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I like this question. Lots of calculations

Clear as mud.

One channel different names...

One approach to making a video longer than it needed to be is to repeat a point over and over…. 👍

Hi. What softwares do you use to make your videos. I want to make videos like this for my students.

Thanks for the nice video. Do you have any recomendations to improve my geometry skills. I am not that good in geometry as I am in other parts of mathematics.

Another solution is this: If we draw a perpendicular line segment from point H to line segment EF, we create a 3-4-5 right triangle (since HK = 5 and AE = 4). Let's name the intersection point which is one of the vertices of this triangle as H'. This is the point where right angle HH'K resides. Let's name the third vertex of the triangle with the vertices K and F as F'. Let's name the third vertex of the triangle which has the common vertex F' with the triangle KFF' as Q'. Then all these 3 triangles (HH'K, KFF' and QQ'F') are similar with the side ratios 3x:4x:5x. Let's name the side of the triangle QQ'F' which is QQ' as b, Q'F' as a and QF' as c. Then b = 3x, a = 4x, c = 5x. The line segment DF' is k*a that is DF'/Q'F' = k. Then k = (a+b+c)/b = 4. Then k*a = 4*a = 4*4x = 16x. Since QC also equals to b then QC+QF'+DF' = 12. 3x+5x+16x = 12. x = 1/2. The area is 12*QC = 12*3x = 12*3*(1/2) = 18. By the way, AH = (k-1)*a = 3*4x = 12x =6. Moreover, the radius of the inner circle of the triangle HDF' also equals to a = 4x = 2

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Another method: Same as up to 1:11 As △APB is right isosceles so ∠ABP = 45° It is already shown that ∠ABH = 60° so ∠HBP = ∠ABH - ∠ABP = 15° By the inscribed angle theorem ∠HPB = ∠HAB = 30° So ∠PHB = 180° - (30° + 15°) = 135° Define PH as x. Apply the Law of Cosines to △PHB: 2² = x² + (√2)² - x(2√2)cos(135°) cos(135°) = -(√2/2) so this rearranges to give the quadratic x² + 2x - 2 = 0 which solves to x = √3 - 1 or -1 - √3. Discard the -ve result to leave x = √3 - 1

Hello y'all! We keep getting comments about TTS (Text-to-speech). Currently, the reason for using this system is to maintain accuracy, consistency, and to save time. We make these videos after work to share with y'all, so it takes about 3-4 days to create one video, and it would take more time to record my voice. We understand that people might not prefer AI voices, but using TTS saves time, allowing us to continuously upload videos. You can refer to the linked video to hear my real voice. In the future, I hope the program will allow me to use my voice. Thank you and have a great 2025! th-cam.com/video/Ai9Ohgb254M/w-d-xo.html

There was one subtle unexplained leap that I couldn’t follow. How do you know that r is half of 6-r?

Solution: R = radius of the large circle = 6, r = radius of the small circles, A = center of the left circle at the top, B = center of the left circle at the bottom, M = center of the large circle, C = point of contact of the left circle at the top. Due to the symmetry and the division of the hexagon, consisting of the centers of the small circles, into 6 equilateral triangles, BMA is an equilateral triangle with side length s = 2r. Now CM = 2r+r = 3r = R = 6 |/3 ⟹ r = 2 ⟹ s = 2*2 = 4 Height of an equilateral triangle = h = √(4²-2²) = √12, Area of ​​an equilateral triangle = F = s*h/2 = 4*√12/2 = 2*√12 The 6 small circles that form a hexagon have, as usual for a hexagon, an opening angle of 120° at the center, which is a third of a circle. All 6 together have an area of ​​6/3 = 2 small circles. Thus, the colored area is = area of ​​6 equilateral triangles - area of ​​2 small circles = 6*2*√12-2*π*2² = 12*√12-8*π = 24*√3-8*π = 8*(3*√3-π) ≈ 16.4365

Solution: R = radius of the large circle = 2, r = radius of the small circles, s = side of the equilateral triangle, M = middle of the line BC. MCO is the famous right-angled triangle with angles 90° - 30° - 60°. This means that MO = R/2 = 2/2 = 1. And according to the Pythagorean theorem: MC = s/2 = √(2²-1²) = √3 |*2 ⟹ s = 2*√3. D = point of contact of the two lower blue circles, E = center of the left, lower, blue circle. Pythagoras for the right-angled triangle EDO: OE² = ED²+DO² ⟹ (R-r)² = r²+(MO+r)² ⟹ (2-r)² = r²+(1+r)² ⟹ 4-4r+r² = r²+1+2r+r² |-4+4r-r² ⟹ r²+6r-3 = 0 |p-q formula ⟹ r1/2 = -3±√(9+3) = -3±√12 = -3±2*√3 ⟹ r1 = -3+2*√3 and r2 = -3-2*√3 [invalid in geometry] ⟹ Area of ​​the 6 small, blue circles = 6*π*(-3+2*√3)² = 6*π*(9-12*√3+12) = 6*(21-12*√3)*π = 18*(7-4*√3)*π ≈ 4.0600

Text-to-Speech ruins every video.

Hi, great challenge. Also, there is another solution: It is easy to see that triangle ADC has a surface of 120 units. And this surface is also given by tyhe formula sin(x) * (1/2) * AD * DC; which can be obtained by the pythagoric formula of it´s sides Thank you

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Very good

So good Thanks 🎉

2x+3x+70=180 x=22 <CAB=2×22=44

Ty

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A different angle chasing route: In △BCD: ∠BCD = 180° - (15° + 40°) = 125° As BC is tangent to circle Q at C then ∠BCQ = 90° and so ∠QCD = 125° - 90° = 35° △CQD is isosceles so ∠QCD = ∠QDC = 35° and so ∠QDK = 40° - 35° = 5° △DQK is isosceles so ∠QKD = ∠QDK = 5° Circle P and circle Q are tangent at K, so P, K and Q are colinear. ∠PKB and ∠QKD are opposite angles so ∠PKB = ∠QKD = 5° △BPK is isosceles so ∠PBK = ∠PKB = 5° and so ∠PBK = 180° - (5° + 5°) = 170° ∠BAK (= x) is drawn from the same chord (BK) as ∠BPK so by the inscribed angle theorem ∠BAK = (∠BPK)/2 = 170°/2 = 85°

Solution: y = -x²+6x+7 This is a quadratic parabola that opens downwards. Zeros of this function: -x²+6x+7 = 0 |*(-1) ⟹ x²-6x-7 = 0 |p-q formula ⟹ x1/2 = 3±√(9+7) = 3±4 ⟹ x1 = 3+4 = 7 and x2 = 3-4 = -1 ⟹ B = (-1;0) and C = (7;0) The highest point = vertex of this parabola lies exactly between the two zeros, so: xH = (-1+7)/2 = 3 ⟹ yH = -3²+6*3+7 = 16 ⟹ A = (3;16) r = radius of the circle = BP = CP = AP. D = point exactly below P = (3;0). Pythagoras: DC²+DP² = CP² ⟹ (7-3)²+(16-r)² = r² ⟹ 4²+16²-32r+r² = r² |-r²+32r ⟹ 272 = 32r |/32 ⟹ r = 8.5

This is useful, as an international student ım not familiar w desmos at all! No need for negative comments

is desmos allowed on the SAT

My newborn baby could do this without a guide

CE=10•1/(1+2)=10/3 BE=vʼ(10²+(10/3)²)=(10/3)vʼ10 BF=CG=10/2=5; EG=5-10/3=5/3 | BP+PQ+EQ=BE | | BF²=BP(BP+PQ) | => PQ=2vʼ15 | EG²=EQ(EQ+PQ) |

I got to the point where I'd calculated EQ = 2√5 A dropped perpendicular from E to OQ gives a right triangle with legs 4 and 2 and hypotenuse 2√5, so ∠OQE = arccos(√5/5) = ∠OQP △QOP is isosceles and so ∠POQ = 180° - 2arccos(√5/5) Using the law of cosines on △QOP gives (PQ)² = 32(1 - cos(∠POQ)) = 32(1 - (3/5)) = 64/5 So PQ = √(64/5) = (8√5)/5

Bro, when you make videos, can you tell if we can use trigonometry or not to solve the problem before giving the solutinons. thangks

another method sin(A/2)=1/3 sin A=2.sin A/2 .cos A/2 BC/sin A=2R thank u sir for this amazing video

OK, just a small word of advice: when you're posing a question involving circle theorems, r usually denotes the radius of the circle, which in this case is 115/2 = 57.5 Of all the letters available to you, why would you choose r to represent a quantity other than the radius? A quantity that is not even easily represented on the diagram?

116

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What a joke 🤣.... ratio of arcs is equal to ratio of the angles

1:55 (6+6+9)/2=21/2 vʼ((21/2)(9/2)(9/2)(3/2))=(27/4)vʼ7

1:18-1:38 r=(a+b-c)/2 => a+b=5

The link for the original video under the logo

78.75

Which program you use

Just draw a right triangle(because the hypo is the diameter of the big circle) FBA where BH is the height, then find the height via Leg*Leg/Hypo, then multiply the height by 2 (the radius theorem when a radius bisects a chord into two equal lengths).

Uh, what does this have to do cosign of an angle, as mentioned in the title? I solved it using slightly calculations, but generally the same way, using the Pythagorean Theorem.

Hi everyone, one of our viewers noticed an error in our previous video, and we have now corrected it. The question has been adjusted for clarity. We hope this updated video is helpful. Thank you!

I did all manner of things after using law of cosines to show CD = 1 as per the video. Brahmagupta's formula to get [ABCD] = 2√3 30-60-90 △ rule to show that the height of △ABE is √3 (if AE is the base). So 2x [△CDE] = (p + 3)√3 - 4√3 = (p - 1)√3 Dropping a perpendicular from D to CE gives another 30-60-90 △ to show that the height of △CDE is √3/2 (with CE as the base), so 2x [△CDE] = q√3/2 Hence p - 1 = q/2 so q = 2p - 2 Intersecting secants theorem shows that p(p + 3) = q(q + 2). Substitute in q = 2p - 2 to give p² + 3p = 4p² - 4p which solves to p = 7/3 and so q = 8/3 Then [△CDE] = (8/3)(√3/2)(1/2) = (2√3)/3

Another 2 methods: 1) Determine that AB = AC = 4√2 as per the video. Extend AO to form a diameter AF. By Thales's theorem, you have a right triangle △ABF with hypotenuse AF = 6. So by Pythagoras BF = 2 The point H (as in the video) gives another right triangle △BHF which is similar to △ABF and so it can be shown that BH = (4√2)/3, and so BC = (8√2)/3 2) ∠OAD = ∠OAE = arctan(1/(2√2)) Law of cosines: BC² = (4√2)² + (4√2)² - 2(4√2)(4√2)cos(2arctan(1/(2√2))) BC² = 64 - (64·⁹/₇) = 128/9 So BC = √(128/9) = (8√2)/3

hi. I did something wrong in my solutin but i dont know what. as BQ is KD because if we rotate averything it´s the same, therefore qc = kc and kqc = 45 deg and bqp = 135 deg, and the sum of the angles of a cuadrilateral equals 360 deg, so x = 45 deg an cos(45 deg) = root(2)/2. what i did wrong and why. Very interesting video and solution.

75 square unit ?

Although the method described here is ultimately intelligible, this is honestly one of the worst mathenatics videos I have ever seen on TH-cam. Having a robotic voice recite an alphabet soup of sector names is just a bad teaching methodology.

2 cm