What is the relation between the radii of these circles?

[@JasonOvalles]

I really appreciate you making videos at a variety of difficulty levels. I always get super excited when I see a problem I can actually solve (like this one!) but also learn a lot from seeing solutions I never thought of myself.

[@kevinmartin7760]

You can certainly determine this visually just by scaling the diagram by a factor of R1/R2, using the intersection of the two common tangent lines as the center of the scaling operation. The angle between the lines will not change, so the scaled large circle 2 will coincide with the original large circle 1, and everything else remains tangent appropriately, so everything moves by one index: scaled small/large circle i coincides with original small/large circle (i-1). Thus the same scale factor relates any of the consecutive radii, the R and r values each form a geometric sequence, and the middle entry of any three consecutive values is the geometric mean of its two neighbours.

[@williamsherwood2239]

Your proof is freaking amazing man Thank you so much.

[@ABCD-bm2hs]

kzbin.info/www/bejne/gpLdfnWhetKSabs

[@goodplacetostop2973]

17:36 I’ve posted a homework on my channel! More soon in the next few days hopefully

[@SlidellRobotics]

11:16: He didn't establish that the centers of the large circles are colinear. The easiest way to do this is to show that they're all on the bisector of the angle at the lower right of the triangle. This week I revisited a problem with only the two largest circles, and asked to solve for the ratio of the height and hypotenuse of the right triangle in terms of the ratio k=R1/R2.

[@padraiggluck2980]

For any incircle in a right triangle the line segments from a common vertex (ie, from an exterior point) to the points of tangency are equal. Draw the radii from the center of the circle to the points of tangency. They meet at right angles. Therefore by SAS the two resulting triangles are congruent. Therefore the line joining the center of the circle and the vertex is the angle bisector. Since this is true for all four of the larger triangles simultaneously with respect to one of the vertices their centers are collinear.

[@SlidellRobotics]

@@padraiggluck2980 Actually, SAS does not apply here, as the right angles are not included between the congruent sides. As an example, consider SSA of 7, 5, 45 degrees. Place the common point of the two sides at the origin and the 7 unit segment along the x axis, ending at (7,0). The third vertex can be at (3,4) or (4,3), producing clearly non-congruent triangles. However, if the angle is a right angle, SSA is enough, because you can use the Pythagorean theorem to get to SSS. You can also prove SSA for obtuse angles (left as an exercise).

[@padraiggluck2980]

@@SlidellRobotics The figure is a ‘kite’. The corresponding radii are equal, the corresponding right angles are equal, and the corresponding edges from the external point are equal. SAS most certainly applies.

[@SlidellRobotics]

@@padraiggluck2980 I acknowledge it is a kite, and you can also prove it using kite-specific theorems (whose easiest proofs use the Pythagorean theorem just as I did). It's still not the SAS theorem I learned as an underclassman in high school (or find when I do a web search today, so things haven't changed THAT much in the past fourty years) because the angle has to be included between the congruent sides for SAS to apply.

[@petersievert6830]

So in other words, r_n is the geometric mean of r_n-1 and r_n+1 . This is a really nice result.

[@ABCD-bm2hs]

kzbin.info/www/bejne/gpLdfnWhetKSabs

[@demenion3521]

that is basically one of the classical ways to determine the geometric mean geometrically

[@jursamaj]

That's a lot to say the obvious: The Rn and rn series are each geometric series.

[@OrbitTheSun]

r1 : r2 = R1 : R2 simply by the self-similarity of the arrangement. Same ratio goes for r2 : r3 and R2 : R3 and so on.

[@mariochavez3834]

The last step of problem was beautiful

[@demenion3521]

so in the end you just have the same similarity relation for the small circles as for any other length in that triangle with r3/r2=r2/r1

[@s.m.m99203]

Hello all Sorry if my question is a straight forward one. On 9:57 how could we ensure that a single line segment would connect the centers of the 4 green circles? Thank you (Edit) Oops yes. They all lie on the same bisector line (same distances from the two sides for each point)

[@ABCD-bm2hs]

kzbin.info/www/bejne/gpLdfnWhetKSabs

[@cycklist]

Is it trivial to know that the line connecting the four centres is straight?

[@perrydimes6915]

I don't think it is

[@wtrmute]

Yes, because the circles are all inscribed in the right triangle, so their centres lie on a straight line connecting the rightmost vertex of the triangle and the leftmost side, which is the bisector of that angle. That is because at every point in that bisector, the straight-line distance to either side of the triangle will be the same (that is, you can inscribe a circle tangent to both sides with centre on that bisector).

[@perrydimes6915]

@@wtrmute very clear explanation, but that is not the same thing as being trivial!

[@Grizzly01]

@@perrydimes6915 It's a fairly fundamental circle theorem

[@deepjyoti5610]

@@perrydimes6915 it is trivial in olympiads,and genral too

[@tahasami597]

Thank for Michael penn

[@ABCD-bm2hs]

kzbin.info/www/bejne/gpLdfnWhetKSabs

[@notacow69]

One thing that occurred me while watching this is that \frac{R_{n+1}}{R_n} has to be less than \frac{1}{n}, for sufficiently large n; otherwise, the sum of the lengths of the radii would be equal or greater than the harmonic series, which diverges. I'm also thinking that one can use field theory and the fact that this is a constructible figure to prove that the ratio between consecutive radii has to be expressible in terms of the constructible numbers. I don't think this forces the relationship to be that of a geometric series; they're just thoughts I had while watching the setup. It certainly led me to conclude almost immediately that the relationship was very likely to be a geometric series.

[@stephenhousman6975]

Kinda interesting that r_2 is the geometric mean of the other two. Visually that does make sense.

[@CM63_France]

Hi, Very nice! The result is not obvious. 13:26 : you can say that directly, because R1, R2, R3, R4 follow a geometrical progression. (and we can guess this will also be the case of r1, r2, r3, but with a different rate)(1) 13:57 : same comment. For fun: 10:00 : "ok, great". (1) : PS : so the relationship between R1, R2, R3 is the same as that between r1, r2, r3, and you get the result.

[@emoore06905]

I missed this - where did you prove the colinearness of the centers of the four circles?

[@emoore06905]

Are they always on the angle bisector?

[@Grizzly01]

He didn't prove it, as it's a pretty fundamental circle theorem. Just as he didn't need to prove Pythagoras' Theorem.

[@ABCD-bm2hs]

kzbin.info/www/bejne/gpLdfnWhetKSabs

[@Grizzly01]

@@ABCD-bm2hs substandard MYD rip-off. Spam. Nothing to do with the subject at hand. Fail.

[@u.v.s.5583]

@@emoore06905 You basically had answered your own question. This is indeed the key observation. You can make a symmetry argument - where else could the centers lie if not on the bisector? Those circles have got to be right in the middle between the sides of the triangle if they touch both sides in one point, right?

[@quadruplay9788]

what is the relationship between a and the angles of the triangle?

[@rorydaulton6858]

Perhaps I am missing something, but this problem (the relationship between r1, r2, and r3) seems trivial to me. If you do a dilation transformation with its center at the lower-right corner of the outer right triangle, we can map any of those small circles onto another of those small circles. I.e. the same transformation maps the first small circle onto the second and the second one onto the third. That means those radii are in proportion: r1/r2 = r2/r3. (That ratio is the constant for the dilation transformation) And that is equivalent to your final answer. It is pretty easy to prove my stated results of the dilation transformations, so why bother with all the rest of this video? (Perhaps you do not like transformation geometry?)

[@anshumanagrawal346]

I don't even understand many of the words you said, so I'm pretty sure it's not trivial XD

[@Zarunias]

Yes, I had the same thoughts. After seeing the illustration I started by seeing that r1/r2=r2/r3. After that I read the question. And then I was confused.

[@bsmith6276]

I had the same line of thought. Why go through all this mass of geometrically derived equations when elementary transformation geometry basically tells you r1/r2=r2/r3 almost immediately.

[@artsmith1347]

I'm with Anshuman Agrawal. Perhaps the first video on your channel will illustrate dilation transformation within the context of transformation geometry. I am not sure what I just said. Did I swear in a foreign language?

[@anshumanagrawal346]

All jokes aside, what is dilation transformation? Is it something a high school maths student can understand?

[@daniellopez3124]

As always, very didactic the explanation and the result, at least for me very unexpected. Great job.

[@ABCD-bm2hs]

kzbin.info/www/bejne/gpLdfnWhetKSabs

[@sp_danger1729]

Could this problem be solved using inversion?

[@anushrao882]

It is obvious by homothety

[@sc0utop547]

It was quite lengthy way,it could have been done in much smaller time

[@ABCD-bm2hs]

kzbin.info/www/bejne/gpLdfnWhetKSabs

[@mikegallegos7]

How come -R1 on left of = doesn't cancel with +R1 on right of = ? Thx

[@SparkDragon42]

because it's not on the same side of the =, if you use numbers instead of R1, saying that they cancel means : a+3=b-3 a=b and if you don't see the issue yet let's continue with this : 2+3=8-3 2=8 do you see the issue now ? to cancel + and - they have to be on the same side of the =, if you want to cancel things on opposite sides of the = they need to be either both positive or both negative

[@ABCD-bm2hs]

kzbin.info/www/bejne/gpLdfnWhetKSabs

[@tylerl.8394]

The vertical side of the triangle simply doesn’t matter here. What matters is the lower right angle and the ratio could be written out in terms of that angle

[@Sgrunterundt]

Why is this video so long? Similar figures means that r_n will be an exponentially decreasing series. Thus r_1/r_2 = r_2/r_3 or r_1*r_3 = (r_2)^2

[@tonyhaddad1394]

Congratualtion mst michael no mistakes 😁 good job

[@ABCD-bm2hs]

kzbin.info/www/bejne/gpLdfnWhetKSabs

[@keymasta3260]

3:23 oh no, not capital b but small b :)

[@ofekn]

you can square it and get r1r3=r2^2 also it make sense because R1R3=R2^2

[@jursamaj]

Yes, both series are geometric series, as was obvious by inspection.

[@ABCD-bm2hs]

kzbin.info/www/bejne/gpLdfnWhetKSabs

[@srivatsav9817]

Hi Sir, thanks for making this video😀😀🙂

[@mikegallegos7]

dang trigonometric heaven, at last lol

[@Qermaq]

I think you said "Artoo" more than C3PO did ;)

[@u.v.s.5583]

You mean FarToo and Creepio?

[@e-pluszak9419]

Exercise for the viewer: prove the final identity in less than 17 minutes Hint: . . . . . . . . . . . . . . . . . . . . . . . . Homothety centred at right corner

[@ryanb1874]

1/4 of angle theta that desctibes R R2 R3. Guessing to lazy on saturday.

[@heliocentric1756]

Fractal.

[@ABCD-bm2hs]

kzbin.info/www/bejne/gpLdfnWhetKSabs

[@u.v.s.5583]

I don't think this is fractal. This is more like a geometric series.