんねる-v4n]

半径=r ∠EOC=θとおくとr+rsinθ=27 r-rcosθ=6 ∴(27/r-1)^2+(1-6/r)^2=1 3/r=tとおくと85t^2-22t+1=(5t-1)(17t-1)=0 ∴t=1/5, 1/17 ∴r=15, 51 51>27より r=15

[@harikatragadda]

Draw the diagonal CG and note that ∠AEC = ∠CGE By Similarity of ∆CEA and ∆CEG, CE/CA = 2R/CE CE² = 12R = CA² + AE² = 6² + (27-R)² (R -33) = ±18 R = 15 and reject 51.

[@PreMath]

Thanks for sharing! Cheers! You are awesome. Keep it up 👍

[@ybodoN]

From the steps in the solution, it looks like G was obtained by drawing the diameter EG. 🧐

[@harikatragadda]

​@@ybodoN Its the line COG with ∠CEG =90°

[@ybodoN]

@@harikatragadda Indeed, it also works with G at this position 💡

[@shmulin8484]

Love from Bangladesh ❤❤ Thank you bro for making this this type video❤❤

[@ybodoN]

Here too the Pythagorean theorem could be replaced by the intersecting chords theorem: CF² = EF ⋅ (2r − EF). Also, we could use the formula r² = ¼ (a² + b² + c² + d²) which would give r² = ¼ (EF² + (2r − EF)² + 2 (AB − r)²). Either way, it looks like we'll always end up with the same quadratic equation.

[@ernestschoenmakers8181]

I did the same.

[@PreMath]

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍

[@WaiWai-qv4wv]

Oh very thanks

[@PreMath]

You’re welcome 😊 You are awesome. Keep it up 👍

[@yurenchu]

Fun fact: for r = 15 , the sides of triangle CFO (with C to the left of F) are (9, 12, 15), making it simiilar to a (3, 4, 5)-triangle (i.e. matching the Pythagorean triple (3, 4, 5) ). for r = 51 , the sides of triangle CFO (with C to the _right_ of F) are (24, 45, 51), making it similar to a (8, 15, 17)-triangle (i.e. matching the Pythagorean triple (8, 15, 17) ). In both instances, the scale factor is 3 (probably a result from the fact that in the diagram, the given parameters (6 and 27) have a greatest common divisor of 3).

[@murdock5537]

Nice! 6 < r < 27 → sin⁡(φ) = (r - 6)/r → cos⁡(φ) = (27 - r)/r → sin^2(φ) + cos^2(φ) = 1 → (r - 33)^2 = (18)^2 → r1 = 15 → r2 = 51 > 27 → r2 ≠ solution btw: sin⁡(φ) = 3/5 → CF = 12 → ∆ CFO = pyth. triple 3(3 - 4 - 5)

[@PreMath]

Thanks for sharing! Cheers! You are awesome. Keep it up 👍

[@Copernicusfreud]

Yay! I solved the problem. The grouping and factoring method seemed too difficult, so I just used the quadratic formula to find the r values. r = 15 for the given problem.

[@PreMath]

Bravo! Thanks for sharing! Cheers! You are awesome. Keep rocking 👍

[@waheisel]

Thanks again PreMath for another fun math puzzle. I would argue that 51 should not be rejected. The figure shows A and B on opposite sides of E but it isn't stated that A can't be between E and B. In that case the horizontal side of the right triangle is r-27 rather than 27-r. The quadratic equation is the same, but the solution r=15 is rejected and r=51 is accepted.

[@MrPaulc222]

I made another attempt at this as I forgot I had done it before. This time I went for extending the lines and using intersecting chords: (r-6)(r-6)=27(2r-27). After some number crunching I ended up with the same quadratic.

[@bigm383]

❤👍😀🥂

[@PreMath]

Thank you! Cheers! 😀 You are awesome. Keep rocking 👍

[@mathswan1607]

r^2=(r-6)^2+(27-r)^2 r^2-66r+765=0 r=15 or r=51(rejected)

[@yurenchu]

(27-r)² + (r-6)² = r² 729 - 54r + r² + r² - 12r + 36 = r² 765 - 66r + r² = 0 765 - 15r - 51r + r² = 0 15(51 - r) - r(51-r) = 0 (15-r)(51-r) = 0 r = 15 OR r = 51 When r = 51 , the line segment of length 27 would be shorter than the radius and therefore (presuming the rightmost point of the line segment is still vertically below the circle's rightmost point) the lowest point of the circle would not be on the line segment, but on the line segment's extension to the left; in other words, point A would be inbetween points E and B. (It would still be a solution, as the vertical distance |AC| between the line segment's leftmost point and the circle would still be 6 , but it's not a solution that matches the drawing.) So the answer is r = 15 .

[@jimlocke9320]

Well done, PreMath! There are other ways to solve it, of course, but I don't see anything more straightforward than what you presented. Factoring is fine if the factors drop out. I prefer using the quadratic formula, which yields the same values for r, 15 and 51, and 51 gets rejected for the same reason given in the video. Factoring produces a quick solution when it works, but it doesn't always work and the quadratic formula will be needed when it doesn't. However, factoring is a very useful tool and it is great that you taught how to solve by factoring!

[@PreMath]

Excellent! Thanks for your feedback! Cheers! 😀 You are awesome. Keep it up 👍

[@bobbyheffley4955]

If the quadratic coefficient is 1 and the linear coefficient is even, you can also complete the square.

[@quigonkenny]

As EB and BD are tangent to Circle O, and ∠EBD is 90°, EB and BD are each the length of the radius r, and ∠OEB is 90°. As EB = r, AE = 27 - r. Let P be a point on OE where CP is perpendicular to OE. By observation, CP = AE = 27 - r, PE = CA = 6, and OP = r - 6. Triangle ∆CPO: c² = a² + b² r² = (r-6)² + (27-r)² r² = r² - 12r + 36 + r² - 54r + 729 r² - 66r + 765 = 0 r² - 51r - 15r + 765 = 0 r(r-51) - 15(r-51) = 0 (r-51)(r-15) = 0 r - 51 = 0 | r - 15 = 0 r = 51 ❌ too big r = 15 ✓

[@aayushw]

Random Guy: "there can't be an interesting, persistent, and frequent youtuber who uploads math videos" Premaths: "Hold my protractor 🥂"

[@PreMath]

Thank you! Cheers! 😀 You are awesome. Keep it up 👍

[@batavuskoga]

First, your explanation are really awesome. You are one of the best But why don't you use the formula (-b+-√b²-4ac) / 2a ? In my opinion this way is much easier. I always solve a quadratic equation with this formula. Keep up the great work.

[@PreMath]

Thank you! Cheers! 😀 You are awesome. Keep it up 👍

[@zdrastvutye]

someone else can write the graphics into this code: 10 l1=6:l2=27:p=l1+l2:q=l1^2+l2^2:dis=p*p-q r1=-sqr(dis)+p:r2=p+sqr(dis):print r1,r2 15 51 > run in bbc basic sdl and hit ctrl tab to copy

[@wmcomprev]

r=51 would also be rejected because that would make the diameter 102. AB is 27, which is a radius & part of a radius. Even if you assumed the radius to be equal to AB (making AE=0), that would still only give a diameter of 54, not 102.

[@yurenchu]

r = 51 is the solution when point A is between E and B, instead of to the left of E.

[@prossvay8744]

(r-6)^2+(27-r)^2=r^2 r=15

[@ilxomjurayev6610]

(27-r)^2=6×(2r-6) Uzbekistan.

[@PreMath]

Uzbekistan ❤️ Thanks for sharing! Cheers! You are awesome. Keep it up 👍

[@predator1702]

👍 🙏

[@PreMath]

Thanks, dear! Cheers! 😀 You are awesome. Keep it up 👍

[@marioalb9726]

Pythagorean theorem: r² = (27-r)² + (r-6)² r² = (27²-54r+r²)+(r²-12r+6²) r² + 66r - 765 = 0 r = 15 cm ( Solved √ )

[@PreMath]

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍

[@soli9mana-soli4953]

√ (r² - (r-6)²) + r = 27

[@DariyonLandellHycheKrattTCl]

This is Interesting Math I Will Study About It Btw Thanks About (A+B) and That Solves My Problems, Ae = 47 - y, square Is O and y - 6 F 27 -r C R

[@PreMath]

Thank you! Cheers! 😀 You are awesome. Keep it up 👍

[@giuseppemalaguti435]

(r-6)^2+(27-r)^2=r^2...r=15

[@PreMath]

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍

[@gelbkehlchen]

Solution: I see a right triangle: Hypotenuse: OC = r Horizontal leg: AB-OD = 27-r Vertical leg: BD-AC = r-6 Pythagoras: r² = (27-r)²+(r-6)² ⟹ r² = 729-54r+r²+r²-12r+36 |-r² ⟹ r²-66r+765 = 0|p-q formula ⟹ r1/2 = 33±√(1089-765) = 33±18 ⟹ r1 = 33+18 = 51 and r2 = 33-18 = 15 ⟹ The radius of the yellow circle must be less than 27, so r2 = 15 is correct.

[@lukeheatley4148]

when you do factoring and grouping you never really explain why you choose to 15 and 51 I just ask myself what 2 numbers add up to 66 and have a product of 765. prime factors of 765 are 3 3 5 & so these need to be combined in to products which add up to 66. so its: 17 x 3 = 51 and 3 x 5 = 15. 51 + 15 = 66 Which gets you to (x - 51)(x - 15) = 0 If i am solviing, say, x² + 4x - 12 = 0, I just look for 2 numbers which have a difference of 4 and product of 12. so its 6 and 2. The original equation contains -12 so one is positive and one is negative, but it's +4x so i need the bigger of 6 and 2 to be positve. The answer is therefore (x + 6)(x - 2) = 0 This rarely takes me more than a few seconds.

[@wackojacko3962]

Factoring and Grouping is hard and really tricky cuz we gotta figure out what gets multiplied to produce the expression given to us. But knowing identities helps immensely like the Difference of Squares the Quadratic Formula The Sum and Difference of Cubes and expressions for (a + b)³ and (a - b)³ ....or just get lazy and let Computer Computer Algebra Systems do all the reductions and factoring for ya! ,...it's what I do! Just kidding! 🙂

[@PreMath]

Excellent! Thanks for your feedback! Cheers! 😀 You are awesome. Keep rocking 👍

[@nahiansparten3041]

when I saw the question I try to solve the question in my mind and within 30 seconds I solved it. And the answer was [13.5]. but but I think was wrong 😢

[@PreMath]

Thank you! Cheers! 😀 You are awesome. Keep it up 👍

[@louislaw5296]

For a math-dummy like me, I can never factoring out the 66 and 765... No clue at all...

[@Zhang158]

Use the determinant formula Roots are -b +-√(b^2-4ac)/2a

[@prossvay8744]

15

[@aleksandrsavuskan7124]

R=15

[@JSSTyger]

r=15