Problem 2. Let $\Omega$ and $\Gamma$ be circles with centres $M$ and $N$, respectively, such that the radius of $\Omega$ is less than the radius of $\Gamma$. Suppose $\Omega$ and $\Gamma$ intersect at two distinct points $A$ and $B$. Line $MN$ intersects $\Omega$ at $C$ and $\Gamma$ at $D$, so that $C, M, N, D$ lie on $MN$ in that order. Let $P$ be the circumcentre of triangle $ACD$. Line $AP$ meets $\Omega$ again at $E\neq A$ and meets $\Gamma$ again at $F\neq A$. Let $H$ be the orthocentre of triangle $PMN$.

Prove that the line through $H$ parallel to $AP$ is tangent to the circumcircle of triangle $BEF$.

Solution 1Solution 2Solution 3Solution 4

Solution 1 I claim that the tangency point is the point $X$ such that $AMXN$ is a parallelogram.

First note that $\angle CAD > 90^\circ$, so we do not run into any configuration issues.

Claim 1: $CE \parallel AD$ and $DF \parallel AC.$
Proof: It suffices to show that $\angle ACE = 180^\circ - \angle CAD.$ Note that $\angle CEA = \angle CBA = 90^\circ - \angle ACD$ and $\angle CAE = 90^\circ - \angle ADC$ by known circumcenter facts. This gives $\angle ACE = 180^\circ - \angle CAD,$ so $CE \parallel AD.$ Similarly, $DF \parallel AC,$ as claimed.

Claim 2: $BX$ bisects $\angle EBF.$
Proof: Since $AB \perp BX,$ it suffices to show that $AB$ is the external bisector of $\angle EBF,$ or that $\angle ABE + \angle ABF = 180^\circ.$ This just follows because $\angle ABE = \angle ACE = 180^\circ - \angle CAD$ and similarly $\angle ABF = \angle CAD.$

Claim 3: $M,E,X$ are collinear.
Proof: We have $MX \parallel AN$ by definition, so it suffices to show that $ME \parallel AN.$ This follows because $\angle MEA = 90^\circ - \angle ACE = \angle CAD - 90^\circ$ and $\angle EAN = \angle FAN = 90^\circ - \angle ADF = \angle CAD - 90^\circ.$

Similarly, $N,F,X$ are collinear.

Claim 4: $BEXF$ is cyclic.
Proof: We have that $\angle FXB = \angle XNM = \angle AMN$ since $XN \parallel AM$, and $\angle FEB = 180^\circ - \angle AEB = \angle ACB = 2 \angle ACD.$ So we must show that $\angle AMN = 2\angle ACD,$ which follows by sum of angles in $\triangle AMC.$

It now suffices to show that $XH \parallel AP.$ Reflecting across the midpoint of $MN,$ this is equivalent to showing that the circumcenter of $\triangle PMN$ lies on line $AP.$ This is true because $\angle MPA = 90^\circ - \angle CAP = \angle ADC = \angle NPY,$ where $Y$ is the foot of the altitude from $P$ to $MN.$ Thus we are done.
Solution 2 Let \( K = AB \cap MN \), and let \( S \) be the midpoint of arc \( MN \) in \( (AMN) \).
Since \( \ell \parallel EF \), it is enough to prove that the ratio of the altitudes from \( B \) and \( H \) to \( EF \) is the same as \( AK \) and the altitude from \( S \) to \( MN \).

Since \( AK \) and \( AP \) are the altitude and the line through the circumcenter, we get
\[ \angle KAP = \angle ACD - \angle ADC. \]
Note that \( \angle MNP = 90^\circ - \angle ADC \) and \( \angle NMP = 90^\circ - \angle ACD \), and in triangle \( PMN \), \( PA \) and \( PH \) are the altitude and the line through the circumcenter, respectively (since \( PM \perp AC \), \( PN \perp AD \), and they are the midpoints of \( AC \) and \( AD \), forming a cyclic quadrilateral with \( A \) and \( P \); then, by homothety at \( P \), \( AP \) passes through the circumcenter of \( \triangle PMN \)). So:
\[ \angle APH = \angle PMN - \angle PNM = \angle ACD - \angle ADC. \]
Hence, the ratio of the altitudes from \( B \) and \( H \) to \( EF \) is \( \frac{AB}{PH} \).
But note that \( \frac{AB}{PH} = \frac{2AK}{PH} \).

Now note that \( S \) is the circumcenter of \( \triangle PMN \), so the altitude from \( S \) to \( MN \) is \( \frac{PH}{2} \).
Then
\[ \frac{AK}{\frac{PH}{2}} = \frac{2AK}{PH} \]as desired.
Solution 3 Claim $: MEPC$ and $NFPD$ are cyclic.
Proof $:$ Note that $\angle MCP = \angle MAP = \angle MAE = \angle MEA$. we prove the other part with same approach.
Claim $: BEF$ and $AMN$ are similar.
Proof $:$ Note that $\angle AMN = \angle ACB = \angle BEF$ and $\angle ANM = \angle ADB = \angle BFE$.

Now note that $\angle NME = \angle EPC = \angle APC = 2\angle ADC = \angle ADB = \angle ANM$. similarly $\angle MNF = \angle AMN$ so if $ME,NF$ meet at $A'$ then $AMA'N$ would be a parallelogram.
Claim $: BEA'F$ is cyclic.
Proof $:$ Note that $\angle EA'N = 180 - \angle EMN - \angle FNM = 180 - \angle ADB - \angle ACB = 180 - \angle ANM - \angle AMN = \angle MAN = \angle EBF$ so $\angle EBF + \angle EA'F = 180$.

Now note that $\angle A'FE = \angle NFA = \angle NAP = \angle NDP = \angle CDP = \angle DCP = \angle MCP = \angle MAP = \angle MEA = \angle A'EF$ so $A'EF$ is isosceles triangle and so proving $A'H \parallel AP$ would suffice. Let $K$ be midpoint of $MN$ and Let $PMN$ meet $AP$ at $T$.
Claim $: T$ is antipode of $P$ in $PMN$.
Proof $:$ Note that $TMN = \angle TPN = \angle APN = \angle ACD$ so $TM \parallel AC$. similarly $TN \parallel AD$ and since $PM \perp AC$ and $PN \perp AD$, we would have $\angle PMT = \angle PNT = 90$.

Now note that it's well-known that $H$ is reflection of $T$ w.r.t $K$. also earlier we proved $AMA'N$ is parallelogram so $A'$ is also reflection of $A$ w.r.t $K$ so $AT \parallel A'H$ which proves what we desired..
Solution 4

Complex bash with $\triangle ACD$ inscribed in the unit circle, so that


$$|a|=|c|=|d|=1$$

$$p=0$$

Then $B$ is the reflection of $A$ over line $CD$, so


$$b = c + d - \frac{cd}a$$

Then


$$m = \frac{ac(b\overline{b}-1)}{b-a-c+ac\overline{b}} = \frac{a(c+d)}{a+d}$$

$$n = \frac{ad(b\overline{b}-1)}{b-a-d+ad\overline{b}} = \frac{a(c+d)}{a+c}$$

And then we find
$$e = m + a^2\overline{m} - a = \frac{a(ad+c^2)}{c(a+d)}$$

$$f = n + a^2\overline{n} - a = \frac{a(ac+d^2)}{d(a+c)}$$

Now we find the coordinate of $H$. Since $HM \perp NP$, we have


$$\frac{h-m}{n-p} \in i\mathbb{R}$$

$$\frac{h-\frac{a(c+d)}{a+d}}{\frac{a(c+d)}{a+c}} \in i\mathbb{R}$$

$$\frac{(a+c)(ah+dh-ac-ad)}{a(a+d)(c+d)} = -\frac{d(a+c)(ac\overline{h}+cd\overline{h}-c-d)}{c(a+d)(c+d)}$$

$$ach+cdh-ac^2-acd = -a^2cd\overline{h}-acd^2\overline{h}+acd+ad^2$$

$$ch(a+d) + acd\overline{h}(a+d) = a(c+d)^2$$

$$\overline{h} = \frac{a(c+d)^2 - ch(a+d)}{acd(a+d)}$$

Similarly, since $HN \perp MP$, we have


$$\overline{h} = \frac{a(c+d)^2 - dh(a+c)}{acd(a+c)}$$

We intersect these to get


$$\frac{a(c+d)^2 - ch(a+d)}{acd(a+d)} = \frac{a(c+d)^2 - dh(a+c)}{acd(a+c)}$$

$$a(a+c)(c+d)^2 - ch(a+c)(a+d) = a(a+d)(c+d)^2 - dh(a+c)(a+d)$$

$$a(c-d)(c+d)^2 = h(c-d)(a+c)(a+d)$$

$$h = \frac{a(c+d)^2}{(a+c)(a+d)}$$

Now let $O$ be the circumcenter of $\triangle BEF$. We have the vectors


$$e' = e-b = \frac{a(ad+c^2)}{c(a+d)} - \frac{ac+ad-cd}a = \frac{d(a-c)(a^2-cd)}{ac(a+d)}$$

$$f' = f-b = \frac{a(ac+d^2)}{d(a+c)} - \frac{ac+ad-cd}a = \frac{c(a-d)(a^2-cd)}{ad(a+c)}$$

Then $\overline{e'} = \frac{e'}{d^2}$ and $\overline{f'} = \frac{f'}{c^2}$. So if $o' = o-b$, we have


$$o' = \frac{e'f'(\overline{e'} - \overline{f'})}{\overline{e'}f'-e'\overline{f'}} = \frac{c^2e' - d^2f'}{c^2-d^2} = \frac{\frac{cd(a-c)(a^2-cd)}{a(a+d)} - \frac{cd(a-d)(a^2-cd)}{a(a+c)}}{c^2-d^2} = \frac{cd(a^2-cd)(d^2-c^2)}{a(a+c)(a+d)(c^2-d^2)} = -\frac{cd(a^2-cd)}{a(a+c)(a+d)}$$

and


$$o = b+o' = \frac{(a+c)(a+d)(ac+ad-cd) - cd(a^2-cd)}{a(a+c)(a+d)} = \frac{a(ac+ad+c^2+d^2)}{(a+c)(a+d)}$$

Now the line through $H$ parallel to $AP$ has equation


$$\frac{z-h}{a-p} \in \mathbb{R}$$

$$\frac{z-\frac{a(c+d)^2}{(a+c)(a+d)}}a \in \mathbb{R}$$

$$\frac{z(a+c)(a+d) - a(c+d)^2}{a(a+c)(a+d)} = \frac{acd\overline{z}(a+c)(a+d) - a^2(c+d)^2}{cd(a+c)(a+d)}$$

$$cdz(a+c)(a+d) - acd(c+d)^2 = a^2cd\overline{z}(a+c)(a+d) - a^3(c+d)^2$$

$$z = \frac{a^2cd\overline{z}(a+c)(a+d) - a(c+d)^2(a^2-cd)}{cd(a+c)(a+d)}$$

So let $Q$ be the reflection of $O$ over the line through $H$ parallel to $AP$. Then


$$q = \frac{a^2cd\overline{o}(a+c)(a+d) - a(c+d)^2(a^2-cd)}{cd(a+c)(a+d)} = \frac{a^2(ac^2+ad^2+c^2d+cd^2) - a(c+d)^2(a^2-cd)}{cd(a+c)(a+d)} = \frac{a(-2a^2cd+ac^2d+acd^2+c^3d+2c^2d^2+cd^3)}{cd(a+c)(a+d)}$$

Let $R$ be the midpoint of $OQ$, so that $R$ is the projection of $O$ onto the line through $H$ parallel to $AP$. Then


$$r = \frac{o+q}2 = \frac{acd(ac+ad+c^2+d^2) + a(-2a^2cd+ac^2d+acd^2+c^3d+2c^2d^2+cd^3)}{2cd(a+c)(a+d)} = \frac{a(-a^2+ac+ad+c^2+cd+d^2)}{(a+c)(a+d)}$$

We want to show that the circumcircle of $\triangle BEF$ is tangent to the line through $H$ parallel to $AP$ at the point $R$. Indeed,


$$o-r = \frac{a(a^2-cd)}{(a+c)(a+d)}$$

and so


$$|o-r| = |o'|$$an

d thus the distance from $R$ to $O$ equals the circumradius of $\triangle BEF$.