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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 1i<jn1 \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[i]. C_k = x_k + y_k i \quad \text{and} \quad M = \{C_1, \dots, C_n\} \subset \mathbb{Z}[i].
Let pp be an arbitrary prime divisor of dd. Let us show that there is a rotation of the RR plane such that RMpZ[i]RM \subset p\mathbb{Z}[i], that is, R(1pM)Z[i]R(\frac{1}{p}M) \subset \mathbb{Z}[i].
We have p2N(z)p^2 \mid N(z) and p2N(zu)p^2 \mid N(z-u) for any z,uMz, u \in M, where N(a+bi)=a2+b2N(a+bi) = a^2 + b^2 means norm. If pzp \mid z for any zMz \in M, then everything is proven. Suppose that there exists vMv \in M such that pvp \nmid v. Then it can be shown that there exists πZ[i]\pi \in \mathbb{Z}[i] such that N(π)=pN(\pi) = p and π2v\pi^2 \mid v. Then for any uMu \in M:
- either pup \nmid u, whence π2u\pi^2 \mid u,
- or pup \mid u and pvup \nmid v-u, from which π2vu\pi^2 \mid v-u and again π2u\pi^2 \mid u.
Apply the following plane transformation, which is a rotation, since N(π/π)=1N(\pi/\pi) = 1:
Rπ:x+yiπˉπ(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 RπMpZ[i]R_{\pi}M \subset p\mathbb{Z}[i] as required.
Let us now consider a decomposition of the number d=p1p2pmd = 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[i]R_1(\frac{1}{p_1}M) \subset \mathbb{Z}[i]. Similarly, there is a rotation R2R_2 such that R2(R1(1p1p2M))Z[i]R_2(R_1(\frac{1}{p_1p_2}M)) \subset \mathbb{Z}[i], and so on, there is a rotation RmR_m such that
Rm(R2(R1(1p1p2pmM)))Z[i]. 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}[i].
Since 1p1p2pmM={A1,,An}\frac{1}{p_1p_2\cdots p_m}M = \{A_1, \dots, A_n\}, the composition of rotations RmR2R1R_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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