A point on the midline of BC.

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A point on the midline of BC.

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Source: 2017 Peru Southern Cone TST P5

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#1 h Apr 5, 2025, 7:21 PM • 3 Y

Y by PikaPika999, KAME06, RANDOM__USER

Let $ABC$ be an acute triangle with circumcenter $O$ . Draw altitude $BQ$ , with $Q$ on side $AC$ . The parallel line to $OC$ passing through $Q$ intersects line $BO$ at point $X$ . Prove that point $X$ and the midpoints of sides $AB$ and $AC$ are collinear.

This post has been edited 1 time. Last edited by EmersonSoriano, Apr 5, 2025, 7:33 PM
Reason: change

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#2 h Apr 5, 2025, 9:26 PM • 1 Y

Y by PikaPika999

Let $A$ move projectively on the circumcircle of $ABC$ . Then $BO$ and $CO$ remain constant and $Q$ moves projectively as it is a projection of $B$ onto a projectively rotating line $AC$ . Now let us consider the point $\infty_{CO}$ (meaning the point at infinity through which $OC$ passes), then $Q\infty_{CO}$ is a projectively moving line, then intersect that line with the constant line $BO$ to obtain a projectively moving point $X$ . Points $D$ and $E$ obviously move projectively. Then essentially, we have the following configuration of projective transformations,

$A \rightarrow Q \rightarrow \mathcal{P}(\infty_{CO}) \rightarrow BO \rightarrow \mathcal{P}(\infty_{BC})$

The last step is to obtain the line through $X$ which is supposed to be the middle line of the triangle. Then we simply consider another projective transformation which maps $A$ to the midline. All that is left is to prove that both of these transformations are equal, to do this we simply need to consider three convinient positions for $A$ .

Case 1 and 2: If $AB \perp BC$ , then the statement is obvious because $X$ is the midpoint of $AC$ . The same for $AC \perp BC$ .
Case 3: When $AB = BC$ the statement is again trivial, due to $X = Q = E$ , where $E$ is the midpoint of $AC$ .

Thus the two projective transformations are equal and thus $X$ always lies on the midline of $\triangle ABC$ .

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#3 h Apr 5, 2025, 9:55 PM

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Quick complex bash: let $\triangle ABC$ be inscribed in the unit circle so that
$|a|=|b|=|c|=1$ $o=0$ $q=\frac{b^2+ab+bc-ac}{2b}$ Now since $X$ is on line $BO$ , we have
$\overline{x} = \frac{x}{b^2}$ And since $QX\parallel CO$ , we have
$\frac{q-x}{c-o} \in \mathbb{R}$ $\frac{b^2+ab+bc-ac-2bx}{2bc} = \frac{abc+b^2c+ab^2-b^3-2acx}{2ab^2}$ $ab^3+a^2b^2+ab^2c-a^2bc-2ab^2x = abc^2+b^2c^2+ab^2c-b^3c-2ac^2x$ $x = \frac{a^2b^2-a^2bc+ab^3-abc^2+b^3c-b^2c^2}{2ab^2-2ac^2} = \frac{a^2b+ab^2+abc+b^2c}{2ab+2ac} = \frac{b(a+b)(a+c)}{2a(b+c)}$ Then
$\frac{a+b}2 - x = \frac{a(a+b)(b+c) - b(a+b)(a+c)}{2a(b+c)} = \frac{c(a-b)(a+b)}{2a(b+c)}$ and so
$\frac{\frac{a+b}2-x}{\frac{a+b}2-\frac{a+c}2} = \frac{c(a-b)(a+b)}{a(b+c)(b-c)}$ which is real. $\blacksquare$

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KAME06

#4 h Apr 5, 2025, 10:30 PM

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Let $M$ and $N$ the midpoints of $AB$ and $AC$ , respectively; $Y \in OM \cap BQ$ and $\angle OCB=\alpha$
(1)We know that $\triangle BOC$ is isosceles, so $\angle BOC= 180^\circ -2\alpha$ . Also, using the Central Angle Theorem: $\angle BAC = \angle MAQ= \angle BAQ= \frac{180^\circ -2\alpha}{2}=90^\circ -\alpha$ .
(2) $90^\circ = \angle BQA \Rightarrow \angle YBA = \angle QBA = 180^\circ - \angle BQA - \angle BAQ=180^\circ - 90^\circ +\alpha-90^\circ=\alpha$ .
(3) $\angle YAQ = \angle BAC - \angle MAY= 90^\circ -\alpha -\alpha = 90^\circ -2\alpha$ ( $\angle MAY = \alpha$ because is well known that $OM$ is the perpendicular bisector of $BA$ , so also $YM$ and $\triangle YBA$ is isosceles).
(4) $\angle YMA = \angle YQA = 90^\circ$ , so $AMYQ$ must be cyclic. Then, $\angle YAQ= \angle YMQ = 90^\circ -2\alpha$ .
(5) $\angle QMA = \angle YMA - \angle YMQ=90^\circ -90^\circ +2\alpha=2\alpha$
(6)We know that $QX \parallel OC$ , so $\angle QXB=180^\circ -\angle BOC=180^\circ-180^\circ+2\alpha=2\alpha$ .
(7)Using (5) and (6), we deduce that $MQXB$ must be cyclic (supplementary opposite angles). Then $\alpha = \angle MAY = \angle MQY = \angle MQB = \angle MXB$
(8) $\angle MXB = \alpha = \angle OBC = \angle XBC$ , so $MX \parallel BC$ , but $MN \parallel BC$ , so $X$ and the midpoints of sides $AB$ and $AC$ are collinear,

This post has been edited 1 time. Last edited by KAME06, Apr 5, 2025, 10:32 PM

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#6 h Apr 5, 2025, 10:49 PM

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Just one ratio and it finishes

Let midpoints of $AB$ and $AC$ be $M_C$ and $M_B$ respectively. Then
$\dfrac{TO}{OB}=\dfrac{TM_B}{M_BQ}=\dfrac{TO}{OC}=\dfrac{TX}{TQ}$ implies line $M_BX$ bisects $\angle QXB$ . Thus $\angle QXM_B=\angle OCB$ . Since $QX\parallel OC$ we also have $\angle M_BQX=M_BCO$ . Thus, $X-M_B-M_C$ as desired.

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#7 h Apr 6, 2025, 4:25 AM • 1 Y

Y by teomihai

By the way, an interesting property about the problem is:

Let $B'$ be antipode of the point $B$ . Then $AM_BXB'$ is cyclic. Also we have $\triangle ABH\sim \triangle QXM_B$ .

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