Can You Find the Radius of the Small Yellow Circle? | Step-by-Step Explanation

[andrewjames]

There's a mistake in the decimal point : (4 + 2sqrt2)e2 = 46.627 Anyway, I enjoy your explanations - I'm 80 and it keeps the grey cells busy!

[princejag]

My day never completes without watching your

[PiyushSharmaMotivation]

There is a slight mistake in the points while calculating.. Instead of 466.27 it should be 46.62

[tylerwright]

One of the best geometric problems you have presented. Thank you so much!

[اممدنحمظ]

تمرين جميل شرح مفصل مرتب . وبارك الله فيكم وعليكم والله يحفظكم ويرعاكم ويحميكم وينصركم جميعا . شكرا جزيلا لكم د تحياتنا لكم من غزة فلسطين .

[kirin]

Decimal point is off by 1 on th step before the end, but corrected on final result.

[p.m.]

Solving this problem with law of cosines for triangle NPM is also really convenient, you calculate cosine of angle NPM using trigonometric equations for finding cos(180-a-b), where cos and sin of angles "a" and "b" are easy to calculate using two rectangular triangles of NXP and MYP where points X and Y are below circles' centres and lay in a distance of respectively 2-r and 4-r to points N and M. Then you plug calculated cosine into law of cosines for traingle NPM and end up with equation of one unkown.

[edziobydlak]

Your method can be used only if the radius of the blue circle is twice as big as the radius of the red circle (the tangent needs to be parallel to the base line). There is a similar method that works for all circles.

[rayrash]

In one of your videos, you had established that the distance between their contacts with the common tangent of two osculating circles is d = 2 \sqrt r_1 \times r_2, where the rs are the radii of the two circles. Applying this directly to this problem thrice leads to your answer. If r is the radius of the small circle, x the distance between the points of contact of the radius 2 circle and the small circle, X that been the points of contact of the small circle and the one with radius 4, we get x = 2 \sqrt (2r), X = 2 \sqrt (4r) and x+X = 2 \sqrt (8). This directly gives r = 0.686.

[alokekumar]

wonderfully explained with the help of simple geometry and algebra thank you very much

[satyapalsingh]

If radii are a, b&c (a<b<c)then use 1/sqrt a=1/sqrt b +1/sqrt c. Now 1/sqrt a=1/sqrt 2 +1/2.Now

[waeelrihan]

The basic idea is great
Make the radius whose value you want to know one of the values in one of the sides of a right-angled triangle
You are a genius, man

[tanyaerskine]

That was really cool to watch. Thank you!

[wwoods]

15:00 You could have taken it a little further before pulling out the calculator: r = 12 - 8√2

[larswilms]

For the right angle triangles you constructed and calculating AB and BC and FM, it helps to rewrite the pythagorean theorem from a^2 +b^2 = c^2 to a^2 = c^2 - b^2. You now have a difference of squares on one side of the equation. This can be written as a^2 = (c-b)(c+b). If c and b are the known adjacent and hypotenuse, like in this example, then the length of the remaining adjacent side is easily determined.

the final answer can be simplified further:
(4 + 2sqrt(2))^2 * R = 32 expand the square
(16 + 2 * 4 * 2 sqrt(2) + 8) R = 32 divide by 8
((2+1) + 2 sqrt(2)) R = 4 divide by (3 + 2 sqrt(2))
R = 4 / (3 + 2 sqrt(2)) multiple by 1 as (3 - 2 sqrt(2)) / (3 - 2 sqrt(2))
R = 4 (3 - 2 sqrt(2)) / ((3 + 2 sqrt2)(3 - 2 sqrt(2)) the denominator is now of the form (A+B)(A-B) = A^2 - B^2 (3 + 2 sqrt2)(3 - 2 sqrt(2) = 9 - 2*4 = 1
R = 4 (3 - 2 sqrt(2)) = 12 - 8 sqrt(2)
R is approximately 0.686

[pedroreis]

Very nice! But It should have been prooved that F, N and A are collinear and therefore you could say that |FM|=|AC|, otherwise, It can't be assumed just by the image. In fact they are collinear because the height in relation to AC of 2 times the radius of the left circle is exactly the radius of the right circle, then when you connect with a perpendicular line (tangent) that line will be parallel to AC and almost |AC|=|FM|

[kutyuvadi]

(4+2sqr(2))^2=46.627 and not 466.27, but the final result is OK. The explanation is correct, although a little too slow. Thanks.

[timothybrown]

Looking at the question gave me anxiety, weird, right? Watching him work through it was satisfying. So much learned in school, stored away and forgotten about.

[deenulazarus]

Still got the result... Loved the explanation... Couldn't stop but wonder how people in olden days used to calculate by hand ... Measuring the sides, lengths and angles

[ROCCOANDROXY]

I generalized the problem using R1 and R2 for the radii of the two larger circles and r for the radius of the smaller inner circle. Your problem was a special case where say R2 = 2R1. Using the Pythagorean theorem and the law of cosines I obtained:
r = (R1 * R2)/(R1 + R2 + 2 * sqrt(R1 * R2))