An acute triangle $ABC$ is inscribed in a circle of radius 1 with centre $O;$ all the angles of $ABC$ are greater than $45^\circ.$ $B_{1}$ is the foot of perpendicular from $B$ to $CO,$ $B_{2}$ is the foot of perpendicular from $B_{1}$ to $AC.$ Similarly, $C_{1}$ is the foot of perpendicular from $C$ to $BO,$ $C_{2}$ is the foot of perpendicular from $C_{1}$ to $AB.$ The lines $B_{1}B_{2}$ and $C_{1}C_{2}$ intersect at $A_{3}.$ The points $B_{3}$ and $C_{3}$ are defined in the same way. Find the circumradius of triangle $A_{3}B_{3}C_{3}.$ Proposed by F.Bakharev, F.Petrov
Problem
Source: Tuymaada 2004, day 1, problem 3. - Authors : F. Bakharev, F. Petrov.
Tags: geometry, circumcircle, parallelogram, congruent triangles, geometry proposed
Erken
30.06.2007 18:02
Let $ C_1C_2$ intersect AC at point B' and $ B_1B_2$ intersect AB at point C'. It is easy to prove that BB' is perpendicular to AC: Let $ \measuredangle C_1BC=a$ then $ \measuredangle A=90^{\circ}-a$ and $ \measuredangle C_1B^{\prime}C=a$, so that $ BB^{\prime}C_1C$ is an inscribed quadriterial and $ \measuredangle BB^{\prime}C=90^{\circ}$. So $ A_3$ is the orthocenter of triangle $ AC^{\prime}B^{\prime}$. The same way $ B_3$ and $ C_3$ are orthocentres of triangles $ BC^{\prime}A^{\prime}$ and $ CA^{\prime}B^{\prime}$ respectively. It is easy to prove that $ A_3B_3C_3$ and $ A^{\prime}B^{\prime}C^{\prime}$ are congruent triangles ($ A^{\prime}B^{\prime}A_3B_3$ is parallelogram) but the circumradius of A'B'C' is $ \frac12$.
Erken
12.10.2007 18:59
Just want to post a picture to this great problem: Image not found P.S.This problem is great,as all geometry problems from Tuymaada. .
snakeaid
03.10.2020 15:09
Toss onto complex plane having $(ABC)$ as the unit circle centered at $O=0$. Here we are using the complex foot formula from $Z$ onto $\overline{AB}$ given by $w=\frac{(a-b)\overline{z}+\overline{a}-\overline{b}+\overline{a}b-a\overline{b}}{2(\overline{a}-\overline{b})}$. Then we compute $b_1=\frac{b^2+c^2}{2b}$ and $b_2=\frac{2abc+2bc^2+b^2c+c^3-ab^2-ac^2}{4bc}$. Now $\frac{b_2-c_2}{b-c}=\frac{b^2+c^2}{4bc} \in \mathbb{R} \implies B_2C_2 \parallel BC \implies$ $\triangle AB_2C_2$ and $\triangle ABC$ are homothetic with center $A$. Notice that $AB_2A_3C_2$ is cyclic with $\overline{AA_3}$ as diameter and let $K$ be its circumcenter. Then from the homothety $K$ lies on $\overline{AO}$. Now we use the circumcenter formula for a triangle $\triangle XYZ$ with $z=0$ which is given by $p=\frac{xy\overline{(\overline{x}-\overline{y})}}{\overline{x}y-x\overline{y}}$. Then using that $b_2-a=\frac{(b+c)^2(c-a)}{4bc}$ we easily compute $k-a=-\frac{a(b+c)^2}{4bc}$, whence $k=-\frac{a(b-c)^2}{4bc}$. Since $K$ is the midpoint of $\overline{AA_3}$, $a_3=2k-a=-\frac{a(b^2+c^2)}{2bc}$. Now $\frac{a-o}{b_3-c_3}=\frac{2bc}{c^2-b^2} \in i\mathbb{R} \implies AO \perp B_3C_3 \implies A_3O \perp B_3C_3 \implies O$ is the orthocenter of $\triangle A_3B_3C_3$. Since $O$ is the orthocenter it's well-known that the circumradius of $\triangle A_3B_3C_3$ is equal to the circumradius of $\triangle A_3OC_3$. Let the circumcentesr of $\triangle A_3B_3C_3$ and $\triangle A_3OC_3$ be $T$ and $P$, respectively. Then we compute $p=-\frac{1}{2b}$ and hence $TA_3=TB_3=TC_3=PO=|p-o|=\frac{1}{2}$, as desired. $\square$