Let $A,B,C$ be fixed points in the plane , and $D$ be a variable point on the circle $ABC$, distinct from $A,B,C$ . Let $I_{A},I_{B},I_{C},I_{D}$ be the Simson lines of $A,B,C,D$ with respect to triangles $BCD,ACD,ABD,ABC$ respectively. Find the locus of the intersection points of the four lines $I_{A},I_{B},I_{C},I_{D}$ when point $D$ varies.
Problem
Source: 11-th Taiwanese Mathematical Olympiad 2002
Tags: geometry, geometric transformation, dilation, ratio, parallelogram, Euler, geometry unsolved
gemath
23.01.2007 16:54
One for all! Problem:Given triangle $ABC$ inscribed $(O)$, $M$ lie on $(O), d$ is Simson's line of $M$ with triangle $ABC$ ,$d'$ is Simson's line of $A$ with triangle $MBC$. Find the locus of intersections of $d$ and $d'$ Proof Lemma (Well know):Steiner's line is dilation of Simson's line $d$ center $M$ ratio $2$, $s=D_{M}^{2}(d)$ and Steiner's line is through orthocenter $H$ of $ABC$.Thus $d$ is through midpoint $K$ of $MH$ $(1)$ Called $H'$ is othorcenter of triangle $MBC$ easily seen $AHH'M$ is paralelogram, thus $K$ is midpoint of $AH$.Apply the lemma we have $d'$ is through $K$ $(2)$. From $(1),(2)$ we have intersection $K$ of $d$ and $d'$ is midpoint of $MH$ thus $K=D_{H}^\frac{1}{2}(M)\Leftrightarrow \{K\}=D_{H}^\frac{1}{2}(\{M\})=D_{H}^\frac{1}{2}((O))$=Euler circle. @N.T.TUAN If you don't post all Taiwan Olympiad yet, you don't want stop Please post all them in one box of olympiad box and post solution for it if you have
N.T.TUAN
23.01.2007 19:04
gemath wrote: @N.T.TUAN If you don't post all Taiwan Olympiad yet, you don't want stop Please post all them in one box of olympiad box and post solution for it if you have I don't understand. I haven't got solution of these problems. I only problem and NOT solution! First, I post all them in box National Olympiad. Second, I post all them in boxes : Algebra, Combinatoric,... That is LAW of this forum. I will check your solution later. Now, I have to sleep
Sprdalo
15.03.2020 21:03
I know it has been more than 13 years since this post was published by @N.T.TUAN, but at the moment I only knew how to do basic calculations (addition only). However, now I have a solution for this via complex plane geometry: Let us have quadrilateral $ABCD$ with a unit circle around it. It is known (if anybody needs proof, I can write it) that Simson's lines (from $A$ to $BCD$, $B$ to $ACD$, $C$ to $ABD$ and $D$ to $ABC$) all intersect in one point, and that point is (I can write the proof of this as well if needed): $x = \frac{a+b+c+d}{2}$. So the geometric place of all points $x$ is circle with center in $\frac{a+b+c}{2}$ and radius $1$, because $|x - \frac{a+b+c}{2}| = |\frac{d}{2}| = \frac{1}{2}$.