Uniqueness of a point with strange properties

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Uniqueness of a point with strange properties

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Source: InfinityDots MO 2 Problem 3

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TacH

#1 h Apr 9, 2018, 3:14 PM • 4 Y

Y by ThE-dArK-lOrD, SAUDITYA, Adventure10, Mango247

Let $A,B,C$ be three distinct points on a line $\ell$ . Prove that for each pair of distinct points $B_1,C_1$ such that $\overleftrightarrow{B_1C_1}$ does not pass through $A$ , and $\overleftrightarrow{B_1C}$ is not parallel to $\overleftrightarrow{C_1B}$ , there is a unique point $A_1$ satisfying:
(i) $A_1$ does not lie on $\overleftrightarrow{B_1C_1}$ ,
(ii) the projections of $A$ onto $\overleftrightarrow{B_1C_1}$ , of $B$ onto $\overleftrightarrow{C_1A_1}$ , and of $C$ onto $\overleftrightarrow{A_1B_1}$ lie on a line not parallel to $\ell$ , and
(iii) the reflections of $A$ over $\overleftrightarrow{B_1C_1}$ , of $B$ over $\overleftrightarrow{C_1A_1}$ , and of $C$ over $\overleftrightarrow{A_1B_1}$ lie on a line not parallel to $\ell$ .

Proposed by TacH

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#2 h Apr 10, 2018, 12:39 AM • 2 Y

Y by Adventure10, Mango247

Very nice geometry! The degenerate cases are a bit annoying to clear up. But the main idea is

This post has been edited 1 time. Last edited by MarkBcc168, Apr 10, 2018, 12:39 AM

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#3 h Apr 10, 2018, 3:42 AM • 3 Y

Y by lminsl, Adventure10, Mango247

Here is the construction of the point.
Denote $H_A$ be the projection of $A$ onto $B_1C_1$ , and $A',B^*,C^*$ be the reflections of $A,B,C$ over $B_1C_1$ . $\omega_1$ is the circle with center $C_1$ and radius $C_1B$ , and $\omega_2$ is the circle with center $B_1$ and radius $B_1C$ . $D$ is the point on ray $A_1B_1$ such that $B_1B^*$ is paralel with $C^*D$ , and $E$ is the point on ray $A_1C$ such that $B^*C$ and $C^*E$ are paralel. $\omega$ is the circle with center $D$ and radius $DE$ .
$\omega_1$ and $\omega$ meet at two points, $B'$ and $X$ . (We can prove that $X$ is the reflection of $B$ over $B_1C_1$ , and $B'$ is the other point.)
$\omega_2$ and $A'B'$ meet at $C'$ . We can construct $A_1$ from here.

Proof:
1. $H_A,H_B,H_C$ are collinear.
$\Longrightarrow$ $H_AH_B$ , $B^*B'$ , $C^*C'$ and $H_AH_C$ are paralel with each other.
2. The point is unique.
This can be done by proving the other way.

P.S. I will try to write the full proof later when I have access to my laptop again.

This post has been edited 1 time. Last edited by MNJ2357, Apr 10, 2018, 3:43 AM

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TacH

#4 h Apr 12, 2018, 8:18 PM • 3 Y

Y by MNJ2357, Adventure10, Mango247

Official Solution. Let $A_\perp, B_\perp, C_\perp$ be the projections of $A$ onto $\overleftrightarrow{B_1C_1}$ , of $B$ onto $\overleftrightarrow{C_1A_1}$ , and of $C$ onto $\overleftrightarrow{A_1B_1}$ , and let $A',B',C'$ be the reflections of $A$ over $\overleftrightarrow{B_1C_1}$ , of $B$ over $\overleftrightarrow{C_1A_1}$ , and of $C$ over $\overleftrightarrow{A_1B_1}$ respectively.

Since the lines in (ii) and (iii) are not parallel to $\ell$ , there exist a spiral similarity $\lambda$ sending $A\to A'$ , $B\to B'$ , and $C\to C'$ .
Let $S$ be the center of $\lambda$ , let $\ell' = \overleftrightarrow{B_1C_1}$ , and let $T=\ell\cap \ell'$ .

Note that there is also a spiral similarity $\chi$ with center $S$ that sends $A\to B$ and $A_\perp\to B_\perp$ .
Let the tangent to $(ABS)$ at $B$ cut $C_1B_\perp$ at $P$ . Since $\angle A_\perp AB=\angle B_\perp BP$ , $\triangle AA_\perp T\sim \triangle BB_\perp P$ .
Therefore $\chi$ sends $T\to P$ . Hence, from $B=AT\cap BP$ , $(BTPS)$ is cyclic.

Now, $S$ is also the center of a spiral similarity $\psi$ that sends $T\to A_\perp$ and $P\to B_\perp$ .
Since $C_1=TA_\perp\cap PB_\perp$ , $(STPC_1)$ is cyclic.

Combining this with the above, we have now shown that $S$ must be on $(BTC_1)$ .
Similarly, $S$ must be on $(CTB_1)$ as well, hence $S$ is unique, and is the intersection of $(BTC_1)$ and $(CTB_1)$ .

Finally, from this unique $S$ , we can construct $A_1$ as the intersection of $B_\perp C_1$ and $C_\perp B_1$ . Clearly this $A_1$ works, and it must be unique as well. $\blacksquare$

Comments
1. The conditions in the problem are needed; if $A$ is on $\overleftrightarrow{B_1C_1}$ then $\ell_1,\ell_2$ found in the above solution will coincide with $\ell$ , and if $\overleftrightarrow{B_1C}$ is parallel to $\overleftrightarrow{C_1B}$ , the locus becomes a line since $B_\perp,C_1,C_\perp,B_1$ lie on the same line.

2. If we remove the not parallel to $\ell$ condition, there may be four additional points $A$ formed, which can be constructed as follows. First, draw a line $\ell'$ through $A_\perp$ parallel to $\ell$ . Then, draw the circle $\Gamma_B$ with diameter $BC_1$ possibly cutting $\ell$ at $X_1,X_2$ , and the circle $\Gamma_C$ with diameter $CB_1$ possibly cutting $\ell$ at $Y_1,Y_2$ . Then the additional points are the points formed by the intersection of $C_1X_i$ and $B_1Y_j$ .

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TacH

#5 h Apr 12, 2018, 8:23 PM • 3 Y

Y by MNJ2357, Adventure10, Mango247

Btw, I would very much appreciate if anyone found a solution involving Complex Number

Since one of our team claimed that he solved it with complex number (I didn't say it was ThE-dArK-lOrD)
and forgot at the moment we were writing the solutions.

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