Let $ABC$ be a scalene acute triangle. Let $B_1$ be the point on ray $[AC$ such that $|AB_1|=|BB_1|$. Let $C_1$ be the point on ray $[AB$ such that $|AC_1|=|CC_1|$. Let $B_2$ and $C_2$ be the points on line $BC$ such that $|AB_2|=|CB_2|$ and $|BC_2|=|AC_2|$. Prove that $B_1$, $C_1$, $B_2$, $C_2$ are concyclic.
Problem
Source: BxMO 2022, Problem 3
Tags: geometry, BxMO
01.05.2022 20:33
Pretty simple First notice that $B_1$ and $C_2$ lie on the perpendicular bisector of $\overline{AB}$and similarly $C_1$ and $B_2$ lie on the perpendicular bisector of $\overline{AC}$ Now $$\measuredangle B_1CC_1 = \measuredangle ACC_1=\measuredangle C_1AC=\measuredangle BAC=\measuredangle BAC_1=\measuredangle B_1BA \implies \odot(BB_1CC_1)$$$$\angle B_1C_1B_2=\angle B_1C_1B - \angle B_2C_1B= \angle ACB - 90 + \angle BAC = 90 - \angle ABC = \angle B_1C_2B_2$$which implies $B_1$, $C_1$, $B_2$, $C_2$ are concyclic as desired $\blacksquare$
07.05.2022 21:47
Let $O$ be the circumcenter of $\triangle ABC.$ Notice $$\measuredangle BOC=2\measuredangle CAB=\measuredangle ABB_1=\measuredangle CC_1A$$so $OBC_1CB_1$ is cyclic. Hence, $$\measuredangle B_2AO=\measuredangle OCB_2=\measuredangle CBO$$so $ABB_2O$ and similarly $AC_2CO$ are cyclic. Thus, $$\measuredangle B_1C_2B_2=\measuredangle OC_2C=\measuredangle AOC=\measuredangle B_1CO=\measuredangle B_1C_1O.$$$\square$
27.09.2022 12:48
BVKRB- wrote: Pretty simple First notice that $B_1$ and $C_2$ lie on the perpendicular bisector of $\overline{AB}$and similarly $C_1$ and $B_2$ lie on the perpendicular bisector of $\overline{AC}$ Now $$\measuredangle B_1CC_1 = \measuredangle ACC_1=\measuredangle C_1AC=\measuredangle BAC=\measuredangle BAC_1=\measuredangle B_1BA \implies \odot(BB_1CC_1)$$$$\angle B_1C_1B_2=\angle B_1C_1B - \angle B_2C_1B= \angle ACB - 90 + \angle BAC = 90 - \angle ABC = \angle B_1C_2B_2$$which implies $B_1$, $C_1$, $B_2$, $C_2$ are concyclic as desired $\blacksquare$ How is the following equal $$\angle B_1C_1B - \angle B_2C_1B= \angle ACB - 90 + \angle BAC$$
02.12.2022 17:27
Another way...too easy with complex bash Let $(ABC)$ be unit circle so that $|a|=|b|=|c|=1$ Since $B_1$ lies on $AC$ we have $b_1+\bar{b_1}ac=a+c\implies \bar{b_1}=\frac{a+c-b_1}{ac}(1)$ On the other hand $B_1A=B_1B \implies |b_1-a|=|b_1-b| \implies (b_1-a)(\bar{b_1}-\bar{a})=(b_1-b)(\bar{b_1}-\bar{b})(2)$ Then using fact $1$ in expression $2$ gives us $b_1=\frac{b(a+c)}{b+c}$ also $\bar{b_1}=\frac{a+c}{a(b+c)}$ After doing similar things for $c_1,b_2,c_2$ we get $c_1=\frac{c(a+b)}{b+c}$ , $\bar{c_1}=\frac{a+b}{a(b+c)}$ $b_2=\frac{a(b+c)}{a+b}$ , $\bar{b_2}=\frac{b+c}{c(a+b)}$ $c_2=\frac{a(b+c)}{a+c}$ , $\bar{c_2}=\frac{b+c}{b(a+c)}$ Now we need to show $\frac{(c_1-b_1)(b_2-c_2)}{(b_2-b_1)(c_1-c_2)} \in \mathbb{R} \iff \frac{(c_1-b_1)(b_2-c_2)}{(b_2-b_1)(c_1-c_2)}=\frac{(\bar{c_1}-\bar{b_1})(\bar{b_2}-\bar{c_2})}{(\bar{b_2}-\bar{b_1})(\bar{c_1}-\bar{c_2})}$ Boring calculation yields $LHS=\frac{a^2(c-b)^2(c+b)^2}{(c^2a+cab-cb^2-a^2b)(c^2b+ca^2-cab-ab^2)}=RHS$ so we are done
03.12.2022 18:08
BVKRB- wrote: First notice that $B_1$ and $C_2$ lie on the perpendicular bisector of $\overline{AB}$and similarly $C_1$ and $B_2$ lie on the perpendicular bisector of $\overline{AC}$ We can show this by the law of sines on triangles $ABC_2$ and $AB_2C$ , then simple angle chasing follows the result.
15.01.2023 17:05
Here is my solution: https://calimath.org/pdf/Benelux2022-3.pdf And I uploaded the solution with motivation to: https://youtu.be/CflsSrErskQ