Advanced geometry - circles inside of a circle

Brilliant circles

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[WesternPuchuu]

I personally think that the proportions are the points worth to learn from this video:
Circle inscribed in semicircle: r/2
Circle inscribed between two of the first mentioned circles: r/3
Circle between the first circle and the radius: r/4
This is valid for every value of r = radius of the big circle.

[ahuman]

I got into a math weighted boarding high school! We had 4 exams and their total score determined do we get in. Only 20 people can get in each year. The exams had quite hard problems and even when the now best finnish mathmetacians didnt get full points when they took them at age 15. I am very happy that I got in and this channel helped me at that. Thank you for wanting to teach others to become better.

[JLvatron]

Oh Newn! I actually got to the 2 triangles for C, but never thought of solving for both X-squared and equating them.
Great puzzle!

[ArloLipof]

You can use René Descartes’ Kissing Circles Theorem to determine a. Let’s call k_1 and k_2 the curvatures of the circles of radius 6 and k_3 the curvature of the big circle. Because of the internal/external tangency with the circle of radius a, we have: k_1 = k_2 = 1/6 and k_3 = -1/12. Thus:
k_4 = k_1 + k_2 + k_3 +2•sqrt(k_1•k_2 + k_2•k_3 + k_1•k_3)
<=> 1/a = 1/6 + 1/6 - 1/12 + 2•sqrt(1/36 - 1/72 - 1/72)
<=> 1/a = 1/4
<=> a = 4

[nomekop]

Wait, you don't need to solve for x squared to get c. By putting the two triangles up, you can see that the short legs add up to the radius of b, so c is half of b, or 3

[hxveelbadtv]

Great video Presh! The problem was really fun as well!!! 🙏

[TheEulerID]

It really should be explained why the construction of the line @2:25 passes through the centre of the smallest circle and touches the perimeter of the large circle at the same point as the small circle. That's because where the tangents of two circles touching like that are always the same, and are thus at 90 degrees to their radius. Hence that line passes through the centre of both the large and small circle (the same would be true of any two circles which are touching - a line which passes through the centre of one and their common point will always, if long enough, pass through the centre of the other).

It might seem obvious to those familiar with the sort of problems, but it ought to have been made explicit as it's absolutely critical to the solving of this question.

I solved the question in a similar way save that my two right angle triangles shared a common line which is the base of the blue triangle. It is, of course, identical. In all, a neat little problem.

I've also been trying to work out if the series of reducing radii 12, 6, 4, 3 has any deeper meaning. They go down in the ratio 2, 1.5, 1.33

[richardgratton]

I guessed 13, just eyeballing the diagram! 😂

[JasonsAccount]

I felt really smart when I got the last step a little quicker because the triangles with sides “c” and “x” are congruent which meant that c=b/2…but I needed a lot of help before that part.

[sans]

b=radius of a circle whose diameter=the radius of the big circle=12
b=12/2=6
the centerpoints of circles a and b can form a triangle with base 12-a and height 6 so our equation will look like this:
(12-a)^2+6^2=(a+6)^2
or a^2-24a+180=a^2+12a+36
-a^2 -a^2
-24a+180=12a+36
+24a +24a
180=36a+36
/36 /36 /36
5=a+1
-1 -1
a=4
lemme get c one sec
radius b extends halfway from the center to the right edge, and since circle c is taking all the space it can get above the diameter (given circle b is still there), the diameter of circle c must take up the remaining space of the diameter of the biggest circle.
=>c=3
a=4
b=6
c=3
a+b+c=4+6+3=13
a+b+c=13

[GaborRevesz_kittenhuffer]

obvs, b = 12/2 = 6. via pythagoras,

(a+6)² = 6² + (12–a)²,

so 36a = 12², i.e. a = 4. place the origin at the center of the largest circle. then the center of the upper circle is (0, 6) and that of the smallest circle is (x, c). so

|(x, c)| = 12–c and |(0, 6)–(x, c)| = c+6.

iow, x² + c² = (12–c)² and x² + (c–6)² = (c+6)².

so x² = -24c + 144 and x² = 24c. so 48c = 144. so c = 3. so a+b+c = 6+4+3 = 13.

[KannaTum]

Excuse me, what apps or programs do you use? Is it free?

[onradioactivewaves]

When you drew the last blue triangle, it should be easy to see that the he and green triangles are the same, or the b-c=c => b=2c

[ИванВ-щв]

How did you prove that the triangles are podpbnye?
And that the hypotenuse passes through the tangent of two circles? And if you conduct the hypotenuse through the tangent of two circles, then why will the radius C cathetus?
If this is so, then the equations can not be solved, but simply divide the radius B / 2 = 6/2 = 3

[salamander]

I started assuming the big circle diameter was 12 (didn't read the question carefully enough...), so got b=3 quickly. Interesting that using the same method to find a as Presh the triangle comes out as a 3 4 5 triangle! They do seem to crop up in unusual places!!

[Grizzly]

Was stuck on how to get c, until it occurred to me that the smallest circle was also tangent to the largest circle, then it clicked.

[prod_EYES]

Honestly i just looked at it and said that b=6 (1/2 the radius of large circle), a=4 (it looked 2/3 the size of the largest inscribed circle) and c=4 (it looked half the size of the said circle). Nice solution

[kokor]

It was easy to get the 'b' circles radius and find it's center to draw it.
Q1- How did you determine the centre of the second circle with 'a' radius and to draw the right triangle? How did you select where the hypotenuse (a+b) of the triangle should meet the horizontal diameter? or the length of the line (12-a)?
Q2- Samequestions for drawing the first triangle for the circle with radius 'c'

[evveliahiri]

I find a and b as you but for c I have different solution (I guess) : If we call center of biggest circle is O and b circle center point B and center of c circle is C. Then If we draw circle such as CB and CO is tangents of that circle so CB equal to CO so 12-c=b+c then 2c=6 c=3 (Sorry for my English)

[keithwood]

I like to copy and paste the starting graphic into powerpoint, then print it out and work on it. But when the entire background is black it makes it hard to work on, and also uses way too much ink. Can you use a white background?