Equal angles

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

Equal angles

G H J

Reveal topic tags

geometric transformation

geometry unsolved

L

G H BBookmark kLocked kLocked NReply

Source: Israel 1995

The post below has been deleted. Click to close.

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

1648 posts

#1 h May 1, 2007, 7:14 PM • 4 Y

Y by Adventure10, Mango247, and 2 other users

I found this problem in the site of Hong Kong. There is a collection with proposed problems which can be solved with inversion. I post one of them. Any other solution is also welcome.

Attachments:

Z K Y

The post below has been deleted. Click to close.

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

1145 posts

#2 h May 1, 2007, 8:04 PM • 5 Y

Y by bestwillcui1, Adventure10, Mango247, and 2 other users

Very nice problem, Stergiu!

Denote $K$ the point where the semicircle cuts the circle $O$ , and $T$ the tangenty point where $AB$ cuts the circle $O$ .

Let's extend the semircircle and denote $D$ the intersection of $AB$ with the circle $H$ .

Thus, Because $AD$ is tangent to the circle $O$ , which is also tangent to the circle $H$ , then $KT$ is bisects the angle $AKD$ . Therefore $Q$ is the midpoint of the arc $\widehat{AD}$ .

Now because $\angle{PBT}=\angle{PKT}=90$ , we have that the quadrilateral $PBTK$ is cyclic, therefore $QB \cdot QP=QT \cdot QK=QC^{2}$ (the power of point $A$ , w.r.t to circle $O$ )

But in the right triangle $APQ$ , $AB$ is the altitude, therefore $AQ^{2}=QB \cdot QP$ , therefore from the above equality we have $QC=QA$ .

Now the problem is dead, because let's note $\angle{BAC}=x$ , therefore $\angle{ACB}=90-x$ , but from $QC=QA$ , we have also $\angle{CAQ}=90-x$ , therefore $\angle{BAQ}=90-2x$ , so $\angle{BQA}=2x$ , but $\angle{PAB}=\angle{BQA}$ , therefore $\angle{PAC}=2x-x=x$ , so $AC$ bisects $\angle{PAB}$ , and thus the problem in solved!

Another way for solving the problem from where we found out that $QC=QA$ , is just considering the circle $O'$ tangent to circle $H$ , $BD$ , and $BP$ , so the construction is basicly simetric, therefore $CQ=AQ=DQ$ , and $PB$ is the bisector of $\angle{APD}$ in triangle $APD$ , therefore $C$ is the incenter of triangle $\triangle{APD}$ , so $AC$ bisects $\angle{PAB}$ .

Generalization:
Let $ABC$ be a triangle, and $D \in (BC)$ such that $\angle{BAD}=\angle{ACB}$ , and consider the circle $H$ tangent to the circumcircle of $ABC$ at $K$ , to $BD$ at $P$ , and to $AD$ at $P'$ . Show that $AP$ is the bisector of $\angle{BAD}$ .

Another enuntiation of the generalization: (***) - own
In a triangle $ABC$ , denote it's circumcircle $w$ , and consider the Thebault circles $k_{1}, k_{2}$ corresponding to the bisector $AD$ of $\angle{BAC}$ . Prove that $k_{1}$ and $k_{2}$ are tangent at $I$ , where $I$ is the incircle of $ABC$ .

Z K Y

The post below has been deleted. Click to close.

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

639 posts

M4RI0

#3 h May 1, 2007, 9:22 PM • 2 Y

Y by Adventure10, Mango247

Here's a link: http://www.mathlinks.ro/Forum/viewtopic.php?t=123139

Z K Y

The post below has been deleted. Click to close.

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

1145 posts

#4 h May 1, 2007, 9:34 PM • 2 Y

Y by Adventure10, Mango247

I guess that now you got your geometric solution, Mario!

This post has been edited 1 time. Last edited by pohoatza, Apr 24, 2008, 4:41 PM

Z K Y

The post below has been deleted. Click to close.

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

639 posts

M4RI0

#5 h May 1, 2007, 9:48 PM • 2 Y

Y by Adventure10, Mango247

Yeah

Finally after a long time, but now I wonder how can it be solved with inversion

:

Z K Y

The post below has been deleted. Click to close.

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

7054 posts

#6 h May 2, 2007, 4:43 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Very nice the Pohoatza's generalization of this cool problem !

Proof. Suppose w.l.o.g. that $B\in (OP)$ and $C\in (PB)$ . Denote : the circles $\mathcal H=C(O,R)$ and $\mathcal O=C(I,r)$ ;

the reflection $A'$ of the point $A$ w.r.t. the diameter $[PQ]$ and the incenter $S$ of the $P$ - isosceles triangle $APA'$ .

Prove easily that $QC=QA=R+\sqrt{R(R-2r)}$ $\implies$ $C\equiv S$ , i.e. the ray $[AC$ bisects the angle $\widehat{PAB}$ .

Indeed, $OC^{2}=OI^{2}-IC^{2}=$ $(R-r)^{2}-r^{2}\implies$ $\left\{\begin{array}{c}OC=\sqrt{R(R-2r)}\\\\ QC=QO+OC\end{array}\right\|$ $\implies$ $\left\{\begin{array}{c}\boxed{QC=R+\sqrt{R(R-2r)}}\\\\ QB=QC-BC\end{array}\right\|$ $\implies$

$\left\{\begin{array}{c}QB=R-r+\sqrt{R(R-2r)}\\\\ PQ=2R\ ,\ QA^{2}=PQ\cdot QB\end{array}\right\|$ $\implies$ $QA^{2}=$ $R^{2}+2R\sqrt{R(R-2r)}+\left(\sqrt{R(R-2r)}\right)^{2}$ $\implies$ $\boxed{QA=R+\sqrt{R(R-2r)}}$ .

This post has been edited 2 times. Last edited by Virgil Nicula, May 2, 2007, 9:00 PM

Z K Y

The post below has been deleted. Click to close.

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

188 posts

#7 h May 2, 2007, 7:58 AM • 4 Y

Y by Adventure10, Mango247, and 2 other users

This is wanted solution with inversion.

We use inversion with center $C$ . Then circle $O$ maps onto a line
$O^{*}$ parallel with $PQ$ , and semicircle $H$ maps into a semi circle $H^{*}$ with center
on line $PQ$ , and diameter $P^{*}Q^{*}$ which touches $O^{*}$ . $A^{*}$ is intersection of the
line $CA$ and $H^{*}$ . Line $AB$ maps into a circle that goes through
$C$ and $A^{*}$ , with center on $CB$ (because $CB\bot AB$ ) and
touches line $O^{*}$ . $B^{*}$ is the second intersection point of that
circle and $PQ$ . Now, we have two equal circles $(AB)^{*}$ and $H^{*}$
(because they have their centers on the same line, and they touch
the line parallel with the line on which their centers are), and
their intersection is $A^{*}$ , so $A^{*}P^{*}=A^{*}B^{*}$ , in other words $\angle A^{*}P^{*}C=\angle CB^{*}A^{*}$ . But $\angle CP^{*}A^{*}=\angle CAP$ and $\angle CB^{*}A^{*}=\angle CAB$ so this is the end of a proof.

Please ask if something isn't quite clear, I'm not so experienced
in writing solutions with inversion so I don't know in how many
details I should go.

Bye

Z K Y

The post below has been deleted. Click to close.

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

1648 posts

#8 h May 2, 2007, 9:03 AM • 2 Y

Y by Adventure10, Mango247

I had forgotten this problem

. In my archiev I have a synthetiacal solution. I think I have sent it to this forum in the past. It is different from those you sent. I had not a solution with inversion.Now I feel happy because we have many solutions to the problem.
Thank you all for your nice solutions. I'll try to send the next problem from this collection.

Babis

Z K Y

The post below has been deleted. Click to close.

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

349 posts

#9 h May 3, 2007, 3:01 AM • 1 Y

Y by Adventure10

This can solved with a lot of computations!

One friend say to try with inversion. I see one nice and easy solution, basically the same of SpongeBob (i think that also the same of the book 'Mathematical Olympiads Around The World, 1995/06')

Z K Y

The post below has been deleted. Click to close.

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

1270 posts

April

#10 h May 8, 2007, 4:30 AM • 3 Y

Y by drEdrE, Adventure10, Mango247

Quote:

$AB$ is a half chord perpendicular to the diameter $PQ$ of the semicircle $C(O,\,R).$ A circle $C(O',\,R')$ is inscribed, toucher $PQ$ at $C.$ Prove that: $AC$ is angle bisector of $\angle PAB$

Proof. We have:
$OO'=OC^{2}+O'C^{2}\Longrightarrow (R-R')^{2}=(QB+R'-R)^{2}+R'^{2}\\ \Longrightarrow 2\cdot QB\cdot R=(QB+R')^{2}\Longrightarrow QB\cdot QQ=(QB+BC)^{2}\Longrightarrow QA^{2}=QC^{2}\\ \Longrightarrow QA=QC\Longrightarrow \angle QCA=\angle QAC\Longrightarrow \angle CAB=\angle PAC$

Attachments:

Z K Y

The post below has been deleted. Click to close.

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

1648 posts

#11 h May 8, 2007, 7:54 PM • 2 Y

Y by Adventure10, Mango247

April wrote:

Quote:

$AB$ is a half chord perpendicular to the diameter $PQ$ of the semicircle $C(O,\,R).$ A circle $C(O',\,R')$ is inscribed, toucher $PQ$ at $C.$ Prove that: $AC$ is angle bisector of $\angle PAB$

Proof. We have:
$OO'=OC^{2}+O'C^{2}\Longrightarrow (R-R')^{2}=(QB+R'-R)^{2}+R'^{2}\\ \Longrightarrow 2\cdot QB\cdot R=(QB+R')^{2}\Longrightarrow QB\cdot QQ=(QB+BC)^{2}\Longrightarrow QA^{2}=QC^{2}\\ \Longrightarrow QA=QC\Longrightarrow \angle QCA=\angle QAC\Longrightarrow \angle CAB=\angle PAC$

April ,

thank you . I did not know this simple and nice solution. Just Pythagorian theorem and ... done.Very nice!

Babis

Z K Y

The post below has been deleted. Click to close.

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

14 posts

#12 h Jan 21, 2019, 6:45 AM • 1 Y

Y by Adventure10

https://artofproblemsolving.com/community/c6h1770754p11619140

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a