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The diagram shows an equilateral triangle. Circles of the same color are congruent. Wherever things look tangent, they are tangent. The green circles are vertically aligned.

enter image description here

Which circles are larger, the green or red?

(There is an intuitive explanation.)

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3 Answers 3

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The

reds

are bigger.

To easily see that

just mirror the figure at the left (right also works) of the triangle's sides and add an auxiliary green circle at the top. This auxiliary circle forms a regular triangle with the centre red circle and its mirror image of side length exactly 5 green radii and one red radius (measuring from centre to centre). As this distance is also spanned by 4 reds their radii can only be larger or the same if the 4 reds form a straight line. But the latter cannot be because then the greys (the left and its mirror image) could not fit. (Each would have to sit precisely in the middle between the two reds it touches. Because the greys also touch each other their radius would have to be exactly double the red. To see this is not possible insert a grey circle or double radius green (or red) circle at the top of the triangle. The grey, by 120° symmetry, will touch the other two. The double green will not because it passes through the centres of the three red and the outer green.)

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    $\begingroup$ That's very nice! Could you please explain the last sentence a bit more? $\endgroup$ Commented Feb 13 at 20:34
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    $\begingroup$ Sorry, I still don't follow. rot13[V gubhtug lbh jbhyq nethr gur sbyybjvat: va gur fvghngvba jurer erq naq terra pvepyrf unir fnzr enqvhf, gur enqvhf bs terl pvepyr fubhyq or gjvpr gung (nf lbh nethrq), ohg gura gur gjb terl pvepyrf vafvqr gur gevnatyr pnaabg gbhpu (ol fbzr fvzcyr pnyphyngvba).] However, your current last sentence says something else. Perhaps I am missing something. $\endgroup$ Commented Feb 13 at 21:30
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    $\begingroup$ rot13(Jul qbrf gur qbhoyr terra pvepyr cnff guebhtu gubfr pragref? Gung’f gur pnfr bayl vs vg’f pbapragevp jvgu gur vaare terra pvepyr. Ohg V qba’g frr jul gung’f gur pnfr.) $\endgroup$ Commented Feb 14 at 9:42
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    $\begingroup$ @Dan Please note that the first part of the answer settles most of the question by showing red >= green. THe grey circles are only needed to rule out red==green, so we can use red and green interchangeably i nthis branch. $\endgroup$ Commented Feb 14 at 18:41
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    $\begingroup$ @Pranay Because the centre of a circle inscribed in a 60° angle is two radii from the cusp of the angle. $\endgroup$ Commented Feb 14 at 18:48
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The larger circles are ...

red

Because

If the red and green circles were the same size, you could move the red side circles up to hug the side, the red middle circle and the green circle above. The line connecting the red circle centers would be perpendicular to the triangle side because because all circles centers would lie on a equilateral triangular grid.

Therefore

If from there you moved the red side circles down touching the triangle side and a grey circle, that would open a gap between the lateral circles and the center red circle. Still assuming equal sizes.

This shows that to make all circles touch as in the picture, you need to increase the red circles slightly and reduce the green circles.

Note: this assumes the picture is accurate in the sense that if the red circles are placed on the grey circles, and the green circles are placed above, then there is indeed a gap between the red and the green circles.

Complement

Why is it that with equally sized circles the red circle can be moved to touch both circles and the triangle?

You can picture the bottom green circle with 6 circles tightly packed around it, the top green one and 5 more circles of the same size. The top, bottom-left and bottom-right of these circles lie on an equilateral triangle parallel to the large triangle. Since the top circle (also green) is tangent to the triangle, the two others should also be.

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    $\begingroup$ rot13(Jura gur erq naq terra pvepyrf unir gur fnzr enqvhf, naq gurl nyy gbhpu rnpu bgure nf qrfpevorq va gur svefg fcbvyre, jul qb gur yngreny erq pvepyrf abg bireync jvgu gur terl pvepyrf?) $\endgroup$ Commented Feb 11 at 22:19
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    $\begingroup$ @Pranay Good question. Addressing that question will make Florian's answer complete. $\endgroup$ Commented Feb 12 at 3:26
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    $\begingroup$ @Dan. I know how to address that issue but not sure if it’s worth writing up an answer when the rest of it is similar to Florian’s answer. $\endgroup$ Commented Feb 12 at 4:48
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    $\begingroup$ I don’t think the update addresses my question. My question was, when the red and green circles have the same size, why do the lateral red circles not overlap with the grey circles? This is important because in the next step you slide the lateral red circles down until they touch the grey circles. This assumes that there was a gap between them to begin with. $\endgroup$ Commented Feb 13 at 14:03
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    $\begingroup$ @FlorianF I agree with Pranay. There is a nice way to resolve the issue. Here is a hint, which refers to this diagram. (rot13) Vs jr qenj n pvepyr gnatrag gb gur ybjrfg terra pvepyr, gur evtug terra pvepyr, naq gur evtug fvqr bs gur gevnatyr, jr pna rnfvyl fubj guvf arj pvepyr jvyy abg gbhpu be vagrefrpg gur gevnatyr'f iregvpny nkvf bs flzzrgel. Gura... $\endgroup$ Commented Feb 13 at 15:16
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Aha! So easy! If we observe very very very carefully, we will find the red circle is 0.0820 times the side length of triangle, and the green circle is 0.0818 times.

enter image description here

Supplementary: The picture is drawn by pure compass-and-straightedge construction. The green circles are easy to find, and the red circles are some harder Apollonius tangent problem. There is a crucial point there, similar to Florian F 's argument, that we have to notice double-line H1-(tangent point)-N1-P1 is longer than straight line H1-P1, where they would be the same line if assuming red and green circle are equally large. So if they were equally large circle N1 cannot be tangent to circle E. We want to make circle H and N1 larger, and then follows Florian F's answer.

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  • $\begingroup$ This is not an "intuitive explanation". Math puzzle answers are generally expected to engage in the question as a puzzle (e.g. "ahah!" moments, elegant logic) instead of as a problem (e.g. case-bashing, detailed calculations) $\endgroup$ Commented Feb 14 at 17:56
  • $\begingroup$ I think Florian F's reasoning is correct. But still brute force is good too! $\endgroup$ Commented Feb 14 at 18:00
  • $\begingroup$ @bobble I did this by pure compass-and-straightedge construction. Then it can be solved, equivalently, by a series of quadratic equations. I just let Geogebra do the calculation. "observe" is a math joke here. $\endgroup$ Commented Feb 14 at 18:04
  • $\begingroup$ Ah I understand you now. I thought this was a math problem, brute force calculation is for math but not puzzles, I admit... $\endgroup$ Commented Feb 14 at 18:07

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