The special Miquel's point from a familiar problem

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

B

No tags match your search

M

The special Miquel's point from a familiar problem

G H J

Reveal topic tags

geometric transformation

geometry proposed

L

G H BBookmark kLocked kLocked NReply

The post below has been deleted. Click to close.

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

25 posts

#1 h Jul 23, 2023, 2:52 PM • 1 Y

Y by sixoneeight

Problem. Let $ABC$ be an acute-angled triangle with $AC > AB$ , let $O$ be its circumcentre. The line through $A$ perpendicular to $BC$ intersects circle $(O)$ again at $T$ . The tangents at $B$ and $C$ of $(O)$ intersect at $S$ . $AS$ intersects $(O)$ at $X \neq A$ . $OB$ intersects $AT$ at $P$ . Let $N$ be the midpoint of $TC$ .
Prove that $T, P, N, X$ are concyclic.

Attachments:

configure.pdf (64kb)

Z K Y

The post below has been deleted. Click to close.

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

168 posts

#2 h Jul 23, 2023, 8:47 PM

Y by

What's $N$ ?

Z K Y

The post below has been deleted. Click to close.

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

1133 posts

#3 h Jul 23, 2023, 10:24 PM • 1 Y

Y by aqwxderf

danil_e wrote:

Let $N$ be the midpoint of $TC$ .

@above (#2)

Z K Y

The post below has been deleted. Click to close.

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

177 posts

#4 h Jul 24, 2023, 7:18 AM

Y by

Hint

Z K Y

The post below has been deleted. Click to close.

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

1 post

#5 h Dec 15, 2023, 8:25 AM

Y by

The line $AX$ is the symmedian of triangles $ABC$ and $XBC$ . If $M$ the midpoint of $BC$ then,
$\angle BXA=\angle CXM \quad (1)$ . Let $\omega$ the circle through the points $T$ , $P$ και $X$ , which intersects
$TD$ at the point $N'$ . We shall prove that $N'$ is the midpoint of $TC$ , or equivalently that $MN' \parallel BT$ .
From the cyclic quadrilateral $PTXN'$ we have that $\angle TPX=\angle TN'X \quad (2)$
Hence, $\angle APX=\angle CN'X$ as complements of the equal angles of $(1)$ .
Also, $\angle PAX=\angle TAX=\angle TCX=\angle N'CX \quad (3)$ as inscribed angles of circumcircle $\Omega$
of triangle $ABC$ subtending the same arc. Hence $\triangle XPA \sim \triangle XN'C$ . Therefore, there is spiral similarity
$S_{p}\;:\;[AP]\;\leftrightarrow \;[CN']$ with centre $X$ and ratio $\frac{CN'}{AP}$ . But then,
there is also the spiral similarity $S^{\prime}_{p}$ with centre also the point $X$ , and ratio $\frac{PN'}{AC}$ , $S^{\prime}_{p}\;:\;[AC]\;\leftrightarrow \;[PN']$ .
Thus $\triangle XAC \sim \triangle XPN'$ .

We shall prove that the quadrilateral $HTXM$ είναis cyclic ( $H$ το the foot of the altitude from the vertex $A$ ).
Indeed, $\angle TXM=\angle TXB+\angle BXM=\angle BAT+\angle AXC=\angle BAH+\angle ABC=90^{\circ}= \angle THM$ .
Hence, $\angle xTH=\angle xTP=\angle HMX=\angle PN'X$ as external of cyclic quadrilaterals $HTXM$ and $PTXN'$ .
If $\{Y\}=PN' \cap BC$ , then since $\angle YMX=\angle YN'X$ , the quadilateral $YXN'M$ είνis cyclic.
From the similarity of the triangles $XPN'$ και $XAC$ (see above) we have $\angle XPN'=\angle XAC=\angle XBC$ , δηλαδή,
the quadrilateral $BPYX$ εis cyclic.
We shall prove that the image of $B$ with the spiral similarity $S^{\prime}_{p}$ is the point $M$ .
Is enough $\angle BXM =\angle AXC$ which is it true since the line $XA$ is the symmedian of triangle $\triangle XBC$ ,
and $\frac{XM}{XB}=\frac{XN'}{XP}=\frac{XC}{XA}$ , which also is it true since $\triangle XBM \sim \triangle XPN' \sim \triangle XAC$ .
Finally we have proved that $S^{\prime}_{p}\;:\;A\;\leftrightarrow \;C$ , $S^{\prime}_{p}\;:\;P\;\leftrightarrow \;N'$ , $S^{\prime}_{p}\;:\;B\;\leftrightarrow \;M$ . Hence, $\triangle ABP \sim \triangle CMN' \; \Rightarrow \angle ABP=\angle CMN' \quad (3)$ .
But the lines $BO$ and $BH$ are isogonal w.r.t sides of the angle $ABT$ (radius and altitude from the vertex $B$ ), hence
$\angle ABP=\angle HBT \quad (4)$ .
From $(3)$ and $(4)$ we have that $\angle CBT=\angle CMN' \Rightarrow MN' \parallel BT$ , which is what we had to prove.

Attachments:

This post has been edited 3 times. Last edited by giannismanik, Dec 15, 2023, 11:01 AM
Reason: correction

Z K Y

The post below has been deleted. Click to close.

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

76 posts

#6 h Dec 15, 2023, 1:17 PM

Y by

Easy one. Let $M$ be the midpoint of $AX$ . Evidently, $\triangle CXM \sim \triangle CBA$ (derived by $\left(\frac{AB}{BX}\right)^2 = \frac{AS}{SX} = \left(\frac{AC}{CX}\right)^2$ , which yields $AB \cdot XC = AC \cdot BX$ , further proved by Ptolemy's Theorem).
Thus, $2 \frac{XC}{AX} = \frac{XC}{XM} = \frac{CB}{AB}$ .

In $\triangle ABP$ , we clearly have $\angle APB = 180° - \angle BAP - \angle BPA = 180° - (90° - \angle ABC) - (90° - \angle ACB) = 180° - \angle BAC$ . Hence, $\frac{AP}{\cos \angle ACB} = \frac{AP}{\sin \angle ABP} = \frac{AB}{\sin \angle APB} = \frac{AB}{\sin \angle BAC}$ .

Therefore, $\frac{\frac{XC}{CN}}{\frac{XA}{AP}} = \frac{\frac{2XC}{TC}}{\frac{XA}{AP}} = \frac{\frac{2XC}{2r \cos \angle ACB}}{\frac{XA}{AP}} = \frac{\frac{2XC}{XA}}{\frac{2r \cos \angle ACB}{AP}} = \frac{\frac{2XC}{XA}}{\frac{2r \sin \angle BAC}{AB}} = \frac{\frac{2XC}{XA}}{\frac{CB}{AB}} = 1$ .

Thus, $\frac{XC}{CN} = \frac{XA}{AP}$ , given that $\angle XCN = \angle XAP$ , we have $\triangle XNC \sim \triangle XPA$ , and hence $\angle XNT = \angle XTP$ , which implies $TPNX$ is cyclic.

Z K Y

The post below has been deleted. Click to close.

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

177 posts

#7 h Dec 16, 2023, 8:36 AM

Y by

Siddharth03 wrote:

Hint

Here's my solution:

Let's do $\sqrt{bc}$ inversion at $T$ in triangle $TBC$ and let's use $'$ to denote the transformed points. We need to show that $P',X',N'$ are collinear.

Note that $N'$ becomes the reflection of $T$ in $B$ . Also, as $ABXC$ is harmonic, we have that $A',X'$ are points on $BC$ satisfying $(B,C;A',X')=-1$ .
Observe that as $\measuredangle P'CT = \measuredangle BPT = \measuredangle BAC = \measuredangle BTC$ we have that $P'C\parallel BT$ and hence as $B$ is the midpoint of $TN'$ , $(P'B,P'C;P'T,P'N') = -1$ .
So, as $(P'B,P'C;P'A',P'X')=-1$ , we get that $P',X',N'$ are collinear 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.

42 posts

#8 h Dec 16, 2023, 9:47 AM

Y by

A very simple solution: Triange ABP ~ Triange ACT then we can ABxCT=APxBC, We get BCxAX=2ABxCX,then Triange XNC ~ Triange XPA

Z K Y

The post below has been deleted. Click to close.

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

1980 posts

#9 h Apr 21, 2025, 5:56 AM

Y by

Very cool cross ratio trick

danil_e wrote:

Problem. Let $ABC$ be an acute-angled triangle with $AC > AB$ , let $O$ be its circumcentre. The line through $A$ perpendicular to $BC$ intersects circle $(O)$ again at $T$ . The tangents at $B$ and $C$ of $(O)$ intersect at $S$ . $AS$ intersects $(O)$ at $X \neq A$ . $OB$ intersects $AT$ at $P$ . Let $N$ be the midpoint of $TC$ .
Prove that $T, P, N, X$ are concyclic.

The map which sends every point $Y \in \overline{AP}$ to a point $Y' \in \odot(ABC)$ such that $TNYY'$ is cyclic and $Y' \ne T$ unless tangency, is projective: upon inversion at $T$ it becomes a perspectivity between lines. Thus since $A \mapsto A, \overline{AP}_{\infty} \mapsto C$ and we want to show $P \mapsto X$ , it suffices to show that $M \mapsto B$ where $M$ is the midpoint of $AP$ , since $(AP; M \infty)=-1$ and $(AX; BC)=-1$ . Now $\triangle BPA \sim \triangle BTC$ hence $\triangle BMP \sim \triangle BNT$ so $\angle BMT=\angle BNT$ , proving $B, M, N, T$ cyclic which implies the claim.

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a