## Problems Problem 1 Let \( M \) be a point inside the triangle \( ABC \). Denote by \( A_1 \), \( B_1 \), and \( C_1 \) the feet of perpendiculars from \( M \) to \( BC \), \( CA \), and \( AB \). Prove that \[ MA\cdot MB\cdot MC\geq (MB_1+MC_1)\cdot(MC_1+MA_1)\cdot (MA_1+MB_1).\] Denote by \( A^{\prime} \), \( B^{\prime} \), \( C^{\prime} \) the feet of perpendiculars from \( A \), \( B \), \( C \) to the opposite sides of the triangle. Let \( a \), \( b \), \( c \) be the lengths of the sides \( BC \), \( CA \), and \( AB \). Since \( MA+MA_1\geq AA^{\prime} \) we have that \( a\cdot MA_1+b\cdot MB_1+c\cdot MC_1=2S_{ABC}=a\cdot AA^{\prime}\leq a\cdot (MA+MA_1) \) which implies that \begin{eqnarray*} a\cdot MA&\geq& b\cdot MB_1+c\cdot MC_1. \quad\quad\quad\quad\quad (1) \end{eqnarray*} Let \( M^{\prime} \) be the point symmetric to \( M \) with the respect to the bisector of the angle \( CAB \). Let \( A_1^{\prime} \), \( B_1^{\prime} \) and \( C_1^{\prime} \) denote the feet of perpendiculars form \( M^{\prime} \) to \( BC \), \( CA \), and \( AB \) respectively. Then \( MA=M^{\prime}A \), \( MB_1=MC_1^{\prime} \) and \( MC_1=MB_1^{\prime} \). Similarly to the previous reasoning we obtain \begin{eqnarray*} a\cdot MA&\geq& b\cdot MC_1+c\cdot MB_1. \quad\quad\quad\quad\quad (2) \end{eqnarray*} Adding (1) and (2) yields \( 2a\cdot MA\geq (b+c)\cdot (MB_1+MC_1) \). Writing the analogous relations for \( 2b\cdot MB \) and \( 2c\cdot MC \) and multiplying them out yields to \( 8abc\cdot MA\cdot MB\cdot MC\geq (a+b)(b+c)(c+a)\cdot (MB_1+MC_1)\cdot(MC_1+MA_1)\cdot (MA_1+MB_1) \). Applying the inequalities \( a+b\geq 2\sqrt ab \), \( b+c\geq 2\sqrt bc \), \( c+a\geq 2\sqrt ca \) to the right-hand side of the last inequality implies the desired result. Let us now investigate the case when the equality holds. The equality holds in (1) if and only if \( AA_1 \) is the altitude. Since we had two analogous inequalities, in order to have the equality \( M \) must be the orthocenter. For \( a+b=2\sqrt ab \) we must have \( a=b \). Similarly if the equality holds then \( b=c \), therefore if the equality holds then \( ABC \) must be isosceles and \( M \) its incenter. It is easy to verify that in this case the equality holds indeed. Problem 2 Prove that \[ \frac1r=\frac1{r_a}+\frac1{r_b}+\frac1{r_c}=\frac1{h_a}+\frac1{h_b}+\frac1{h_c}\geq\frac1{l_a}+\frac1{l_b}+\frac1{l_c}\geq \frac1{m_a}+\frac1{m_b}+\frac1{m_c} \geq \frac2R.\] From \( S=pr=(p-a)r_a=(p-b)r_b=(p-c)r_c \) we get \( \frac1r=\frac pS= \frac{p-a+p-b+p-c}S=\frac{p-a}S+\frac{p-b}S+\frac{p-c}S=\frac1{r_a}+\frac1{r_b} +\frac1{r_c} \). On the other hand \( \frac1r=\frac pS=\frac{a+b+c}{2S}=\frac a{2S}+\frac b{2S}+\frac c{2S}= \frac1{h_a}+\frac1{h_b}+\frac1{h_c} \). This way the first two equalities are proved. From \( 1/h_a\geq 1/l_a\geq 1/m_a \) the first two inequalities follows easily. All that remains to be proved is \[ \frac1{m_a}+\frac1{m_b}+\frac1{m_c}\geq \frac 2R. \] Using the inequality between arithmetic and harmonic mean we get \[ \frac1{m_a}+\frac1{m_b}+\frac1{m_c}\geq\frac9{m_a+m_b+m_c}.\] Applying the result of Problem 1 from Inequalities in triangle geometry yields the desired result. Problem 3 Find all real numbers \( \alpha \) that satisfy: If \( a \), \( b \), \( c \) are sides of a triangle then \[ a^2+b^2+c^2\leq \alpha(ab+bc+ca).\] For each triangle there are positive real numbers \( x \), \( y \), \( z \) such that \( a=y+z \), \( b=x+z \), \( c=x+y \). We need to find all real numbers \( \alpha \) such that \[ (x+y)^2+(y+z)^2+(z+x)^2\leq \alpha((x+y)(y+z)+(y+z)(z+x)+(z+x)(x+y))\] holds for all real numbers \( x \), \( y \), \( z \). After simplifying the previous inequality becomes equivalent to \( 2x^2+2y^2+2z^2+2xy+2yz+2zx\leq \alpha(x^2+y^2+z^2+3xy+3yz+3zx) \). We immediately get that \( \alpha\geq 2 \). Indeed, if \( \alpha< 2 \), then we can take \( y=z \) and obtain \( (2-\alpha)x^2+ 4xy+ 6y^2-6\alpha xy \leq 5\alpha2y^2 \) which can’t hold if \( x\rightarrow \infty \). On the other hand, if \( \alpha\geq 2 \) the inequality becomes true since the right-hand side is bigger than \( 2(x^2+2y^2+2z^2+3xy+3yz+3zx) \). Problem 4 Let \( A_1 \), \( B_1 \), \( C_1 \) be the intersections of the internal angle bisectors of the angles \( \angle A \), \( \angle B \), \( \angle C \) of \( \triangle ABC \). Denote by \( d_a \) the distance from \( A_1 \) to the side \( AB \). Similarly we define \( d_b \) and \( d_c \). If \( h_a \), \( h_b \), and \( h_c \) are the lengths of the altitudes of \( \triangle ABC \) corresponding to the vertices \( A \), \( B \), and \( C \), prove that \[ \frac{d_a}{h_a} +\frac{d_b}{h_b}+ \frac{d_c}{h_c}\geq\frac32.\] Since \( 2S=ah_a=(b+c)d_a \) we get \( \frac{d_a}{h_a}=\frac a{b+c} \) and the inequality becomes equivalent to \( \frac a{b+c}+\frac b{c+a}+\frac c{a+b}\geq \frac 32 \). This holds for all real numbers \( a \), \( b \), \( c \) (not only for the sides of the triangle). To prove this let us use the notation \( p=b+c \), \( q=c+a \), \( r=a+b \). Then \( a=(q+r-p)/2 \) etc. and the inequality now becomes equivalent to \[ \frac qp+\frac pq+\frac qr+\frac rq+\frac pr+\frac rp\geq 6\] which is true since \( \frac qp+\frac pq\geq 2 \) etc. Problem 5 A triangle \( ABC \) and three positive real numbers \( \alpha \), \( \beta \), and \( \gamma \) are given. Using a straight-edge and compass construct a point \( M \) in the plane for which \( \alpha MA+\beta MB+\gamma MC \) is minimal. Assume that \( \alpha\geq \beta+\gamma \). Then \( \alpha MA+\beta MB+\gamma MC\geq \beta(MA+MB)+\gamma (MA+MC)\geq \beta AB+\gamma AC=\alpha AA+\beta AB+\gamma AC \) and the required point is \( A \). Similarly we can consider the cases \( \beta\geq \alpha+\gamma \) and \( \gamma\geq \alpha+\beta \). Therefore, assume that \( \alpha \), \( \beta \) and \( \gamma \) are sides of a triangle. Without loss of generality we may assume that \( \alpha=1 \) (otherwise we can rescale the initial triangle \( ABC \)). Construct a triangle \( XYZ \) with sides \( YZ=1 \), \( XY=\gamma \), \( XZ=\beta \). Consider the rotational homothety with center \( B \), coefficient \( \gamma \) and angle \( \angle XYZ \). Denote by \( M^{\prime} \) and \( C^{\prime} \) the images of \( M \) and \( C \). Then we have \( M^{\prime}BM\sim XYZ \) and \( BC^{\prime}C\sim BCM \). We also have \( MM^{\prime}=\beta MB \) and \( M^{\prime}C^{\prime}=\gamma MC \). From \( AC^{\prime}\leq AM+MM^{\prime}+M^{\prime}C^{\prime}=\alpha MA+\beta MB+\gamma MC \) we conclude that the given sum has the minimum when \( M \) and \( M^{\prime} \) belong to \( AC^{\prime} \). The point \( C^{\prime} \) can be constructed without the knowledge of \( M \), and the point \( M \) can be easily found on the segment \( AC^{\prime} \) (we can construct a line through \( B \) at an angle \( \angle XZY \) to \( AC^{\prime} \), it intersects \( AC^{\prime} \) at \( M \)). Problem 6 An equilateral triangle \( ABC \) is partitioned into \( n \) convex polygons. If no line intersects more than \( 40 \) of the given polygons, is it possible for \( n \) to be greater than a million? Yes. Initially "cut" the three angles of the triangle to obtain the partition into three triangles and one hexagon. Each line can cut at most \( 2 \) of the triangles and the hexagon. Then "cut" the six angles of the hexagon to obtain \( 9 \) triangles and one \( 12 \)-gon. Now each line can cut at most \( 2 \) of the new \( 6 \) triangles. We repeat the procedure \( 19 \) times. Each line can cut at most \( 1+2\cdot 19=39 \) of the polygons. However, the total number of polygons is equal to \( 1+3+6+12+24+\cdot + 3\cdot 2^{18}=1+3\cdot (2^{19}-1)> 2\cdot 2^{19}=2^{20}=1024\cdot 1024> 1000000 \). Problem 7 Denote by \( h \) the longest altitude of the triangle whose circumradius and inradius are \( R \) and \( r \) respectively. Prove that \[ R+r\leq h.\] When does the equality hold? Let \( ABC \) be the given triangle, \( O \) its circumcenter, and \( A^{\prime\prime} \), \( B^{\prime\prime} \), \( C^{\prime\prime} \) the midpoints of arcs \( BC \), \( CA \), \( AB \) that don’t contain \( A \), \( B \), \( C \). Let \( A_1 \), \( B_1 \), and \( C_1 \) denote the midpoints of \( BC \), \( CA \), and \( AB \). According to the incircle-excircle theorem (part (c)), we have \( A^{\prime\prime}A_1=\frac{r_a-r}2 \) hence \( OA_1= R-\frac{r_a-r}2 \). Similarly we find \( OB_1 \) and \( OC_1 \) which implies that \( OA_1+OB_1+OC_1=3R+\frac32r-\frac12(r_a+r_b+r_c) \). From the Theorem 1 in Inequalities in triangle geometry (the last equality) we further get \( OA_1+OB_1+OC_1=3R+\frac32r- 2R-\frac12r=R+r \). Assume that \( h_c \) is the longest altitude. Then \( c \) is the shortest side and \[ R+r=OA_1+OB_1+OC_1=\frac{OA_1\cdot c+OB_1\cdot c+OC_1\cdot c}{c}\leq \frac{OA_1\cdot a+OB_1\cdot b+OC_1\cdot c}{c}=\frac{2S_{ABC}}c=h_c.\] The equality holds if and only if \( ABC \) is equilateral. |

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