interesting 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

No tags match your search

M

interesting problem

G H J

Reveal topic tags

geometric transformation

geometry unsolved

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.

77 posts

#1 h Nov 19, 2014, 5:00 PM • 2 Y

Y by ricarlos, Adventure10

Given $\triangle ABC$ , circumcircle $(O)$ , incircle $(I)$ . $(I)$ tangents $BC,CA,AB$ at $D,E,F$ . $DG \perp EF$ . $AO$ cuts $(O)$ at $A'$ . Prove that $A',I,G$ is collinear.

Attachments:

Z K Y

The post below has been deleted. Click to close.

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

2311 posts

#2 h Nov 19, 2014, 5:10 PM • 5 Y

Y by mathandyou, BlackSelena, ATimo, Adventure10, and 1 other user

My solution:

Let $I'$ be the reflection of $I$ in $O$ .
Let $I_a, I_b, I_c$ be three excenters of $\triangle ABC$ .

Easy to see $\triangle DEF$ and $\triangle I_aI_bI_c$ are homothetic .

Since $I, O$ is the orthocenter, nine point center of $\triangle I_aI_bI_c$ , respectively .
so we get $I'$ is the circumcenter of $\triangle I_aI_bI_c$ ,
hence $\triangle DEF \cap G \cap I$ and $\triangle I_aI_bI_c \cap A \cap I'$ are homothetic ,
so we get $GI \parallel AI'$ . ... $(1)$
Since $AIA'I'$ is a parallelogram ,
so we get $IA' \parallel AI'$ . ... $(2)$

From $(1)$ and $(2)$ we get $G, I, A'$ are collinear .

Q.E.D

Z K Y

The post below has been deleted. Click to close.

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

9767 posts

jayme

#3 h Nov 20, 2014, 5:04 AM • 1 Y

Y by Adventure10

Dear Mathlinkers,
you can look at
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=499885
and we have more.

Sincerely
Jean-Louis

Z K Y

The post below has been deleted. Click to close.

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

2311 posts

#4 h Nov 21, 2014, 8:24 AM • 6 Y

Y by buratinogigle, bgn, kaapi, Adventure10, Mango247, and 1 other user

Remark

From this probelm we can get the interesting property of incenter and orthocenter :

Let $I, H$ be the incenter, orthocenter of $\triangle ABC$ , respectively .
Let $D, E, F$ be the tangent point of $(I)$ with $BC, CA, AB$ , respectively .
Let $T$ be the projection of $D$ on $EF$ .

Then $DT$ bisect $\angle HTI$ .

Z K Y

The post below has been deleted. Click to close.

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

4145 posts

#5 h Nov 21, 2014, 2:06 PM • 5 Y

Y by kaito_shinichi, ATimo, UltraInstinct, snakeaid, Adventure10

Let $M \equiv AI \cap EF$ be the midpoint of $\overline{EF}.$ Inversion WRT $(I)$ takes 9-point circe of $\triangle DEF,$ passing through $M,G,$ into $(O)$ and takes $EF$ into the circle $\odot(IEAF)$ with diameter $\overline{AI}$ $\Longrightarrow$ $G$ goes to the 2nd intersection $X$ of $\odot(IEAF)$ with $(O),$ i.e. $I,G,X$ are collinear and $\angle AXI=90^{\circ}$ $\Longrightarrow$ $IGX$ cuts $(O)$ again at the antipode $A'$ of $A.$

Z K Y

The post below has been deleted. Click to close.

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

2311 posts

#6 h Nov 24, 2014, 2:42 PM • 4 Y

Y by buratinogigle, Adventure10, Mango247, and 1 other user

Another solution:

Let $T=(AI) \cap (ABC), X=EF \cap BC$ .

Since $(X,D;B,C)=-1$ ,
so $DG$ bisect $\angle BGC$ ,
hence we get $\triangle GFB \sim \triangle GEC$ . ... $(*)$
Since $T$ is the Miquel point of $BCEF$ ,
so we get $\triangle TBF \sim \triangle TCE$ ,
hence combine with $(*)$ we get $\frac{TE}{TF}=\frac{EC}{FB}=\frac{EG}{FG}$ . ie. $TG$ bisect $\angle FTE$

Since $I$ is the midpoint of arc $EF$ in $(TEF)$ ,
so we get $T, G, I$ are collinear and $\angle ITA=\angle GTA=90^{\circ}$ ,
hence $GI$ pass through the antipode $A'$ of $A$ in $(ABC)$ .

Q.E.D

Z K Y

The post below has been deleted. Click to close.

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

824 posts

andria

#7 h Sep 8, 2015, 6:37 PM • 4 Y

Y by NHN, kaapi, Adventure10, Mango247

mathandyou wrote:

Given $\triangle ABC$ , circumcircle $(O)$ , incircle $(I)$ . $(I)$ tangents $BC,CA,AB$ at $D,E,F$ . $DG \perp EF$ . $AO$ cuts $(O)$ at $A'$ . Prove that $A',I,G$ is collinear.

Since $G(FDBC)=-1,\angle DGF=90\Longrightarrow \angle BGF=\angle CGE\Longrightarrow \triangle BGF\sim \triangle CGE\Longrightarrow \frac{FG}{EG}=\frac{BF}{CE}=\frac{BD}{CD}$ $\bigstar$ . Let $\odot(AI)\cap \odot(\triangle ABC)=R$ . since $R$ is the center of spiral similarity that sends $BF$ to $CE$ we have that $\frac{RF}{RE}=\frac{BF}{CE}=\frac{BD}{CD}$ Hence $RD$ is bisector of $\angle BRC$ so combine with $\bigstar$ and $\triangle REF\sim \triangle RCB$ we get that $RG$ is bisector of $\triangle ABC\Longrightarrow R,G,I$ are collinear $\Longrightarrow \angle GRA=90$ .
DONE

TelvCohl wrote:

Remark

From this probelm we can get the interesting property of incenter and orthocenter :

Let $I, H$ be the incenter, orthocenter of $\triangle ABC$ , respectively .
Let $D, E, F$ be the tangent point of $(I)$ with $BC, CA, AB$ , respectively .
Let $T$ be the projection of $D$ on $EF$ .

Then $DT$ bisect $\angle HTI$ .

Let $H'$ be the reflection of $I$ in $EF$ obviously $H'$ is orthocenter of $\triangle AEF$ so $HH'$ is steiner line of complete quadrilateral $CEFB$ hence since $R$ is miquel point of $CEFB$ the reflection $R'$ of $R$ in $EF$ lies on $HH'$ . On the other hand since $RI\cap EF=T\Longrightarrow R',T,H'$ are collinear thus $H',T,H$ are collinear $\Longrightarrow \angle HTF=\angle ITE\Longrightarrow TD$ is angle bisector of $\angle ITH$ .
DONE

Attachments:

Z K Y

The post below has been deleted. Click to close.

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

695 posts

#8 h Sep 8, 2015, 7:21 PM • 2 Y

Y by Adventure10, Mango247

Let $\triangle D'E'F'$ be the orthic triangle of $\triangle DEF$ (here we are relabeling the point $G$ ). Let $OI$ cut $DD'$ at $H$ and let $H'$ be the reflection of $H$ in $EF.$

The inversion w.r.t. $(I)$ sends $A, B, C$ to the midpoints of $\overline{EF}, \overline{FD}, \overline{DE}$ , respectively. Consequently, this inversion swaps $(O)$ and the nine-point circle of $\triangle DEF.$ Thus if $N$ is the nine-point center of $\triangle DEF$ , then $I, O, N$ are collinear, implying that $H \equiv OI \cap DD'$ is the orthocenter of $\triangle DEF.$ Then a homothety with center $H$ and ratio $\tfrac{1}{2}$ shows that $IH' \parallel ND'.$ But notice that $E'F'$ and $BC$ are both antiparallel to $EF$ w.r.t. $\angle FDE.$ Therefore, $E'F' \parallel BC$ and it follows by symmetry that $\triangle D'E'F'$ and $\triangle ABC$ have parallel sides. Thus they are homothetic, and there exists a homothety that swaps $ND'$ and $OA.$ It follows that $IH' \parallel OA.$

Let $P_{\infty}, Q_{\infty}$ denote the points at infinity on $DD', OA$ , respectively. Then from $IH' \parallel OA$ and $IP_{\infty} \parallel IA$ , we have $-1 = I\left(H', H; P_{\infty}, D'\right) = \left(Q_{\infty}, O; A, ID' \cap OA\right).$ Therefore, $ID' \cap OA$ is the reflection of $A$ in $O$ , as desired. $\square$

Z K Y

The post below has been deleted. Click to close.

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

27 posts

#9 h Sep 14, 2015, 12:32 PM • 2 Y

Y by Adventure10, Mango247

TelvCohl wrote:

Remark

From this probelm we can get the interesting property of incenter and orthocenter :

Let $I, H$ be the incenter, orthocenter of $\triangle ABC$ , respectively .
Let $D, E, F$ be the tangent point of $(I)$ with $BC, CA, AB$ , respectively .
Let $T$ be the projection of $D$ on $EF$ .

Then $DT$ bisect $\angle HTI$ .

Hi. Can i see your solution for that?

Z K Y

The post below has been deleted. Click to close.

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

9767 posts

jayme

#10 h Aug 21, 2020, 12:10 PM

Y by

Dear Mathlinkers,

http://jl.ayme.pagesperso-orange.fr/Docs/68.%201.%20Incentre%20et%20orthocentres.pdf p. 21...

Sincerely
Jean-Louis

Z K Y

The post below has been deleted. Click to close.

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

125 posts

#11 h Oct 22, 2020, 9:02 AM

Y by

Pretty easy inversive solution, but took me a while to realize. Let $\Gamma$ and $\omega$ be the circumcircle and the incircle of $\triangle ABC$ , respectively. Let $A'I$ intersect $EF$ at $T$ and $\Gamma$ at $K$ . Let $M$ be midpoint of $EF$ . Since $\angle TMI=\angle IKA=90^{\circ}$ , we have that $\triangle IMT \sim \triangle IKA \implies IK\cdot IT=IM \cdot IA$ . Since $M$ is the inverse of $A$ wrt $\omega$ , $IM \cdot IA=r^2$ , so $T$ is the inverse of $K$ wrt $\omega$ . Since $\Gamma$ gets sent to the nine-point circle of $\triangle DEF$ under the inversion about $\omega$ we're done.

Z K Y

The post below has been deleted. Click to close.

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

67 posts

#12 h Jan 25, 2025, 12:49 PM • 3 Y

Y by trigadd123, HotSinglesInYourArea, Calamarul

Also easy with complex:

Take the incircle to be the unit circle and WLOG suppose $\overline{e}=f\implies ef=1$ .

We have $i=0$ , $a=\dfrac{2ef}{e+f}=\dfrac{2e}{e^2+1}$ and $o=\dfrac{2def(d+e+f)}{(d+e)(e+f)(f+d)}=\dfrac{2de(de+e^2+1)}{(e^2+1)(d+e)(de+1)}$ .

Since $O$ is the midpoint of $[AA']\implies$
$\implies a'=2o-a=\dfrac{2de(de+e^2+1)}{(e^2+1)(d+e)(de+1)}-\dfrac{2e}{e^2+1}=...=\dfrac{2e(d^2e+de^2+d-e)}{(e^2+1)(d+e)(de+1)}$ .

$G$ is the projection of $D$ on $EF\implies g=\dfrac{1}{2}(d+e+f-ef\overline{d})=\dfrac{d^2e+de^2+d-e}{2de}$

Now points $G, I, A'$ are collinear iff $\dfrac{g-i}{\overline{g-i}}=\dfrac{a'-i}{\overline{a'-i}}\iff \dfrac{g}{a}\in \mathbb{R}\iff \dfrac{\dfrac{d^2e+de^2+d-e}{2de}}{\dfrac{2e(d^2e+de^2+d-e)}{(e^2+1)(d+e)(de+1)}}=\dfrac{(e^2+1)(d+e)(de+1)}{4de^2}\in \mathbb{R}$ , which is true by conjugating, so we're done.

This post has been edited 1 time. Last edited by Double07, Jan 25, 2025, 12:49 PM

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a