Assume three circles mutually outside each other with the property that every line separating two of them have intersection with the interior of the third one. Prove that the sum of pairwise distances between their centers is at most $2\sqrt{2}$ times the sum of their radii. (A line separates two circles, whenever the circles do not have intersection with the line and are on different sides of it.) Note. Weaker results with $2\sqrt{2}$ replaced by some other $c$ may be awarded points depending on the value of $c>2\sqrt{2}$ Proposed by Morteza Saghafian
Problem
Source: 7th Iranian Geometry Olympiad (Advanced) P3
Tags: geometry, IGO, iranian geometry olympiad
MathsLion
04.11.2020 12:46
I have proved the claim for c=4 on the competition. Does anyone know will it be enough for some points?
Eliot
05.11.2020 16:53
Here is a kindly BUMP
CyclicConcaveTriangle
12.11.2020 18:12
bump this
Mr.C
12.11.2020 18:47
okay just want to add saying the solution here without a diagram is kinda hard (atleast for me) so i will just write some steps of my solution. .
you may change the place of circles untill there exists one line tangent to all $3$ circles
you may change the place of circles again untill there exists $2$ lines tangent to all $3$ circles
now see the shape a triangle with the circles $I_B,I_C$ drawn and another circle with its center on the line $I,I_A$
try to make the line $I,I_A$ better (i mean for example the last circle cant have its center at $I$ try to have the actuall segment which the center moves on
now prove that if we know the bounding is actually true for the third circle at the points $A,B$ then it is true for all the point on the segment $AB$.
now you only need to solve for $2$ points . so calculate like your life depends on it !!!
M.4096
13.11.2020 11:54
suppose O1 , O2 , O3 are centers of these three circles of radius R1 , R2 , R3 respectively. Let AB and CD be common internal tangents of O1 and O2 such that A , D are on circle O1. WLOG , suppose that D , B , O3 are in the same side of line O1O2. Let AB , CD and O1O2 intersect at point A3. O3A3D + O3A3B ≥ DA3B > 90 WLOG , we can consider O3A3D > 45 Let perpendicular line from O3 to A3D , intersect A3D at point Q , O3Q = O3A3. sin ( O3A3D ) > O3A3.√(1/2) Then we know that R3 ≥ O3Q > O3A3.√(1/2) in the same way , we have : R1 > O1A1.√(1/2) R2 > O2A2.√(1/2) R3 > O3A3.√(1/2) Then : R1 + R2 + R3 > √(1/2)(O1A1 + O2A2 + O3A3) 2√2 (R1 + R2 + R3) > 2 (O1A1 + O2A2 + O3A3) O1A3/A3O2 . O2A1/A1O3 . O3A2/A2O1 = R1/R2 . R2/R3 . R3/R1 = 1 Then by Ceva's theorem , O3A3 , O2A2 and O1A1 are concurrent and we have : 2 (O1A1 + O2A2 + O3A3) > O1O2 + O2O3 + O1O3 and we are done.
Modesti
14.12.2020 05:51
M.4096 wrote: O3A3D + O3A3B ≥ DA3B > 90 Sorry if I'm being dumb, but why is $\angle DA_3B>90^{\circ}$?
Emo916math
05.06.2021 19:46
What is the equality case?
NakanoItsuki26
25.07.2021 05:11
Sorry guys but can anyone present a whole perfect solution with image (using Latex) :3 'Cause this one is interesting but I've not seen anyone solved this yet