10 concurrent lines

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Source: Brazil National Olympiad 2019 #6

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#1 h Nov 14, 2019, 3:39 PM • 5 Y

Y by UK2019Project, MrK2, Ruy, Adventure10, Mango247

Let $A_1A_2A_3A_4A_5$ be a convex, cyclic pentagon with $\angle A_i + \angle A_{i+1} >180^{\circ}$ for all $i \in \{1,2,3,4,5\}$ (all indices modulo $5$ in the problem). Define $B_i$ as the intersection of lines $A_{i-1}A_i$ and $A_{i+1}A_{i+2}$ , forming a star. The circumcircles of triangles $A_{i-1}B_{i-1}A_i$ and $A_iB_iA_{i+1}$ meet again at $C_i \neq A_i$ , and the circumcircles of triangles $B_{i-1}A_iB_i$ and $B_iA_{i+1}B_{i+1}$ meet again at $D_i \neq B_i$ . Prove that the ten lines $A_iC_i, B_iD_i$ , $i \in \{1,2,3,4,5\}$ , have a common point.

This post has been edited 2 times. Last edited by pablock, Nov 14, 2019, 4:14 PM

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#2 h Nov 14, 2019, 8:28 PM • 2 Y

Y by Adventure10, Mango247

Let $\ell$ be the radical axis of $\omega_1$ and $\omega_3$ .

https://i.imgur.com/DEARLG9.png

Lemma 1: $\ell$ is $B_2D_2$ .

Proof: $B_2 \in \ell$ , because $B_2A_1 \cdot B_2A_2 = B_2A_3 \cdot B_2A_4$ . Let $X$ be the intersection of $B_2D_2$ and $A_2A_3$ . Using the power of the point $X$ with the circles $\Omega_2$ and $\Omega_3$ , we have $XA_2 \cdot XB_1 = XD_2 \cdot XB_2$ and $XA_3 \cdot XB_3 = XD_2 \cdot XB_2$ . Therefore, $XA_2 \cdot XB_1 = XA_3 \cdot XB_3 \implies X \in \ell$ , thus, $\ell$ is the line $B_2X$ , i.e., the line $B_2D_2$ .

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#3 h Nov 16, 2019, 2:49 AM • 1 Y

Y by Adventure10

This link is useful: https://artofproblemsolving.com/community/q1h1813119p12086815

With the above link, we know that the intersection point of the $B_iD_i$ 's exists, call this point $P.$ Then from Lemma $1$ above, we know that $P$ has equal powers w.r.t. $\omega_1$ and $\omega_3$ . With five similar results, we're done.

$\square$

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#4 h Nov 16, 2019, 9:24 AM • 1 Y

Y by Adventure10

Why not ask for more?

- Let $\Gamma_A = (A_i)$ . The points $C_i$ too belong to a circle $\Gamma_C$ .
- The concurrency point of the problem is precisely the insimilicenter of $\Gamma_A$ and $\Gamma_C$ .

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#5 h Nov 19, 2019, 6:09 PM • 1 Y

Y by Adventure10

All you need is that $C_1C_2C_3C_4C_5$ is cyclic and the concurrency point is going to be the centre of inversion that sends circle
$(A_1A_2A_3A_4A_5)$ $\to$ $(C_1C_2C_3C_4C_5)$ .
Just notice that $C_i$ is the miguel point of quadrilateral $A_iA_{i+1}B_{i+2}A_{i+4}$ and that points
$B_iC_iA_{i+4}C_{i+3}B_{i+2}$ all belong on a circle . The rest is simple angle chasing to show that $C_iC_{i+1}C_{i+2}C_{i+3}$ is cyclic.

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#6 h Nov 20, 2019, 2:30 AM • 2 Y

Y by Adventure10, Mango247

mela_20-15 wrote:

All you need is that $C_1C_2C_3C_4C_5$ is cyclic and the concurrency point is going to be the centre of inversion that sends circle
$(A_1A_2A_3A_4A_5)$ $\to$ $(C_1C_2C_3C_4C_5)$ .
Just notice that $C_i$ is the miguel point of quadrilateral $A_iA_{i+1}B_{i+2}A_{i+4}$ and that points
$B_iC_iA_{i+4}C_{i+3}B_{i+2}$ all belong on a circle . The rest is simple angle chasing to show that $C_iC_{i+1}C_{i+2}C_{i+3}$ is cyclic.

I don't think $C_1C_2C_3C_4C_5$ being cyclic is enough for the problem. Why are $A_i$ and $C_i$ correlated though this inversion?

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#7 h Nov 29, 2019, 11:37 AM • 1 Y

Y by Adventure10

Quote:

mela_20-15 wrote:
All you need is that $C_1C_2C_3C_4C_5$ is cyclic and the concurrency point is going to be the centre of inversion that sends circle
$(A_1A_2A_3A_4A_5)$ $\to$ $(C_1C_2C_3C_4C_5)$ .
Just notice that $C_i$ is the miguel point of quadrilateral $A_iA_{i+1}B_{i+2}A_{i+4}$ and that points
$B_iC_iA_{i+4}C_{i+3}B_{i+2}$ all belong on a circle . The rest is simple angle chasing to show that $C_iC_{i+1}C_{i+2}C_{i+3}$ is cyclic.

I don't think $C_1C_2C_3C_4C_5$ being cyclic is enough for the problem. Why are $A_i$ and $C_i$ correlated though this inversion?

Well given that $C_1C_2C_3C_4$ is cyclic by means of angle chasing you show that $A_1C_1A_3C_3$ is cyclic and then $A_1C_1,A_2C_2,A_3C_3$ concur at a radical center.

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#8 h Jan 13, 2022, 1:58 AM

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Let $\omega_i=(A_iB_iB_{i+1})$ for all $i=1,2,3,4,5$ . Since $A_1A_2A_3A_4A_5$ is cyclic, then $B_3= A_4A_5 \cap A_2A_3$ satisfies $B_3A_4.B_3A_5= B_3A_3.B_3A_2 \implies Pow_{\omega_2}B_3=Pow_{\omega_4}B_3 \implies B_3$ lies on the radical axis of $\omega_2, \omega_4 (\star)$ . Let $E_3=A_4C_4 \cap A_3C_3$ and define $E_1,E_2,E_4,E_5$ similarly. Furthermore, let $T= B_3D_3 \cap A_3A_4 \implies TA_4.TB_4=TB_3.TD_3=TA_3.TB_2 \implies Pow_{\omega_2}T=Pow_{\omega_4}T \implies T$ lies on the radical axis of $\omega_2,\omega_4$ . From $(\star)$ , $B_3T$ is the radical axis of $\omega_2,\omega_4$ . Since $D_3 \in B_3T$ , $B_3D_3$ is the radical axis of $\omega_2,\omega_4$ . Similarly, $B_iD_i$ is the radical axis of $\omega_{i-1},\omega_{i+1}$ for all $i=1,2,3,4,5$ , and $B_iD_i, A_iC_i, A_{i+1}C_{i+1}$ are concurrent by radical axis theorem on $\omega_{i-1},\omega_i,\omega_{i+1}$ . Therefore, it is enough to prove that all $A_iC_i$ are concurrent. $(\spadesuit)$

Now, observe that $\angle A_1C_2B_2= \angle A_1C_2B_2+ \angle A_2C_2B_2= \angle A_1B_1A_2+ \angle A_4A_3B_1= 180º - \angle A_1B_4B_2 \implies B_4A_1C_2B_2$ is cyclic. Moreover, $\angle B_4C_5A_1= \angle B_4C_5A_5 + \angle A_5C_5A_1= \angle B_5A_4B_2 + \angle A_5B_5A_1= \angle B_5A_4B_2+ \angle A_4B_5B_2= 180º - \angle B_4B_2A_1 \implies A_1B_2B_4C_5$ is cyclic, so $A_1C_2B_2B_4C_5$ is cyclic $(\heartsuit)$ .
Notice that $\angle A_2A_5C_5= \angle C_5A_5A_4+ \angle A_4A_5A_2$ , which from $(\heartsuit)$ is equals to $\angle C_5C_2B_2+ 180º - \angle A_2A_3A_4$ . On the other hand, $\angle A_2C_2C_5= \angle A_2C_2B_2- \angle C_5C_2B_2= \angle A_2A_3A_4- \angle C_5C_2B_2$ , so $\angle A_2C_2C_5+ \angle A_2A_5C_5= 180º \implies A_2C_2C_5A_5$ is cyclic. Similarly, $A_{i+2}C_{i+2}C_iA_i$ is cyclic for all $i=1,2,3,4,5$ , and let these circles be $\Gamma_i$ . Thus, by radical center theorem on $\Gamma_i, \omega_i, \omega_{i+1}$ , we have that $A_iC_i, A_{i+1}C_{i+1}, A_{i+2}C_{i+2}$ are concurrent for all $i$ , so all the $A_iC_i$ are concurrent $\implies$ from $(\spadesuit)$ , we are done.

$\blacksquare$

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