Tangent circles

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

Tangent circles

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

Source: MEMO 2022 T6

The post below has been deleted. Click to close.

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

676 posts

#1 h Sep 2, 2022, 12:04 PM

Y by

Let $ABCD$ be a convex quadrilateral such that $AC = BD$ and the sides $AB$ and $CD$ are not parallel. Let $P$ be the intersection point of the diagonals $AC$ and $BD$ . Points $E$ and $F$ lie, respectively, on segments $BP$ and $AP$ such that $PC=PE$ and $PD=PF$ . Prove that the circumcircle of the triangle determined by the lines $AB, CD, EF$ is tangent to the circumcircle of the triangle $ABP$ .

This post has been edited 1 time. Last edited by a_507_bc, Apr 2, 2023, 9:04 AM

Z K Y

The post below has been deleted. Click to close.

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

1388 posts

#2 h Sep 2, 2022, 2:03 PM

Y by

This is the best geo from this exam, in my opinion. Here goes my solution (more like a sketch):
Let $CD$ intersect $AB$ at $R$ , $EF$ meets $AB$ at $T$ and $CD$ meets $EF$ at $S$ . Obviously $DCEF$ is a cyclic isosceles trapezoid; let $X$ be its center. We claim that $X$ is the desired tangency point. Note that $\angle EXF= 2\angle ECF= \angle EPF$ , so $X$ is on $(PEF)$ and $XE=XF$ . By a well-known spiral config (since $BE=AF$ ), we have that $X$ lies on $(ABP)$ as well (and $X$ is on the perpendicular bisector of $AB$ ). In addition, note that $SX$ is the bisector of $\angle DSF$ . Thus, in order to prove $X$ is on $(SRT)$ , we need to prove that $X$ is on the perpendicular bisector of $RT$ , or equivalently, $RB=AT$ , but this can be derived easily with a bit of a length bash using the equal segments: twice Menelaus for $\triangle ABP$ (which I will skip). Now, the tangency easily follows since the triangles $AXB$ and $RXT$ are isosceles, done.

This post has been edited 3 times. Last edited by VicKmath7, Sep 2, 2022, 2:17 PM

Z K Y

The post below has been deleted. Click to close.

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

176 posts

#3 h Sep 3, 2022, 7:55 AM

Y by

Clearly, $DCEF$ is a cyclic isosceles trapezoid. Let $T$ be its center and denote its circumcircle $\omega$ . Furthermore, let $Q=CD\cap EF$ , $X=AB\cap EF$ , and $Y=AB\cap CD$ .

By symmetry, we must have $Q$ , $P$ , and $T$ to be colinear. Note that $\angle FTC=2\angle FDC=2\angle QDF=\angle FYC$ so $T\in (FYC)$ . Also, $\angle DTC=2\angle DFC=\angle DPC$ implying $T\in (DPC)$ . We conclude that $T$ is the Miquel point of complete quad $\{QF,FP,PD,DQ\}$ . Now from the condition, $\mathrm{pow}(A,\omega)=AF\cdot AC=BE\cdot BD=\mathrm{pow}(B,\omega)$ so $TA=TB$ . It follows that $\bigtriangleup ATF\cong\bigtriangleup BTE$ by sss. Now we have $\bigtriangleup EFT\sim\bigtriangleup BTA$ so $\angle ATB=\angle FTE=\angle FPE=\angle APB$ , hence $T\in (APB)$ . $T$ is now the Miquel point of complete quad $\{YA,AP,PD,DY\}$ so $T\in (XYQ)$ . Finally, we must have $\angle YXT=\angle TQC=\angle TQE=\angle TQX=\angle TYX$ so $\bigtriangleup XTY$ is isosceles. We conclude that the circumcircles of $\bigtriangleup APB$ and $\bigtriangleup XYQ$ are tangent at $T$ .

Z K Y

The post below has been deleted. Click to close.

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

256 posts

#4 h Sep 3, 2022, 8:12 AM • 1 Y

Y by mkomisarova

Enjoyable problem.

Define the following points $AB \cap CD=X$ , $EF \cap CD=Y$ and $EF \cap AB =Z$ . Moreover, let $T$ be the point on $\odot(ABP)$ such that $AT=TB$ and $T$ lies on the same side as $P$ with respect to the line $AB$ . We proceed in several steps.

Step 1: Quadrilateral $EFCD$ is cyclic. Moreover, $T$ is the center of $\odot(EFCD)$ .
Proof: Since $PC=PE$ and $PD=PF$ , it is easy to see that quadrilateral $ECDF$ is isosceles trapezoid; therefore a cyclic quadrilateral. Since $AC=BD$ , we can deduce that:
$BE = BD -PE-PD = AC-PC-PF =AF$ Now note that this combined together with $AT=TB$ and $\angle PBT = \angle PAT$ gives us $\triangle FAT = \triangle EBT$ . This implies 2 things. First of all, $\angle BET = \angle AFT \implies \angle PET = \angle PFT$ , which implies that quadrilateral $EPTF$ is cyclic. Second of all, $TE=TF$ as the corresponding elements in equal triangles. Suppose that $\angle EPF = \angle ETF = 2\alpha$ . Since $PF=PD$ , then $\angle PDF = \angle EDF = \alpha$ . Therefore; $\angle ETF = 2\angle EDF$ and $TE=TF$ . This implies that $T$ is the centre of $\odot(EFCD)$ and that points $Y,P,T$ are collinear as they lie on the perpendicular bisector of $FD$ .

Step 2: $TX=TZ$ .
Proof: From Menelaus theorem on the $\triangle ABP$ and the points $E,F,Y$ we have:
$1=\frac{AZ}{BZ} \cdot \frac{BE}{PE} \cdot \frac{PF}{AF} \implies \frac{BZ}{AZ} = \frac{PF}{PE}$ Similarly, from the Menelaus theorem on the $\triangle ABP$ and the points $C,D,X$ we have:
$1 = \frac{AX}{BX} \cdot \frac{BD}{PD} \cdot \frac{PC}{AC} \implies \frac{AX}{BX} = \frac{PD}{PC}$ We conclude that:
$\frac{AX}{BX} = \frac{BZ}{AZ} \implies \frac{AB}{BX}=\frac{AB}{AZ} \implies BX=AX$ .
Now remember that $AT=BT$ and $\angle TBA = \angle TAB \implies \angle TAZ = \angle TBX$ meaning that $\triangle TAZ = \triangle TBX \implies TZ=TX$ as needed.

Step 3: Quadrilateral $XYTZ$ is cyclic.
Proof: Note that $\angle EPF = \angle ETF=\angle ATB$ . Moreover, $AT=TB$ and $ET=TF$ , meaning that $\angle TEZ = \angle TBZ$ . This gives us that quadrilateral $ETZB$ is cyclic; therefore $\angle BZT = \angle PET = \angle PFT = \beta$ . Note that $\angle CTF = 180^{\circ}-2\beta$ . Since $T$ is the center of $\odot(ECDF)$ , then $\angle CTF = 2\angle CDF \implies \angle CDF = \angle ECY = 90^{\circ}-\beta$ . Consequently, we have that
$\angle FYT = \angle DYT = \angle PET = \angle XZT = \angle TXZ$ which means that quadrilateral $XYZT$ is cyclic as desired.

This finished the problem because $\triangle ATB$ and $\triangle XTZ$ are isosceles therefore they obviously their circumcircles are tangent to each other at the point $T$ .

Attachments:

This post has been edited 1 time. Last edited by Kimchiks926, Sep 3, 2022, 8:16 AM
Reason: typo

Z K Y

The post below has been deleted. Click to close.

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

91 posts

#6 h May 9, 2023, 11:30 AM

Y by

solution

Attachments:

Z K Y

The post below has been deleted. Click to close.

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

82 posts

gambi

#7 h Aug 25, 2023, 10:21 PM

Y by

Let $K$ be the center of the Spiral Similarity mapping $BE$ to $AF$ . It then holds $K\in (FPE)$ .
$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(24.636363636363637cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -3.9472727272727264, xmax = 20.68909090909091, ymin = -4.256363636363638, ymax = 6.216363636363635; /* image dimensions */ pen ttffqq = rgb(0.2,1.,0.); pen ffttcc = rgb(1.,0.2,0.8); pen ffwwqq = rgb(1.,0.4,0.); /* draw figures */ draw((4.1,2.18)--(7.64,0.72)); draw((6.28,1.98)--(4.065842881782181,-1.14421322189209)); draw((4.1,2.18)--(4.065842881782181,-1.14421322189209)); draw((4.065842881782181,-1.14421322189209)--(7.64,0.72)); draw((7.64,0.72)--(6.28,1.98)); draw((6.28,1.98)--(4.1,2.18)); draw((4.118521610378965,3.9825461550900787)--(4.1,2.18)); draw((4.065842881782181,-1.14421322189209)--(4.0473212714032165,-2.9467593769821643)); draw((4.0473212714032165,-2.9467593769821643)--(5.54254346298305,2.663231791648056)); draw((4.118521610378965,3.9825461550900787)--(6.28,1.98)); draw((6.28,1.98)--(5.283638032759958,1.6918329017430684)); draw((7.64,0.72)--(4.8061707280050046,-0.09959808294209078)); draw(circle((2.2492427280505414,0.5367348942123785), 3.9201809633626055), ttffqq); draw((6.169216759995953,0.49645618539245184)--(5.54254346298305,2.663231791648056)); draw(circle((4.463986805369431,0.5139778475906414), 1.705319971970585), ffttcc); draw(circle((6.532515637088907,3.056373531023778), 2.585568194915991), ffwwqq); /* dots and labels */ dot((4.1,2.18),linewidth(2.pt) + dotstyle); label("$A$", (3.78,2.270909090909089), NE * labelscalefactor); dot((7.64,0.72),linewidth(2.pt) + dotstyle); label("$C$", (7.816363636363637,0.6345454545454531), NE * labelscalefactor); dot((6.28,1.98),linewidth(2.pt) + dotstyle); label("$D$", (6.270909090909092,2.1618181818181803), NE * labelscalefactor); dot((4.065842881782181,-1.14421322189209),linewidth(2.pt) + dotstyle); label("$B$", (3.68909090909091,-1.42), NE * labelscalefactor); dot((5.896607281177382,1.4390263755596107),linewidth(2.pt) + dotstyle); label("$P$", (5.561818181818183,1.2890909090909075), NE * labelscalefactor); dot((4.8061707280050046,-0.09959808294209078),linewidth(2.pt) + dotstyle); label("$E$", (4.907272727272729,-0.32909090909091054), NE * labelscalefactor); dot((5.283638032759958,1.6918329017430684),linewidth(2.pt) + dotstyle); label("$F$", (4.907272727272729,1.5072727272727258), NE * labelscalefactor); dot((5.54254346298305,2.663231791648056),linewidth(2.pt) + dotstyle); label("$Z$", (5.707272727272729,2.7618181818181804), NE * labelscalefactor); dot((4.118521610378965,3.9825461550900787),linewidth(2.pt) + dotstyle); label("$X$", (3.7072727272727284,3.8890909090909074), NE * labelscalefactor); dot((4.0473212714032165,-2.9467593769821643),linewidth(2.pt) + dotstyle); label("$Y$", (3.7254545454545465,-2.783636363636365), NE * labelscalefactor); dot((6.169216759995953,0.49645618539245184),linewidth(2.pt) + dotstyle); label("$K$", (6.234545454545456,0.5618181818181803), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

Notice
$BE=BD-EP-PD=AC-PC-FP=AF$ Hence such Spiral Similarity is indeed a rotation, meaning that $KE=KF$ . Also,
$2\measuredangle FDE=2\measuredangle FDP=\measuredangle FPE=\measuredangle FKE$ These two results combined mean that $K$ is precisely the circumcenter of $CDFE$ , and so $\triangle KEF\sim \triangle KDC$ .

By the definition of $K$ , we get that $YBEK$ is cyclic.

We also get that $K$ is the center of the Spiral Similarity mapping $BA$ to $EF$ , so
$\triangle KBA\sim \triangle KEF\sim \triangle KDC$
Thus $K$ is the center of the Spiral Similarity mapping $BA$ to $DC$ , and so $AXCK$ and $BXDK$ are cyclic. Since quadrilaterals $BXDK$ and $YBEK$ are both cyclic, then $K$ is also the center of the spiral similarity mapping $BE$ to $XZ$ , from where we get that $K\in (XYZ)$ .

The first definition of $K$ was as the center of the Spiral Similarity mapping $BE$ to $AF$ , so $K\in (ABP)$ .

Finally, let $\ell$ be the line tangent to $(ABPK)$ at $K$ . Then,
$\measuredangle KBA-\measuredangle (\ell,KX)= \measuredangle (\ell,KA)-\measuredangle (\ell,KX)= \measuredangle XKA=\measuredangle XCA \\ =\measuredangle ZCP \\ =\measuredangle KPC-\measuredangle KZC \\ =\measuredangle KBA-\measuredangle KYX$
We get $\measuredangle (\ell,KX)=\measuredangle KYX$ , so $\ell$ is tangent to $(KYX)$ . This way we have proven that circles $(ABP)$ , $(XYZ)$ are both tangent at $K$ and the common tangent line is $\ell$ .

Z K Y

The post below has been deleted. Click to close.

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

Amiralizakeri2007

45 posts

Amiralizakeri2007

#8 h Apr 11, 2024, 9:44 PM

Y by

Quite straightforward!
Simple observation that $AF=BE$ motivates us to Let $P$ be the center of spiral sim taking $BE$ to $AF$ , but we do this a little bit differently:

Note that $DCEF$ is an isosceles trapezoid by POP and congruency.
Let us denote $O$ to be the center of the $DCEF$ .
Note that $\angle FOE=2\angle FCE=\angle EPF$ , thus $FOPE$ and similarly $CPOD$ are cyclic. Also since $T,P,O$ all lie on the perpendicular bisector of $CE$ , we get that $O,P,T$ are collinear.

Since: $\angle OEF=\angle OPF=\angle ODT$ $\implies$ $DOET$ is cyclic

Also since $OE=OF$ , $AF=EB$ , $\angle OFA=\angle OEB= 180-\angle OFP$ we have $\triangle OFA \cong \triangle OEB$ thus $\angle AOB=\angle APB$ thus $O$ is the center of spiral sim taking $BE$ to $AF$ as desired from the start.
Now we have: $\angle OBA=\angle OEF=\angle ODT=\angle ODR$ $\implies$ $DRBO$ is cyclic.
Since $QBRTDE$ is complete by Miquel point theorem $(EDT),(DBR),(QTR)$ concur; Since $DOET$ and $DOBR$ are cyclic wed get that $(QTR)$ passes through $O$ .

Obviously $O$ $\in$ $(APB)$ as proven before, Let the line tangent to $(APB)$ at $O$ be $\ell$ .
Finally $\angle PO\ell=\angle PBO=\angle DBO=\angle DRO=\angle TRO$ $\implies$ $\ell$ is tangent to $(QTR)$ at $O$ and we are done.

Attachments:

This post has been edited 1 time. Last edited by Amiralizakeri2007, Apr 11, 2024, 9:45 PM

Z K Y

The post below has been deleted. Click to close.

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

CrimsonBlade273

61 posts

CrimsonBlade273

#9 h Dec 26, 2024, 2:36 PM

Y by

Fantastic problem! Solved with math_holmes15

Solution

Z K Y

The post below has been deleted. Click to close.

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

8 posts

#10 h Dec 26, 2024, 2:40 PM

Y by

Was really fun to solve

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a