collinear wanted, altitudes , circumcircle and orthocenter related

Art of Problem Solving

AoPS Online

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

collinear wanted, altitudes , circumcircle and orthocenter related

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

Source: Almaty Olympiad 2016 p3 , Kazakhstan

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

30629 posts

#1 h Sep 27, 2018, 8:59 PM • 2 Y

Y by Adventure10, Mango247

The altitudes of $A {{A} _ {1}}$ , $B {{B} _ {1}}$ and $C {{C} _ {1}}$ are drawn in the acute triangle $ABC$ . To the circumcircle of the triangle $ABC$ , the tangents at the points $A$ and $C$ intersecting at the point $Q$ are drawn. A straight line passing through the midpoint of the side $AC$ and the orthocenter of the triangle $ABC$ intersects the line ${{A}_{1}} {{C} _ {1}}$ at the point $F$ . Prove that the points $Q$ , ${{B} _ {1}}$ , and $F$ lie on the same line.

This post has been edited 1 time. Last edited by parmenides51, Jun 12, 2019, 10:44 PM
Reason: typo corrected, mentioned in following comments

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

2125 posts

#2 h Jun 12, 2019, 10:17 AM • 3 Y

Y by tworigami, Adventure10, Mango247

Extremely Beautiful Problem!!! ^ $1C_1$ should be replaced by $A_1C_1$

Almaty Olympiad 2016 P3 wrote:

The altitudes of $A {{A} _ {1}}$ , $B {{B} _ {1}}$ and $C {{C} _ {1}}$ are drawn in the acute triangle $ABC$ . To the circumcircle of the triangle $ABC$ , the tangents at the points $A$ and $C$ intersecting at the point $Q$ are drawn. A straight line passing through the midpoint of the side $AC$ and the orthocenter of the triangle $ABC$ intersects the line ${{B} _{1}} {{C} _ {1}}$ at the point $F$ . Prove that the points $Q$ , ${{B} _ {1}}$ , and $F$ lie on the same line.

Solution: Let $\Delta DEF$ be the orthic triangle WRT $\Delta ABC$ . Let $G$ be the midpoint of $AC$ . Let $GQ \cap BC=M$ . Let $\stackrel{\longrightarrow}{GH}$ $\cap$ $\odot (ABC)$ $=$ $L$ $\implies$ $L$ $\in$ $\odot (BFHD)$ . Let $DF$ $\cap$ $LH$ $=$ $K$ $\implies$ $\angle FLH$ $=$ $\angle FBH$ $=$ $\angle GCH$ $\implies$ $FGCL$ is cyclic
$LH \cdot HG = FH \cdot HC = AH \cdot HD \implies ALDG \text{ is cyclic}$ $\angle GMD=90^{\circ}-C=\angle DAG \implies M \in \odot (ALDG)$ , And
$\angle LMQ=\angle LMA+90^{\circ}-C=\angle LGA+90^{\circ}-C=180^{\circ}-\angle LFC+90^{\circ}-C = 90^{\circ}-C+\angle LBF+90^{\circ}-A=B+\angle LBF=\angle LCQ$ Hence, $LMCQ$ is cyclic. Now Reflect $H$ over $AC$ to $H'$ . Let $BK,BG \cap \odot (ABC)=S,U$ $\implies$ $-1=(L, ~H' ; ~S, ~U)$ $\implies$ $Q \in$ $LH'$ . Let $W$ be the midpoint of $GQ$ and since, $G$ $\in$ $LH$ $\implies$ $W \in LE$
$-1=(BL ~ \cap ~AC , ~E ; ~A , ~C) \overset{B}{=} (L, ~H; ~F, ~D) \overset{D}{=} (L , ~H ; ~K , ~G) \overset{E}{=} (W, \infty_{HH'} ~; KE ~ \cap ~GQ , ~G)$ Hence, $KE \cap GQ=Q$ $\implies$ $Q,E,K$ ( $Q,B_1,F$ ) collinear

This post has been edited 1 time. Last edited by AlastorMoody, Jun 12, 2019, 4:14 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

237 posts

ABCCBA

#3 h Jun 12, 2019, 1:53 PM • 3 Y

Y by AlastorMoody, Adventure10, Mango247

parmenides51 wrote:

The altitudes of $A {{A} _ {1}}$ , $B {{B} _ {1}}$ and $C {{C} _ {1}}$ are drawn in the acute triangle $ABC$ . To the circumcircle of the triangle $ABC$ , the tangents at the points $A$ and $C$ intersecting at the point $Q$ are drawn. A straight line passing through the midpoint of the side $AC$ and the orthocenter of the triangle $ABC$ intersects the line ${{} \color{red}{A_1}} {{C} _ {1}}$ at the point $F$ . Prove that the points $Q$ , ${{B} _ {1}}$ , and $F$ lie on the same line.

Typo corrected
See: Orthocenter
For the generalization see Two circles Touching Internally (a)

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

439 posts

RC.

#4 h Jun 12, 2019, 2:16 PM • 2 Y

Y by Adventure10, Mango247

Let M denote the midpoint of $AC, QM \cap A_1C_1= N$ . $BD \cap A_1C_1 = D$ . It suffices to prove that $\dfrac {QM}{MN} = \dfrac{B_1H}{HD}$
Let $B_1C_1 \cap QB \cap QC = \Delta \Rightarrow \Delta \sim \Delta A_1B_1C_1$ and clearly, we have $M$ is the incenter of $\Delta$ hence done.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

466 posts

#5 h Jun 12, 2019, 3:08 PM • 2 Y

Y by Adventure10, Mango247

Let $M$ be the midpoint of $\overline{AC}$ and define $P= (ABC) \cap (BH)$ so that $M, H, F, P$ collinear. Let $X = BP \cap A_1C_1 \cap AC$ . Observe that inversion about $B$ with radius $\sqrt{BH \cdot BB_1}$ swaps $X, P$ so $(XFP)$ is fixed. Also, $\angle XPF=90^{\circ}$ so $(XF) \perp EF$ . Thus by inversion $(XF) \perp (ABC)$ and so $F$ lies on the polar of $X$ . But so do $B_1$ and $Q$ and we are done.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

187 posts

#6 h Jun 12, 2019, 3:45 PM • 2 Y

Y by Steve12345, Adventure10

parmenides51 wrote:

The altitudes of $A {{A} _ {1}}$ , $B {{B} _ {1}}$ and $C {{C} _ {1}}$ are drawn in the acute triangle $ABC$ . To the circumcircle of the triangle $ABC$ , the tangents at the points $A$ and $C$ intersecting at the point $Q$ are drawn. A straight line passing through the midpoint of the side $AC$ and the orthocenter of the triangle $ABC$ intersects the line ${{} {1}} {{C} _ {1}}$ at the point $F$ . Prove that the points $Q$ , ${{B} _ {1}}$ , and $F$ lie on the same line.

We replace points $A_1, B_1, C_1$ with $D, E, F$ and we shall use an $A$ -labelled diagram. Let $T = DQ \cap EF$ , $H$ be the orthocentre of $\triangle ABC$ and $M$ be the midpoint of $BC$ . We will prove that $T, H,M$ are collinear.

First note that $QB \parallel DF$ since $\angle{FDB} = \angle{BAC} = \angle{QBD}$ . Similarly, we have $QC \parallel DE$ . Let $P = QB \cap EF, R = QC \cap EF$ . Then, $\triangle PQR$ is homothetic to $\triangle FED$ with centre $T$ .

It is enough to prove that $M$ is the incentre of $\triangle PQR$ since $H$ is the incentre of $\triangle DEF$ . Note that $\angle{PBF} = \angle{ACB} = \angle{PFB}$ . So, $PB = PF$ . Also, $MB = MF$ . Thus, $PM$ bisects $\angle{BPF}$ . Similarly, $RM$ bisects $\angle{CRE}$ . Thus, $M$ is the incentre of $\triangle ABC$ and so the claim follows. $\blacksquare$

Attachments:

This post has been edited 2 times. Last edited by MathInfinite, Jun 12, 2019, 3:49 PM

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a