omg theyre out!! ok ok something something cyclic something something radical axis theorem

This is really trivial even for a #1.

Solution

This is the first geometry problem where I felt no need to write a summary for in a long while. This is also the first geometry problem where a terrible diagram was enough to solve in a long while.

Solution

Weird and easy.

solution

spartacle wrote: Weird and easy.

solution

madlad

why are people's solutions so strange lol, a simple angle chase kills

dchenmathcounts wrote: why are people's solutions so strange lol, a simple angle chase kills you omitted the best part

Oh hey, this would fit perfectly in a lecture for my Geo 2 class in a few days! Note that \[ \angle EAC = \angle ABC = \angle EDC \]so quadrilateral $EADC$ is cyclic. Then the three requested lines are pairwise radical axes of $\odot(BDFE)$, $\odot(ABCF)$, and $\odot(AECD)$, hence concurrent. [Though there are some minor config issues to note: since D lies on segment BC, E lies on the same side of AB as C, yielding the first angle equality.]

a1267ab wrote: Let $\Gamma$ be the circumcircle of $\triangle ABC$. Let $D$ be a point on the side $BC$. The tangent to $\Gamma$ at $A$ intersects the parallel line to $BA$ through $D$ at point $E$. The segment $CE$ intersects $\Gamma$ again at $F$. Suppose $B$, $D$, $F$, $E$ are concyclic. Prove that $AC$, $BF$, $DE$ are concurrent.

Storage

jj_ca888 wrote: dchenmathcounts wrote: why are people's solutions so strange lol, a simple angle chase kills you omitted the best part oh yeah, and radical axes djmathman wrote: Oh hey, this would fit perfectly in a lecture for my Geo 2 class in a few days! Note that \[ \angle EAC = \angle ABC = \angle EDC \]so quadrilateral $EADC$ is cyclic. Then the three requested lines are pairwise radical axes of $\odot(BDFE)$, $\odot(ABCF)$, and $\odot(AECD)$, hence concurrent. [Though there are some minor config issues to note: since D lies on segment BC, E lies on the same side of AB as C, yielding the first angle equality.] Completely agreed wrt this being a wonderful teaching problem - there's generally "wishful thinking" involved in the motivation of a lot of Radical Center problems and this is a super easy problem that also shows that off. wait i think the config issues self-resolve b/c everything lies on the same side? Yea maybe "issues" was the wrong word -- it's just worth noting that nothing fails because the first angle chase would break if $\color{red}{E}$ were not in the place it were. And admittedly I knew the problem had to do with radical axes so I was kinda spoiled on that, but it still doesn't change the fact that this is a modern Olympiad problem with a really simple radical center solution! ~dj

I don't care if this has been already posted Let $G = AA \cap BC$ We have that $\measuredangle{ACB}=\measuredangle{GAB}=\measuredangle{AED} \implies \text{AECD is cyclic }$ Conclude by radical axes on $\odot(ABC)$ ,$\odot(BFDE)$ and $\odot(AECD)$ .

Firstly notice that the condition that $(BFDE)$ is cyclic is equivalent to $\overline{CA}=\overline{CB}$ since $\angle BAC=\angle BFC=180^\circ -\angle BFE=180^\circ-\angle BDE=\angle ABC$ $\qquad \square$ Now notice that by radical axis theorem observe that if $G=\odot(ABD)\cap \odot(AEF)$ then $G \in \overline{AC}$. Now clearly since $\overline{CA}=\overline{CB}$ hence we must have that $G \in \overline{BF}$. Now to show that $G \in \overline{DE}$ we just need to show that there exists a spiral similiarity centred at $A$ s.t. $\overline{BD} \mapsto \overline{EF}$ but this is trivial since $\angle ABD=\angle ABC=\angle AFE$ and $\angle DAB=\angle GBA=\angle FBA=\angle EAF$. $\qquad \blacksquare$

Proof without words:

Attachments:

If $BDFE$ is cyclic, then \[ \measuredangle{CBA}=\measuredangle{BDE}=\measuredangle{BFE}=\measuredangle{BFC}=\measuredangle{BAC} \]so $\triangle{ABC}$ is $C$-isosceles. In particular, $AB\parallel DE\parallel CC$. In what follows, let $P=AC\cap DE$ and $Q=AC\cap BF$.

Approach 1 (Ping-Pong Lemma)

Approach 2 (Pascal)

Approach 3 (Angle Chasing)

Approach 4 (Radical Axis)

Approach 5 (Bary)

Approach 6 (Complex)

These approaches are listed in the order that I found them. And yes, approaches 1 and 2 are equivalent.

Radical axis

oops radical axis exists The condition $BDEF$ cyclic is equivalent to $\angle ABC = \angle EDC = \angle BFC$ so $AC = BC$. Let $P_{\infty}$ be the point at infinity along $\overline{AB}$; take a projective transformation fixing $\Gamma$ and sending $AC \cap BF$ to the center of $\Gamma$. Then $CEAP_{\infty}$ becomes a parallelogram and the conclusion is obvious.

An alternate finish without the use of Radical Axis. $G = ED \cap BF$. We prove that $G \in AC$ $\angle EDB = CAE = B \Longrightarrow AECD$ is cyclic. $\angle AGB + \angle CGB = C + \angle CED + 180 - C - \angle FBC = 180$ $A, G, C$ are collinear hence, proving the statement.

a1267ab wrote: Let $\Gamma$ be the circumcircle of $\triangle ABC$. Let $D$ be a point on the side $BC$. The tangent to $\Gamma$ at $A$ intersects the parallel line to $BA$ through $D$ at point $E$. The segment $CE$ intersects $\Gamma$ again at $F$. Suppose $B$, $D$, $F$, $E$ are concyclic. Prove that $AC$, $BF$, $DE$ are concurrent. Nice Problem I'm sad not to solve it at the actual contest Solution: $Claim$: $ADCE$ is cyclic $Pf$ : $\angle ABC= \angle EDC = \angle EAC$ We have that $ABCF$ and $EBDF$ are cyclic quadrilaterals Now, the Radical Axis for $\odot (ADCE)$ and $\odot (ABCF)$ is $AC$ The Radical Axis for $\odot (ADCE)$ and $\odot (EBDF)$ is $ED$ The Radical Axis for $\odot (ABCF)$ and $\odot (EBDF)$ is $BF$ Therefore, by Radical Axis theorem in $\odot (ADCE)$ , $\odot (EBDF)$ and $\odot (ABCF)$ we have that $AC$ , $ED$ and $BF$ are concurrent And we won a pizza

Well, this is surprisingly very easy even for problem 1. We just can observe that $\angle EDC=\angle CBA=\angle CAE$ since $DE\parallel AB$ and EA is tangent to $\Gamma$. We conclude that $DCEA$ is cyclic. Now we can apply the radical axis theorem on the circles $(ADCE)$, $(ABC)$ and $(BDFE)$. And we are done!

easy but nice synthetic Note $\angle EDC=\angle ABC = \angle EAC,$ implying $CDAE$ is cyclic, then by radical axis on $(CDAE), (ABCF),$ and $(BDFE),$ it follows $AC,BF,DE$ concur.

Just notice that $\angle CAE = \angle CBA = \angle CDE$. Thus $A,E,D,C$ are concyclic, which implies the desired concurrency.

Solution

Most prolly the easiest geo in APMO ever. $A E$ is tangent to $(\triangle A B C) \Rightarrow \angle E A C=\angle A B C= \angle E D C \Rightarrow AECD$ is concyclic. $AC,BF,DE$ are pairwise radical axes, hence they are concurrent at the radical center of $(\triangle A B C),(\triangle A C E),(\triangle B D E)$.

Ah, is this is legendary G0? Let $X = ED \cap BF$ then $\angle EXF = \angle ABF = \angle EAF$ so $X = ED \cap (EAF)$. Let $Y = ED \cap AC$ then $\angle FEY = \angle FBC = \angle FAC = \angle FAY$ so $Y = ED \cap (EAF)$. Hence $X = Y$ so $DE, BF, AC$ concur.

$AC \cap DE=K$ $\angle FAC=\angle FBC=\angle FBD=\angle FED \implies A,E,F,K$ are cyclic. $\angle ABC=\angle AFE=\angle AKE \implies A,B,K,D$ are cyclic and $DK \parallel AB$ so $ABDK$ is an isosceles trapezoid. Let $\angle ACE=\alpha$ $\angle EAF=\angle EKF=\alpha \implies \angle AKF=\angle B +\alpha$ $\angle EAC=\angle EDC=\angle B \implies A,D,C,E$ are cyclic. $\angle ADE=\alpha \implies \angle ADC=\angle B+\alpha,\angle AKB=\angle BDA=180-\angle B-\alpha$ So $B,K,F$ are collinear.

very good problem and little bit easy for p1 in APMO Let $$AC \cap DE=M$$and we have power $$AE^2=EF.EC$$from angle chasing $$\angle EAC=\angle EMA$$it means $$AE=ME$$and $$ME^2=EF.EC$$so we found $$\angle MFC=\angle BAM$$and $ABCF$ cyclic so $$B,M,F$$collinear

Did not spot radical axis -_- Let $P = AC \cap DE$. The following angle chase \[ \angle BAC = \angle BFC = 180 - \angle BFE = 180 - \angle EDB = \angle EDC = \angle ABC \]shows that $\triangle ABC$ is isosceles. Since $PD \parallel AB$, $ABDP$ is an isosceles trapezoid and in particular $\angle PBC = \angle DAC$. Furthermore, since $\angle EAC = \angle ABC = \angle EDC$, $AECD$ is cyclic. We have $\angle DAC = \angle DEC$. Hence, \[ \angle PBC = \angle DAC = \angle DEC = \angle DEF =\angle FBC, \]so $F,P,B$ are collinear.

The only angle-based observation we make is that \[ \angle EAC = \angle ABC = \angle EDC, \]so $ADCE$ is cyclic. Thus we conclude by the radical axis theorem on $(ABC)$, $(BDFE)$, and $(ADCE)$, whence the pairwise radical axes are $AC$, $BF$, $DE$.

By (degenerate) Reim on collinear points $B, D, C$ and $E, F, C$ we have $\overline{AB} \parallel \overline{DE} \parallel \overline{CC}$. Done by Pascal's on $AACCFB$.

$\angle EAC = \angle ABC = \angle EDC$ so $EADC$ is cyclic, $AC$, $BF$ and $DE$ turn out to be radical axises so they concur at the radical center

We define $X$ as $BF \cap DE$. Claim: $X$ lies on $AC$. Proof: Notice that $\angle ACB=\angle AEX$, and $\angle BAX=\angle AXE$. Thus we must have $\angle EAX=\angle ABC$. However $\angle ABC=\angle EAC$. Thus we have $\angle EAX=\angle EAC$ implying $A,X,C$ are collinear. Our proof is thus complete.

bruh so trivial i thought i fakesolved Note $\angle EDC = \angle ABC = \angle EAC,$ so $E, A, D, C$ are concyclic, and then radical axis theorem on the three circles $(ADCE), (AFCB), (BDFE)$ implies the result.

mannshah1211 wrote: bruh so trivial i thought i fakesolved Note $\angle EDC = \angle ABC = \angle EAC,$ so $E, A, D, C$ are concyclic, and then radical axis theorem on the three circles $(ADCE), (AFCB), (BDFE)$ implies the result. me too I'm surprised that everyone found such a trivial solution but the solution on the website is like 10 lines (with no reasoning) Click here to see the solution on the website

$\angle{EAC}=\angle{ABC}=\angle{EDC}\implies A,E,C,D$ are concyclic. $DE, BF, AC$ are radical axes of $(ADCE),(BDFE); (BDFE), (ABC); (ABC), (ADCE)$, they are concurrent by radical axis theorem.