assume that ABC is acute traingle and AA' is median we extend it until it meets circumcircle at A". let $AP_a$ be a diameter of the circumcircle. the pependicular from A' to $AP_a$ meets the tangent to circumcircle at A" in the point $X_a$; we define $X_b,X_c$ similary . prove that $X_a,X_b,X_c$ are one a line.
Problem
Source: iran2004(geo exam)
Tags: geometry, circumcircle, projective geometry, modular arithmetic, Euler, geometry proposed
10.09.2004 23:46
From Desargue's theorem, it follows that we need to show that the triangles formed by the tangents to the circumcircle in $A\",B\",C\"$ on the one hand, and by the perpendiculars from $A'$ to $AO$ and the like on the other hand are perspective. Let the perpendiculars through $B',C'$ to $BO,CO$ respectively intersect at $T$, and let the tangents to the circumcircle in $B\",C\"$ intersect at $S$. I'll show that $ST$ passes through $G$ (the centroid). Let $P$ be $B\"C\"\cap BC$. $SG$ is the polar of $P$ wrt $(O)$, so we need to show that $T$ lies on this polar. A quick angle chase shows that $T$ is, in fact, the intersection of the tangents through $B',C'$ to the nine-point circle. We thus need to show that if $T'$ is the intersection pt of the tangents in $B,C$ to $(O)$, then $T'$ lies on the polar of $P$ wrt $(O)$, but this is just a degenerate case of Pascal's theorem.
11.09.2004 15:19
The solution of the problem is a strightforward consequence of the following lemma RL110904 and an opportune iterate application of Desargues theoreme. Lemma RL110904 Given a ciclic quadrialteral ABCD, let K be the intersection of AC and BD and L be the intersection between the tangent to c trhough C and D. If R and S are two point on BD and AC respectively s.t. RS//AB, then the intersection of the perpendiculars from R to the diameter through B and from S to the diameter through A stay on line KL. Proof. this lemma derive from the well-known fact that (which is in turn consequence of Pascal theorem) the diagonals of two dual quadrilaterals are concurrents.
23.05.2009 11:36
Let me restate the problem so that it can fit my solution: "Let $ ABC$ be triangle inscribed in circumcircle $ (O)$. Denote $ A_1,B_1,C_1$ the intersections by the medians from $ A,B,C$ and $ (O)$, respectively. Let $ A_3,B_3,C_3$ be the midpoints of $ BC,CA,AB$, respectively. The perpendicular from $ A_3$ to $ AO$ intersects the tangent from $ A_1$ wrt $ (O)$ at $ X_a$. Define the same for $ X_b,X_c$. Prove that $ X_a,X_b,X_c$ are collinear." Proof: Lemma $ 1$: Let $ AB$ be the chord of $ (O)$. Denote $ C$ is an arbitry point on the greater arc $ AB$. Let $ M$ be the midpoint of $ AB$. Denote $ E$ the intersection by $ CM$ and $ (O)$, $ D$ by the intersection of $ CO$ and $ (O)$. The perpendicular from $ M$ to $ CD$ intersects the tangent from $ E$ wrt $ (O)$ at $ K$. Let $ KN$ be another tangent form $ K$ to $ (O)$ ($ N\in (O)$). Then $ N,M,D$ are collinear. Indeed, let $ S$ be the projection of $ M$ onto $ CD$. Denote $ X$ by the pole of $ MS$ wrt $ (O)$. Since $ K\in$ the polar of $ X$ wrt $ (O)$ hence the polar of $ K$ passes $ X$, which implies that $ X\in NE$, which also implies $ X\equiv CD\cap NE$. Let $ Y\equiv NE\cap MS\Longrightarrow (XYEN) = (XSDC) = (XSCD) = - 1\Longrightarrow YS,EC,DN$ are concurrent. But $ M$ already is the intersection of $ CE,YS$, thus $ M\in ND$. Lemma $ 2$: Let $ ABC$ be a triangle inscribed in circumcircle $ (O)$, $ H$ is its orthocenter. Denote $ A_3,B_3,C_3$ the midpoints of $ BC,CA,AB$, respectively. $ A_1,B_1,C_1$ be the intersections of $ AA_3,BB_3,CC_3$ with $ (O)$, respectively. Denote $ A_0,B_0,C_0$ be the second intersections of circumcircles of diameters $ AH,BH,CH$ with $ (O)$, respectively. Prove that: $ A_1A_0,B_1B_0,C_1C_0$ are concurrent. Indeed, let $ D,E,F$ be the projections of $ H$ onto $ BC,CA,AB$, respectively. Consider the inversion through pole $ H$, power $ \overline {HD}.\overline {HA} = \overline {HE}.\overline {HB} = \overline {HF}.\overline {HC} = k^2$. It is very well- known that $ \mathcal {I}(H,k^2): A_3\mapsto A_0$, $ B_3\mapsto B_0$, $ C_3\mapsto C_0$. Therefore, it is so obvious that $ \{B_3,C_3,B_0,C_0\}$ is the set of concyclic points. Also, $ (C_3C_1,C_3B_3)\equiv (C_3C,C_3B_3)\equiv (CC_1,CB)\equiv (B_1C_1,B_1B_3) \pmod \pi$, which implies $ \{C_1,C_3,B_3,B_1\}$ is a set of concyclic points. Now, we have $ B_1C_1,B_3C_3,B_0C_0$, respectively are the radical axes of $ \{(B_1C_1B_3C_3), (O)\}$, $ \{(B_1C_1B_3C_3),(B_3C_3B_0C_0)\}$, $ \{(B_0C_0B_3C_3),(O)\}$. Therefore $ B_1C_1,B_3C_3,B_0C_0$ are concurrent, let $ Y_1$ be the intersection of these lines. We have $ \overline {Y_1B_3}.\overline {Y_1C_3} = \overline {Y_1B_0}.\overline {Y_1C_0}\Longrightarrow \mathcal {P}_{(Y_1/(A_3B_3C_3))} = \mathcal {P}_{(Y_1/(O))}$. With the same define and argument for $ Y_2,Y_3$. We conclude that $ Y_1,Y_2,Y_3$ are collinear and further the line contains them is the radical axes of the Euler circle and $ (O)$. We have $ Y_1\equiv B_0C_0\cap B_1C_1$, $ Y_2\equiv A_0B_3\cap A_1B_1$, $ Y_3\equiv A_0C_0\cap A_1C_1$, $ Y_1,Y_2,Y_3$ are collinear therefore $ \triangle A_0B_0C_0$ and $ \triangle A_1B_1C_1$ are $ 2$ perspective triangles. Then by Desgaurte theorem, $ A_1A_0,B_1B_0,C_1C_0$ are concurrent. Back to our problem. Let $ X_aA_0,X_bB_0,X_cC_0$ let other tangents form $ X_a,X_b,X_c$, respectively to $ (O)$. Let $ A',B',C'$ be the intersections of $ AO,BO,CO$, respectively with $ (O)$. Then by lemma $ 1$, we conclude that $ A'A_0,B'B_0,C'C_0$ passes through $ A_3,B_3,C_3$, respectively, moreover, they also concurrent at $ H$- the orthocenter of $ \triangle ABC$. Thus, $ A_0,B_0,C_0$ now also play roles that they are respectively the second intersections of circumcircles of diameters $ AH,BH,CH$ with $ (O)$. Apply lemma $ 2$, we conclude that $ C_0C_1,A_0A_1,B_0B_1$ concurrent. But, $ C_0C_1,A_0A_1,B_0B_1$, respectively are the pole of $ X_c,X_a,X_b$ wrt $ (O)$. We conclude $ X_a,X_b,X_c$ are collinear. Our proof is completed then.
25.05.2009 05:48
Yep, very nice problem. Take a look at a very nice solution by Nguyen Minh Ha here.