Problem: Given a tetrahedron ABCD, let D1 be the centroid ò the triangle ABC and let A1,B1,C1 be the intersection points of the lines parallel to DD1 and passing through the points A,B,C with the opposite faces of the tetrahedron. Prove that the volume of the tetrahedron ABCD is one-third the volume of the tetrahedron A1B1C1D1. Does the result remain true if the point D1 is replaced with any point inside the triangle ABC?
Solution: We shall prove that the statement is valid in the general case, for an arbitrary point D1 inside $\triangle{ABC}$. Since D1 belongs to the plane ABC, there are real numbers a,b,c such that $(a+b+c)\vec{DD_1}=a\vec{DA}+b\vec{DB}+c\vec{DC}$. Since $AA_\parallel DD_1$, it holds that $\vec{AA_1}=k\vec{DD_1}$ for some $k\in \mathbb{R}$. Now it is easy to get $\vec{DA_1}=-\frac{b\vec{DA}+c\vec{DC}}{a},\vec{DB_1}=-\frac{a\vec{DA}+c\vec{DC}}{b}$, and $\vec{DC_1}=-\frac{a\vec{DA}+b\vec{DB}}{c}$. This implies: $\vec{D_1A_1}=-\frac{a^2\vec{DA}+b(a+2b+c)\vec{DB}+c(a+b+2c)\vec{DC}}{a(a+b+c)},\vec{D_1B_1}=-\frac{a(2a+b+c)\vec{DA}+b^2\vec{DB}+c(a+b+2c)\vec{DC}}{b(a+b+c)}$, and $\vec{D_1C_1}=-\frac{a(2a+b+c)\vec{DA}+b(a+2b+c)\vec{DB}+c^2\vec{DC}}{c(a+b+c)}$.
By using $6V_{D_1A_1B_1C_1}=\big |[\vec{D_1A_1},\vec{D_1B_1},\vec{D_1C_1}]\big |$ and $6V_{DABC}=\big |[\vec{DA},\vec{DB},\vec{DC}]\big |$ we get
$V_{D_1A_1B_1C_1}=\frac{S}{6abc(a+b+c)^3}=3V_{DABC}$, where $S=\begin{align*}\begin{vmatrix} a^2&b(a+2b+c)&c(a+b+2c)\\ a(2a+b+c)&b^2&c(a+b+2c)\\ a(2a+b+c)& b(a+2b+c)&c^2\end{vmatrix}\end{align*}$.
Thứ Bảy, 23 tháng 3, 2019
IMO 1964 - Problem 5
Problem: Five points are given in the plane. Among the lines that connect these five points, no two coincide and no two are parallel or perpendicular. Through each point we construct an altitude to each of the other lines. What is the maximal number of intersection points of these altitudes (excluding the initial five points)>
Solution: Let us first compute the number of intersection points of the perpendiculars passing through two distinct points B and C. The perpendiculars from B to the lines through C other than BC meet all perpendiculars from C, which counts to 3.6=18 intersection points. Each perpendicular from B to the 3 lines not containing C can intersect at most 5 of the perpendicular passing through C, which counts 3.5=15 intersection points. Thus there are 18+15=33 intersection points corresponding to B,C.
It follows that the required total number is at most 10.33=330. But some of these points, namely the orthcenters of the triangles with vertices at the given points, namely the orthocenters of the triangles with vertices at the given points, are counted thrice. There are 10 such points. Hence the maximal number of intersection points is 330-2.10=310.
Remark. The jury considered only the combinaorial part of the problem and didn't require an example in which 310 points appear. However, it is "easily" verified that, for instance, the set of points A(1,1),B(e,$\pi$),C($e^2,\pi^2$,D($e^3,\pi^3$),E($e^4,\pi^4$)) works.
Solution: Let us first compute the number of intersection points of the perpendiculars passing through two distinct points B and C. The perpendiculars from B to the lines through C other than BC meet all perpendiculars from C, which counts to 3.6=18 intersection points. Each perpendicular from B to the 3 lines not containing C can intersect at most 5 of the perpendicular passing through C, which counts 3.5=15 intersection points. Thus there are 18+15=33 intersection points corresponding to B,C.
It follows that the required total number is at most 10.33=330. But some of these points, namely the orthcenters of the triangles with vertices at the given points, namely the orthocenters of the triangles with vertices at the given points, are counted thrice. There are 10 such points. Hence the maximal number of intersection points is 330-2.10=310.
Remark. The jury considered only the combinaorial part of the problem and didn't require an example in which 310 points appear. However, it is "easily" verified that, for instance, the set of points A(1,1),B(e,$\pi$),C($e^2,\pi^2$,D($e^3,\pi^3$),E($e^4,\pi^4$)) works.
Thứ Sáu, 22 tháng 3, 2019
IMO 1964 - Problem 4
Problem: Each of 17 students talked with every other student. They all talked about three different topics. Each pair of students talked about one topic. Prove that there are three students that talked about the same topic among themselves.
Solution: Let us call the topics T1,T2,T3. Consider an aarbitrary student A. By the pigeonhole principle there is a topic, say T3, he discussed with at least 6 other students. If two of these 6 students discussed T3, then we are done.
Suppose now that the 6 students discussed only T1 and T2 and choose one of them, say B. By the pigeonhole principle he discussed one of the topics, say T2, with three of these students. If two of these three students also discussed T2, then we are done. Otherwise, all the three students discussed only T1, which completes the task.
Solution: Let us call the topics T1,T2,T3. Consider an aarbitrary student A. By the pigeonhole principle there is a topic, say T3, he discussed with at least 6 other students. If two of these 6 students discussed T3, then we are done.
Suppose now that the 6 students discussed only T1 and T2 and choose one of them, say B. By the pigeonhole principle he discussed one of the topics, say T2, with three of these students. If two of these three students also discussed T2, then we are done. Otherwise, all the three students discussed only T1, which completes the task.
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