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Selection Tests of the Belarusian Team to the IMO

Geometry Difficulty 8.4 Shortlist Prove it Belarus

There are nn points A1,…,AnA_1, \dots, A_n on the plane with rational coordinates, all pairwise distances between them being integer. Prove that there exist points B1,…,BnB_1, \dots, B_n on the plane with integer coordinates such that ∣BiBj∣=∣AiAj∣|B_iB_j| = |A_iA_j| for all 1≤i<j≤n1 \le i < j \le n.

Solution

Let dd be the least common multiple of the denominators of all coordinates of the points A1,…,AnA_1, \dots, A_n. After homothety with center at the origin OO and coefficient dd, these points will go to points C1,…,CnC_1, \dots, C_n, respectively, with integer coordinates. Let us switch to complex numbers and work with Gaussian integers. Let us introduce the following notation:
Ck=xk+ykiandM={C1,…,Cn}⊂Z∗. C_k = x_k + y_k i \quad \text{and} \quad M = \{C_1, \dots, C_n\} \subset \mathbb{Z}*.
Let pp be an arbitrary prime divisor of dd. Let us show that there is a rotation of the RR plane such that RM⊂pZ∗RM \subset p\mathbb{Z}*, that is, R(1pM)⊂Z∗R(\frac{1}{p}M) \subset \mathbb{Z}*.
We have p2∣N(z)p^2 \mid N(z) and p2∣N(z−u)p^2 \mid N(z-u) for any z,u∈Mz, u \in M, where N(a+bi)=a2+b2N(a+bi) = a^2 + b^2 means norm. If p∣zp \mid z for any z∈Mz \in M, then everything is proven. Suppose that there exists v∈Mv \in M such that p∤vp \nmid v. Then it can be shown that there exists π∈Z∗\pi \in \mathbb{Z}* such that N(π)=pN(\pi) = p and π2∣v\pi^2 \mid v. Then for any u∈Mu \in M:
- either p∤up \nmid u, whence π2∣u\pi^2 \mid u,
- or p∣up \mid u and p∤v−up \nmid v-u, from which π2∣v−u\pi^2 \mid v-u and again π2∣u\pi^2 \mid u.
Apply the following plane transformation, which is a rotation, since N(π/π)=1N(\pi/\pi) = 1:
Rπ:x+yi→πˉπ(x+yi), R_{\pi}: x + yi \rightarrow \frac{\bar{\pi}}{\pi}(x + yi),
Each number from MM can be represented as π2(a+bi)\pi^2(a + bi), which is sent by the rotation RπR_{\pi} to the number πˉπ(a+bi)=p(a+bi)\bar{\pi}\pi(a + bi) = p(a + bi). Thus, we get RπM⊂pZ∗R_{\pi}M \subset p\mathbb{Z}* as required.
Let us now consider a decomposition of the number d=p1p2⋯pmd = p_1p_2 \cdots p_m into prime factors (not necessarily different). According to what has been proven, there is a rotation R1R_1 such that R1(1p1M)⊂Z∗R_1(\frac{1}{p_1}M) \subset \mathbb{Z}*. Similarly, there is a rotation R2R_2 such that R2(R1(1p1p2M))⊂Z∗R_2(R_1(\frac{1}{p_1p_2}M)) \subset \mathbb{Z}*, and so on, there is a rotation RmR_m such that
Rm(…R2(R1(1p1p2⋯pmM)))⊂Z∗. R_m \left( \dots R_2 \left( R_1 \left( \frac{1}{p_1 p_2 \cdots p_m} M \right) \right) \right) \subset \mathbb{Z}*.
Since 1p1p2⋯pmM={A1,…,An}\frac{1}{p_1p_2\cdots p_m}M = \{A_1, \dots, A_n\}, the composition of rotations Rm∘⋯∘R2∘R1R_m \circ \dots \circ R_2 \circ R_1 sends the set A1,…,AnA_1, \dots, A_n to the set of integer points B1,…,BnB_1, \dots, B_n with the required properties.

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