I inverted about $A$ with $AC$ as radius.

See Equal angles for the synthetic solution and generalization at #2 (the solution also works for the generalization)

I took $\omega$ as the circle tangent to $AB$ at $P$, $CD$ at $Q$ and $\Gamma$ at $T$. Then, by Shooting Lemma, $AC^2=AT \cdot AQ=AP^2$, so $AT=AP$. Suppose the other tangent to $\omega$ from $A$ meets $CD$ at $K'$. Then $AC$ is equal to the semiperimeter of $\triangle ADK'$, which gives $K'=K$. Done .

Let $M$ be the midpoint of $AB$, $s$ the semiperimeter of $\triangle ADK$ and $AD = k$. Set $D$ as origin, $A = (-k, 0)$, $K$ and $C$ are on the $y$-axis. Now, \[AB\cdot AD = AC^2 \implies AB = \frac{s^2}{k}\]\[AM = \frac{s^2}{2k} \implies M = \left(\frac{s^2}{2k} - k, 0\right)\]Let $I$ be the $A$-excenter of $\triangle AKD$, $I = (s-k, s-k)$. Now, \begin{align*} |MI|^2 &= \left(\frac{s^2}{2k} - s\right)^2 + (s - k)^2\\ &= \left(\frac{s^2}{2k}\right)^2 + s^2 - \frac{s^3}{k} + s^2 - 2sk + k^2\\ &= \left|\frac{s^2}{2k} - s + k\right|^2 \\ &= |MA - (s - k)|^2 = |r - r_A|^2 \end{align*}So, done.

Let the excircle of $ \Delta ADK$ be $\omega$. Let $AD$ and $AK$ be tangent to $\omega$ at X and Y. Note that semiperimeter=AX=AY. Hence AX=AY=AC. This actually forced me to see the circle centred at A having radius AC (=AY=AX). Voila! I see orthogonal circles and I invert at A with radius AC(=AY=AX). This fixes C, Y and X and $\omega$. Note that B maps to a point on the line AB s.t if the image of B is $B^*$, then $\angle{AB^*C}=90$. But this point is unique AND it's D. So, $\Gamma$ maps to the line $CB^*$ or $CD$. But $CD$ is tangent to $\omega$ by the question. Hence, the preimage of $CD$ or $\Gamma$ is tangent to $\omega$. Reposted, don't complain dudes. Edit:Thanks to #31. It's typo-free now.

Let $CD \cap \odot (ABC) = C^*$, and $\gamma $ be A-excircle of $\Delta AK'D$ ($K' \in \overline{CD}$) which touches $\odot (ABC)$ at $T$, $AB$ at $H$, $CD$ at $T_A$ and has center at $I_A$, then it's clear that, $I_AHDT_A$ is a square and , $\angle ABC=90^{\circ}-\angle DCB=\angle ACD=\angle ABC^*$, hence, $AB$ bisects $\angle CBC^*$ Hence, $T-T_A-A \implies AH^2=AT_A \cdot AT =AC^2=AC^{*2} \implies \boxed{AH=AC}$ and recall that $AH=AD+DT_A$ $= \text{ semi-perimeter of} $ $ \Delta AK'D$ , forces, $\boxed{K\equiv K'}$

Coordinates

TheDarkPrince wrote: Coordinates How?

And, what do you mean by inverted?

Math-wiz wrote: And, what do you mean by inverted? There is this whole transformation of circles and lines that we do, and we do it about a circle. This thing is called inversion. It's a very useful transformation.

Math-wiz wrote: TheDarkPrince wrote: Coordinates How? Notice the square with $I_a$. Then 4 page bash is enough (but I'm sure messed it up in the test )

Casey Rules!! Let $T,R,S$ be the points where $\Gamma$ touches $AB,CK,AK$.We claim that $AB\cdot CR = AC\cdot BT + BC\cdot$ Let $AB,BC,CA$ be denoted by $c,a,b$.Now $CR = {CD+DR} =\frac{ab}{c} + b - c + \frac{a^2}{c}$,$BT = c - b$,$AS = b$. Therefore $AB\cdot{CR} = ab + bc - c^2 + a^2 = ab + bc -b^2 = ab + b(c-b) = AC\cdot{BT} + AS\cdot{BC}$ as claimed. Now by converse of casey's theorm on point circles $A,B,C$ and circle $\Gamma$ we are done .

Vrangr wrote: Let $M$ be te midpoint of $AB$, $s$ the semiperimeter of $\triangle ADK$ and $AD = k$. Set $D$ as origin, $A = (-k, 0)$, $K$ and $C$ are on the $y$-axis. Now, \[AB\cdot AD = AC^2 \implies AB = \frac{s^2}{k}\]\[AM = \frac{s^2}{2k} \implies M = \left(\frac{s^2}{2k} - k, 0\right)\]Let $I$ be the $A$-excenter of $\triangle AKD$ $I = (s-k, s-k)$. Now, \begin{align*} |MI|^2 &= \left(\frac{s^2}{2k} - s\right)^2 + (s - k)^2\\ &= \left(\frac{s^2}{2k}\right)^2 + s^2 - \frac{s^3}{k} + s^2 - 2sk + k^2\\ &= \left|\frac{s^2}{2k} - s + k\right|^2 \\ &= |MA - (s - k)|^2 = |r - r_A|^2 \end{align*}So, done. I too did the same thing...

Let the ex touch point on side AK , AD, KD be X, N, T respectively. We know that AC=AX=AN=s, so I take inversion at A with radius AC. Let the circle( ABC ) be denoted by ß. And the A-excircle of AKD be denoted by #. Therefore after inversion CD goes to ß and ß comes to CD. Let M belong to ß such that after inversion T gets mapped to M. Since # is tangent to CD at T, therefore after inversion #' is tangent to (CD)' which is ß, at M. So now # (which is (TNX ) and tangent at T), goes to #' which is (MNX) and tangent at M. Due to inversion we know that Angle AMX = angle AX'M' =angle AXT (since inversion fixes X and inverse of M is T) . Again , angle AXT = angle TNX ( AX is tangent to #). Therefore AMX = TMX = TNX (= AXT). This implies MXTN is cyclic . Therefore # and #' are the same circle and since #' is tangent to ß, therefore the excircle # is also tangent to ß ( at M). Please draw the diagram if you can't understand the solution.. :)my first solution in AoPS.

After inversion power of point can also be applied.

^ That is exactly what I did.

Let $\Omega$ be the bigger circle, and $\omega$ be the smaller one (the A-excircle). Let $CD$ hit $\Omega$ at $N$, let $AK$ meet $\omega $ at $P$, let $CD$ meet $\omega $ at $M$. So $AC=AN=AP$. Let $l$ be the line tangent to $\Omega$ at $A$. Since $AC=AN$, we infact have $l\parallel CN$. Thus the homothety $h:\Omega\rightarrow\omega$ takes $A $ to $M$. Let $AM$ meet $\omega$ at $X$. Now power of point says $AP^2=AM\cdot AX\implies AN^2=AM\cdot AX$. This gives $\triangle AMN\sim \triangle ANX$, whence $\angle AXN=\angle ANM=\angle ACN\implies X\in \Omega$. Thus $X$ is an intersection of the circles and $h:\Omega\rightarrow \omega$ maps $X$ to $X$, thus making $X$ the point of tangency of $\Omega$ and $\omega$. $\blacksquare$

I was about to bash but time ran out as I uselessly spent time on P4

Construct the circle tangent to $CD,AB, \Gamma$ at $M,N,P$ respectively. Since $CD$ is parallel to tangent from $A$ to $\Gamma$, it follows that homothety from $P$ which maps constructed circle to $\Gamma$ also maps $N$ to $A$, hence $A,N,P$ are collinear and if $CD \cap \Gamma=E$ then $A$ is the midpoint of arc $CE$, and $\angle ACE=\angle AEC=\angle APC$, or $AC$ is tangent to $(CNP)$ or $AC^2=AN \cdot AP=AM^2$, hence $M$ is a point on $AD$ such that $AM$ is equal to half-perimeter of $\triangle ADK$, and so $M$ is $A$-excircle touchpoint, as desired.

amar_04 wrote: Was Easy... Let $\omega$ be the $A-$ excircle of $\triangle AKD$ and let $\omega$ be tangent to $CD$ at $Q$ and let $\omega\cap AQ=T$ and let $\omega$ be tangent to $BD$ at $R$. Let $s$ be the semiperimeter of $\triangle AKD$. So, $\frac{s}{2}=AR=AC$. So, by PoP we get $AQ.AT=AR^2$. Let $AQ\cap\Gamma=T'$. Then by angle chasing we get $\angle CBA=\angle CT'A=\angle ACQ\implies AQ.AT'=AC^2=AR^2$ as $AC=AR$. Hence, $T'\equiv T$. Hence, $\omega$ and $\Gamma$ are tangent at point $T$. $\blacksquare$. Why do you think this completes the proof? Indeed, what you just proved was $A-Q-T$ are collinear. But how will you show that there is not other common point for the two circles other than $T$. Sorry if this was trivial

Jafet98 wrote: Solution. Let $E$ be the point where the $A$-excircle of $\bigtriangleup ADK$ (say $\omega$) touches $AB$. Since $AE=s_{\bigtriangleup ADK}=AC$, we discover that the circle $\gamma$ centered at $A$ with radius $AC$ passes through $E$ and is orthogonal to $\omega$. Let $f$ be the inversion respect to $\gamma$. Observe that $f(\Gamma)=CD$ and $f(\omega)=\omega$; hence, it suffices to show that $\omega$ and $CD$ are tangent to each other, which is obvious by definition of $\omega$. We are done. Why is $f(\omega)=\omega$?

math_pi_rate wrote: I took $\omega$ as the circle tangent to $AB$ at $P$, $CD$ at $Q$ and $\Gamma$ at $T$. Then, by Shooting Lemma, $AC^2=AT \cdot AQ=AP^2$, so $AT=AP$. Suppose the other tangent to $\omega$ from $A$ meets $CD$ at $K'$. Then $AC$ is equal to the semiperimeter of $\triangle ADK'$, which gives $K'=K$. Done . What is shooting lemma? P.S Sorry for triple post

Red-Apple wrote: Jafet98 wrote: Solution. Let $E$ be the point where the $A$-excircle of $\bigtriangleup ADK$ (say $\omega$) touches $AB$. Since $AE=s_{\bigtriangleup ADK}=AC$, we discover that the circle $\gamma$ centered at $A$ with radius $AC$ passes through $E$ and is orthogonal to $\omega$. Let $f$ be the inversion respect to $\gamma$. Observe that $f(\Gamma)=CD$ and $f(\omega)=\omega$; hence, it suffices to show that $\omega$ and $CD$ are tangent to each other, which is obvious by definition of $\omega$. We are done. Why is $f(\omega)=\omega$? Because $\omega$ is orthogonal to the circle of inversion.

Red-Apple wrote: math_pi_rate wrote: I took $\omega$ as the circle tangent to $AB$ at $P$, $CD$ at $Q$ and $\Gamma$ at $T$. Then, by Shooting Lemma, $AC^2=AT \cdot AQ=AP^2$, so $AT=AP$. Suppose the other tangent to $\omega$ from $A$ meets $CD$ at $K'$. Then $AC$ is equal to the semiperimeter of $\triangle ADK'$, which gives $K'=K$. Done . What is shooting lemma? P.S Sorry for triple post If a circle A is tangent to a circle B at a point P and a chord C of A is also tangent to B at Q then P, Q, and midpoint of arc C are collinear.

Let $\odot(ABC)=\omega$ and $A-\text{excircle}$ of $\triangle AKD$ be $\gamma$. Let $AR\cap\gamma=T$ and $AR\cap\omega=T'$. Note that $AC=AP=AQ$, hence a circle with Center at $A$ and with radius $AC$ passes through $\{P,Q\}$. Now by an Inversion $\Psi_{\odot{AC}}$ we get that $\Psi_{\odot{AC}}: R\mapsto T$ as $\gamma$ and $\omega$ are orthogonal. Again notice that $\Psi_{\odot{AC}}:R\mapsto AR\cap \omega= T'$ as $\omega$ passes through the Center of the Circle of Inversion. Hence, $$\begin{cases} \Psi_{\odot{AC}}:R\mapsto T \\ \Psi_{\odot{AC}}:R\mapsto T'\end{cases}\implies T'\equiv T$$ Hence, $\omega$ and $\gamma$ are internally tangent at the point $T$.

Consider the circle with center $A$ and radius $AC$. Obviously, this circle passes through $C,M,N,C'$ where $A,D,C'$ are collinear and $C'\in (ABC)$ and $M,N$ are contact points of excircle with $AK,AD$. Now, invert about this circle. Clearly, excircle maps to itself (dur to orthogonality) and $(ABC)$ maps to $BC$ with is tangent to excircle. QED

It's easy to show that $K$ is unique (fix $C$, then the excircle gets bigger as $K$ approaches $C$). Let $K'$ be the point where the excircle of $ADK'$ opposite $A$ is tangent to $\Gamma$. By a well-known lemma, $AC$ equals the tangent from $A$ to a circle inscribed in the segment created by line $CD$. But it's also well-known that the tangent from a vertex $A$ to the excircle opposite $A$ of triangle $AXY$ has length $s$, where $s$ is the semi perimeter of $AXY$. This immediately finishes the problem, since $K'$ works and it also satisfies the conditions of $K$.

Take $A=(-1,0),O=(0,0),C=(\cos\alpha,\sin\alpha),B=(1,0),D=(\cos\alpha,0)$ Computing,$AC=\sqrt{(\cos\alpha+1)^2+\sin^2\alpha}=\sqrt{2+2\cos\alpha}=2\cos\frac{\alpha}{2}$ By property of excentre, we have $OX=AX-OX=AC-OX=2\cos\frac{\alpha}{2}-1$ and, $TX=DX=OX-OD=2\cos\frac{\alpha}{2}-1-\cos\alpha$ By Pythagoras theorem, $OT=\sqrt{OX^2+XT^2}=\sqrt{\left(2\cos\frac{\alpha}{2}-1\right)^2+\left(2\cos\frac{\alpha}{2}-1-\cos\alpha\right)^2}$ We need to prove that this is equal to $OT=R-r=1-\left(2\cos\frac{\alpha}{2}-1-\cos\alpha\right)$ Squaring and expanding both sides, this is equivalent to, $4\cos^2\frac{\alpha}{2}=2+2\cos\alpha$,which is trivial.

Another Solution. Let $P$ be the touch point of the excircle on $AB$. Let $r$ be the radius of the excircle and $R$ be the radius of the bigger circle. WLOG let $AC=AP=1$. Let $O_1$ be the center of the excircle. We need to prove that $OO_1^2=(R-r)^2$. Applying Pythagoras on $\triangle OO_1P$, we get that $ OO_1^2=r^2+R^2+1-2Rr$ hence we need to prove that $2Rr-2R+1=0$. Let $\angle OAC=x$ also, $\angle DOC= 180-2x$ Then, $AD= \cos x$, $DO=-R \cos 2x$, $R=AO=AD+DO= \cos x + -R \cos 2x \implies R= \frac{\cos x}{1+ \cos 2x}$ Also, $r=AP-AD=1- \cos x$ $\implies 2Rr-2R+1= \frac{2\cos x (1 - \cos x)}{1+ \cos 2x}- \frac{2 \cos x}{1+ \cos 2x}+1 = 0$ After a simple computation.

Well,this looks like someone has combined EGMO Lemma 2.17 and 4.33/8.15 and rotated the resultant figure by a right angle .Although HBCSE in those days didn't knew thousands had already swallowed evan chen's book lol.

Consider the inversion $\Psi$ which is centered about the circle $(A,AC)$. Then $\Psi(C)=C$,$\Psi(B)=D$ and $\Psi(A)=P_{\infty}$. So $\Psi((ABC))=CD$ Let the $A$-excircle of $\Delta ADK$ be $\Gamma$. Now we need to show $\Psi(\Gamma)$ is tangent to $\Psi((ABC))$ I.e. to $CD$. Now let $AK$ and $AD$ be tangent to $\Gamma$ at $L$ and $M$,respectively. We know $AL=AM=s_{\Delta ADK}$. Hence by the given condition, $L,M \in (A,AC)$ This proves $\Gamma$ is orthogonal with respect to the circle of inversion. Inversely speaking, $\Psi(\Gamma)=\Gamma$ which is clearly tangent to $CD$, hence the result. $\blacksquare$

How come they put such a trivial problem in an MO?

Easy Let $K'$ be a point such that incircle of $ADK'$ is tangent to $\Gamma$. Let incircle of $ADK'$ be $\Omega$. Let $\Omega$ touch $CD$ at $T$, touch $AK'$ at $S$ and touch $\Gamma$ at $P$. It's well known that $P,T,A$ are collinear and it's also well known that $AC = AS$ and also we know $AS$ equals semi perimeter of $ADK'$ so we have $ADK$ and $ADK'$ have same perimeter so $K$ is $K'$. we're Done.

Too easy for P5 if you know that AC=AT, where T is tangency point of excircle ADK and AD, then inversion around A with radii AC.