$ABC$ is an equilateral triangle with side length $11$ units. Consider the points $P_1,P_2, \dots, P_10$ dividing segment $BC$ into $11$ parts of unit length. Similarly, define $Q_1, Q_2, \dots, Q_10$ for the side $CA$ and $R_1,R_2,\dots, R_10$ for the side $AB$. Find the number of triples $(i,j,k)$ with $i,j,k \in \{1,2,\dots,10\}$ such that the centroids of triangles $ABC$ and $P_iQ_jR_k$ coincide.
Problem
Source: RMO Maharashtra and Goa 2016, P6
Tags: counting, Mass Points, geometry
anantmudgal09
11.10.2016 10:53
Remarks: The official wording is different, but as I can't attach a diagram, this is modified. The solution remains same even if the triangle is not equilateral, simply consider an affine mapping of the two.
ThE-dArK-lOrD
11.10.2016 11:19
Let complex coordinates of $A,B,C$ be $a,b,c$. We get that complex coordinate of $P_i\in BC$ is $\frac{b(i)+c(11-i)}{11}$ for all $i=1,2,...,10$. Similarly argument also work for $Q_i,R_i$s. Note that centroids of $\triangle{ABC},\triangle{P_iQ_jR_k}$ coincide $\iff \frac{a+b+c}{3}=\frac{a(11-j+k)+b(11-k+i)+c(11-i+j)}{11\times 3}$. So, we have to count $(i,j,k)\in [10]^3$ that $11=11-k+i=11-i+j=11-j+k\iff i=j=k$. Hence, the answer is $10$.
Ankoganit
11.10.2016 11:52
anantmudgal09 wrote: Remarks: The official wording is different, but as I can't attach a diagram, this is modified. Hello, you can use this Asymptote diagram (click on it to view code): [asy][asy]pair A,B,C; pen lolcatz=fontsize(3.5); //change the fontsize in above line if needed size(6cm); //change the size of the whole figure in the above line defaultpen(lolcatz); B=(0,0); C=(1,0); A=rotate(60,B)*C; for(int i=1; i < 11; ++i) { dot(B+((C-B)/11)*i); label("${}^{\displaystyle P}$"+ string(i),B+((C-B)/11)*i,2*S); } for(int i=1; i < 11; ++i) { dot(C+((A-C)/11)*i); label("${}^{\displaystyle Q}$"+ string(i),C+((A-C)/11)*i,2*E); } for(int i=1; i < 11; ++i) { dot(A+((B-A)/11)*i); label("${}^{\displaystyle R}$"+ string(i),A+((B-A)/11)*i,2*W); } dot(A);dot(B);dot(C);draw(A--B--C--cycle);label("$A$",A,2*N);label("$B$",B,2*SW);label("$C$",C,2*SE);[/asy][/asy]
WizardMath
11.10.2016 13:06
Or what is easier, barycentric coordinates. Ankoganit wrote: anantmudgal09 wrote: Remarks: The official wording is different, but as I can't attach a diagram, this is modified. Hello, you can use this Asymptote diagram (click on it to view code): [asy][asy]pair A,B,C; pen lolcatz=fontsize(3.5); //change the fontsize in above line if needed size(6cm); //change the size of the whole figure in the above line defaultpen(lolcatz); B=(0,0); C=(1,0); A=rotate(60,B)*C; for(int i=1; i < 11; ++i) { dot(B+((C-B)/11)*i); label("${}^{\displaystyle P}$"+ string(i),B+((C-B)/11)*i,2*S); } for(int i=1; i < 11; ++i) { dot(C+((A-C)/11)*i); label("${}^{\displaystyle Q}$"+ string(i),C+((A-C)/11)*i,2*E); } for(int i=1; i < 11; ++i) { dot(A+((B-A)/11)*i); label("${}^{\displaystyle R}$"+ string(i),A+((B-A)/11)*i,2*W); } dot(A);dot(B);dot(C);draw(A--B--C--cycle);label("$A$",A,2*N);label("$B$",B,2*SW);label("$C$",C,2*SE);[/asy][/asy] You need not write the code. draw it up in Geogebra, then export the code to asymptote.
e_plus_pi
03.10.2018 22:55
This was actually a fun problem to work upon.
We place $\triangle ABC$ in the Cartesian Plane with $A = (0, \dfrac{11\sqrt{3}}{2}) , B= (\dfrac{-11}{2},0), C= (\dfrac{11}{2},0)$.
Then, by section formula,
$P_{i} = (\dfrac{11-2i}{2},0) ; Q_{i} = (\dfrac{11-i}{2}, \dfrac{\sqrt{3}i}{2}) ; R_{i} = (\dfrac{-i}{2}, \dfrac{\sqrt{3}(11-i)}{2})$.
From here, we obtain two equations after considering the centroid $G$ of $\triangle P_{i}Q_{j}R_{k}$.
(1) $\rightarrow 2i+j+k = 22$
(2) $\rightarrow j+11-k = 11$
Thus, all triangles $P_{i}Q_{11-i}R_{11-i}$ have the same centroid as $\triangle ABC$. Since there are $10$ values for $i$, the answer is $\boxed{10}$.