Problem:
A circle is divided into 432 congruent arcs by 432 points. The points are colored in four colors such that some 108 points are colored Red, some 108 points are colored Green, some 108 points are colored Blue, and the remaining 108 points are colored Yellow. Prove that one can choose three points of each color in such a way that the four triangles formed by the chosen points of the same color are congruent.
Solution:
Let R,G,B,Y denote the sets of Red, Green, Blue, Yellow points, respectively, and let r,g,b,y denote a generic Red, Green, Blue, Yellow point, respectively. For integers 0≤
k≤431, let Tk denote the counterclockwise rotation of (432360k) degree around the center of the circle. Furthermore, for a set S, let ∣S∣ denote the number of elements in S.
First, we claim that there is some index i1 such that ∣Ti1(R)∩G∣≥28. Indeed, for each k, set Tk(R)∩G consists of all Green points that are the images of Red points under the rotation Tk. Hence the sum
s1=∣T0(R)∩G∣+∣T1(R)∩G∣+⋯+∣T431(R)∩G∣
is equal to the number of pairs of points (r,g) such that g=Tk(r) for some k. On the other hand, for each r and each g, there is a unique rotation Tk with Tk(r)=g, form which it follows that s1=1082=11664. Clearly, ∣T0(R)∩G∣=∣R∩G∣=0 (because R∩G=∅ ). By the Pigeonhole principle, there is some index i1 such that
∣Ti1(R)∩G∣≥⌈431s1⌉=⌈43111664⌉=⌈27.06…⌉=28
establishing our claim. Let RG denote the set Ti1(R)∩G, and let rg denote a generic point in RG.
Second, we claim that there is some index i2 such that ∣Ti2(RG)∩B∣≥8. Again, for each k, set Tk(RG)∩B consists of all Blue points that are the images of the points in RG under the rotation Tk. Hence the sum
s2=∣T0(RG)∩B∣+∣T1(RG)∩B∣+⋯+∣T431(RG)∩B∣
is equal to the number of pairs of points (rg,b) such that b=Tk(rg) for some k. On the other hand, for each rg and each b, there is a unique rotation Tk with Tk(rg)=b, form which it follows that s2≥28⋅108=3024. Clearly, RG is a subset of B, which is disjoint with B, so ∣T0(RG)∩B∣=0. Furthermore, T432−i1(Ti1) is the identity transformation, implying that T432−i1(Ti1(R))=R and T432−i1(RG) is a subset of R which is disjoint with B. In particular, ∣T432−i1(RG)∩B∣=0. By the Pigeonhole principle, there is some index i2 such that
∣Ti2(RG)∩B∣≥⌈430s2⌉≥⌈4303024⌉=⌈7.0325…⌉=8
establishing our claim. Let RGB denote the set Ti2(RG)∩B, and let rgb denote a generic point in RGB.
Third, we claim that there is some index i3 such that ∣Ti3(RGB)∩Y∣≥3. We repeated our previous process one more time. We note that
s3=∣T0(RGB)∩Y∣+∣T1(RGB)∩Y∣+⋯+∣T431(RGB)∩Y∣≥8⋅108=864
and
∣T0(RGB)∩Y∣=∣T432−i2(RGB)∩Y∣=∣T432−i2−i1(RGB)∩Y∣=0
By the Pigeonhole principle, there is some index i3 such that
∣Ti3(RGB)∩Y∣≥⌈429s3⌉≥⌈429864⌉=⌈2.01…⌉=3
establishing our claim.
Let y1,y2,y3 be three points in Ti2(RGB)∩Y. Then
(y1,y2,y3),(b1,b2,b3)=T432−i3(y1,y2,y3),(g1,g2,g3)=T432−i3−i2(y1,y2,y3),(r1,r2,r3)=T432−i3−i2−i1(y1,y2,y3)
are twelve points that we are looking for.
This problem and solution were suggested by Gregory Galperin.
The problems on this page are the property of the MAA's American Mathematics Competitions