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Balti Tee 2021 · Ülesanne 4

Algebra

Let Γ\Gamma be a circle in the plane and SS be a point on Γ\Gamma. Mario and Luigi drive around the circle Γ\Gamma with their go-karts. They both start at SS at the same time. They both drive for exactly 6 minutes at constant speed counterclockwise around the track. During these 6 minutes, Luigi makes exactly one lap around Γ\Gamma while Mario, who is three times as fast, makes three laps.

While Mario and Luigi drive their go-karts, Princess Daisy positions herself such that she is always exactly in the middle of the chord between them. When she reaches a point she has already visited, she marks it with a banana.

How many points in the plane, apart from SS, are marked with a banana by the end of the race?

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Without loss of generality, we assume that Γ\Gamma is the unit circle and S=(1,0)S = (1, 0). Three points are marked with bananas: (i) After 45 seconds, Luigi has passed through an arc with a subtended angle of 45∘45^\circ and is at the point (2/2,2/2)(\sqrt{2}/2, \sqrt{2}/2), whereas Mario has passed through an arc with a subtended angle of 135∘135^\circ and is at the point (−2/2,2/2)(-\sqrt{2}/2, \sqrt{2}/2). Therefore Daisy is at the point (0,2/2)(0, \sqrt{2}/2) after 45 seconds. After 135 seconds, Mario and Luigi's positions are exactly the other way round, so the princess is again at the point (0,2/2)(0, \sqrt{2}/2) and puts a banana there. (ii) Similarly, after 225 seconds and after 315 seconds, Princess Daisy is at the point (0,−2/2)(0, -\sqrt{2}/2) and puts a banana there. (iii) After 90 seconds, Luigi is at (0,1)(0, 1) and Mario at (0,−1)(0, -1), so that Daisy is at the origin of the plane. After 270 seconds, Mario and Luigi's positions are exactly the other way round, hence Princess Daisy drops a banana at the point (0,0)(0, 0). We claim that no other point in the plane, apart from these three points and SS, is marked with a banana. Let t1t_1 and t2t_2 be two different times when Daisy is at the same place. For n∈{1,2}n \in \{1, 2\} we write Luigi's position at time tnt_n as a complex number zn=exp⁡(ixn)z_n = \exp(ix_n) with xn∈]0,2π[x_n \in ]0, 2\pi[. At this time, Mario is located at zi3z_i^3 and Daisy at (zi3+zj)/2(z_i^3 + z_j)/2.

According to our assumption we have (z13+z1)/2=(z23+z2)/2(z_1^3 + z_1)/2 = (z_2^3 + z_2)/2 or, equivalently, (z1−z2)(z12+z1z2+z22+1)=0(z_1 - z_2)(z_1^2 + z_1z_2 + z_2^2 + 1) = 0. We have z1≠z2z_1 \neq z_2, so that we must have z12+z1z2+z22=−1z_1^2 + z_1z_2 + z_2^2 = -1. We proceed with an observation of the structure of Γ\Gamma as a set of complex numbers. Suppose that z∈Γ∖{S}z \in \Gamma \setminus \{S\}. Then z+1+z−1∈Γz+1+z^{-1} \in \Gamma if and only if z∈{i,−i,−i,i}z \in \{i, -i, -i, i\}. For a proof of the observation note that z+1+z−1=z+1+zˉz+1+z^{-1} = z+1+\bar{z} is a real number for every z∈Cz \in \mathbb{C} with norm ∣z∣=1|z| = 1. So it lies on the unit circle if and only if it is equal to 1, in which case the real part of zz is equal to 0, or it is equal to −1-1, in which case the real part of zz is equal to −1-1. We apply the observation to the number z=z1/z2z = z_1/z_2, which satisfies the premise since z+1+zˉ=−z1ˉ⋅z2ˉ∈Γz+1+\bar{z} = -\bar{z_1} \cdot \bar{z_2} \in \Gamma. Therefore, one of the following cases must occur. (i) We have z=±iz = \pm i, that is, z1=±iz2z_1 = \pm iz_2. Without loss of generality we may assume z2=iz1z_2 = iz_1. It follows that −1=z12+z1z2+z22=iz12-1 = z_1^2 + z_1z_2 + z_2^2 = iz_1^2, so that z1=exp⁡(iπ/4)z_1 = \exp(i\pi/4) or z1=exp⁡(5iπ/4)z_1 = \exp(5i\pi/4). In the former case (z1,z2)=(exp⁡(iπ/4),exp⁡(3iπ/4))(z_1, z_2) = (\exp(i\pi/4), \exp(3i\pi/4)), which matches case (1) above. In the latter case (z1,z2)=(exp⁡(5iπ/4),exp⁡(7iπ/4))(z_1, z_2) = (\exp(5i\pi/4), \exp(7i\pi/4)), which matches case (2) above. (ii) We have z=−1z = -1, that is, z2=−z1z_2 = -z_1. It follows that −1=z12+z1z2+z22=z12-1 = z_1^2 + z_1z_2 + z_2^2 = z_1^2, so that z1=iz_1 = i or z1=−iz_1 = -i. This matches case (3) above.

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