Solution:
This solution uses almost exclusively complex numbers. As in other solutions, we reduce to showing that \(H\) , \(S:= (D H M)\cap (E H N)\) , and \(T:= B M\cap C N\) are collinear; this is all of the synthetic information we shall use. (If one computes \(T = \frac{3A + D + E}{5}\) using means other than complex numbers, the solution becomes much shorter.)

We use complex numbers with \(A = a\) , \(D = d\) , and \(E = e\) on the unit circle. Note that \(H = a + d + e\) , \(M = \frac{a + d}{2}\) , and \(B = 2d - e\) . We will make great use of the "arbitrary line intersection formula," which says that the intersection between lines \(W X\) and \(Y Z\) can be written explicitly as

\[\frac{(\overline{{u}} x - w\overline{{x}})(y - z) - (w - x)(\overline{{y}} z - y\overline{{z}})}{(\overline{{w}} - \overline{{x}})(y - z) - (w - x)(\overline{{y}} - \overline{{z}})}.\]

We first use this to find \(T = t\) . We compute

\[b - m = (2d - e) - \frac{a + d}{2} = \frac{3d - a - 2e}{2}\] \[\overline{{b}} - m = \frac{3a e - d e - 2a d}{2a d e}\] \[\overline{{b}} m - b\overline{{m}} = \frac{2e - d}{d e}\cdot \frac{a + d}{2} -(2d - e)\cdot \frac{a + d}{2a d}\] \[\qquad = \frac{a(a + d)(2e - d) - (2d - e)(a + d)e}{2a d e} = \frac{(a + d)(2a e + e^{2} - a d - 2d e)}{2a d e}.\]

If \(\mathcal{E}\) is some expression, we use the notation \(\mathcal{E} - \{\sim \}\) to denote \(\mathcal{E}\) minus the expression formed by swapping \(d\) and \(e\) in \(\mathcal{E}\) . We now have

\[t = \frac{(\overline{{b}}m - b\overline{{m}})(c - n) - \{\sim\}}{(\overline{{b}} - \overline{{m}})(c - n) - \{\sim\}}\] \[\quad = \frac{(a + d)(2a e + e^{2} - a d - 2d e)(3e - a - 2d) - \{\sim\}}{(3a e - d e - 2a d)(3e - a - 2d) - \{\sim\}}\] \[\quad = \frac{[a^{3}(d - 2e) + a^{2}(3d^{2} - 7d e + 5e^{2}) + a(2d^{3} - d^{2}e - 3d e^{2} + 3e^{3}) + d e(4d^{2} - 8d e + 3e^{2})] - \{\sim\}}{[a^{2}(2d - 3e) + a(4d^{2} - 11d e + 9e^{2}) + d e(2d - 3e)] - \{\sim\}}\] \[\quad = \frac{a^{3}(3(d - e)) - a^{2}(2(d^{2} - e^{2})) + a(-(d^{3} - e^{3}) + 2d e(d - e)) + d e(d^{2} - e^{2})}{a^{2}(5(d - e)) - a(5(d^{2} - e^{2})) + d e(5(d - e))}\] \[\quad = \frac{3a^{3} - 2a^{2}(d + e) - a(d^{2} - d e + e^{2}) + d e(d + e)}{5(a^{2} - a(d + e) + d e)}\] \[\quad = \frac{(a - d)(a - e)(3a + d + e)}{5(a - d)(a - e)} = \frac{3a + d + e}{5}.\]

(The factorization in the last line can be motivated by noting that the expression, while cubic in \(a\) , is only quadratic in \(d\) . When written out as a polynomial in \(d\) , each coefficient is divisible by \(a - e\) ; by symmetry, the numerator is divisible by \(a - d\) as well, and the factorization follows.)

Computing \(S = s\) is slightly harder, as it is the intersection of two circles rather than of two lines. We get around this by noting that \(\{h,d,m,s\}\) are concyclic if and only if \(\{0,h - d,h - m,h - s\}\) are concyclic, which happens if and only if \(\{\frac{1}{h - d},\frac{1}{h - m},\frac{1}{h - s}\}\) are collinear. (One can see this by inversion, or just by writing out the cross- ratio in the special case when one of the points is zero.) Thus \(\frac{1}{h - s}\) is the intersection of the line through \(w:= \frac{1}{h - d}\) and \(x:= \frac{1}{h - m}\) and the line through \(y:= \frac{1}{h - e}\) and \(z:= \frac{1}{h - n}\) . We compute

\[w - x = \frac{1}{a + e} -\frac{1}{\frac{a + d}{2} + e} = -\frac{a - d}{(a + e)(a + d + 2e)}\] \[\overline{w - x} = \frac{ae^{2}(a - d)}{(a + e)(2ad + ae + de)}\] \[\overline{w} x - w\overline{x} = \frac{ae}{a + e}\cdot \frac{2}{a + d + 2e} -\frac{1}{a + e}\cdot \frac{2ade}{2ad + ae + de}\] \[\qquad = 2ae\frac{(2ad + ae + de) - d(a + d + 2e)}{(a + e)(a + d + 2e)(2ad + ae + de)}\] \[\qquad = \frac{2ae(a - d)(d + e)}{(a + e)(a + d + 2e)(2ad + ae + de)}.\]

Using the line intersection formula, we have

\[\frac{1}{h - s} = \frac{(\overline{w} x - w\overline{x})(y - z) - \{\sim\}}{(\overline{w} - \overline{x})(y - z) - \{\sim\}}\] \[= \frac{[\frac{2ae(a - d)(d + e)}{(a + e)(a + d + 2e)(2ad + ae + de)}] - \{\sim\}}{[\frac{ae^{2}(a - d)}{(a + e)(2ad + ae + de)}] - \{\sim\}}\] \[= 2(d + e)\frac{[e(ad + 2ae + de)] - \{\sim\}}{[e^{2}(a + d + 2e)(ad + 2ae + de)] - \{\sim\}}\] \[= 2(d + e)\frac{[a(de + 2e^{2}) + de^{2}] - \{\sim\}}{[a^{2}(de^{2} + 2e^{3}) + a(d^{2}e^{2} + 5de^{3} + 4e^{4}) + de(de^{2} + 2e^{3})] - \{\sim\}}\] \[= \frac{2(d + e)(a(2(e^{2} - d^{2})) + de(e - d))}{(a^{2} + de)(2(e^{3} - d^{3}) + de(e - d)) + a(4(e^{4} - d^{4}) + 5de(e^{2} - e^{2}))}\] \[= \frac{2(d + e)(2a(d + e) + de)}{(a^{2} + de)(2d^{2} + 3de + 2e^{2}) + a(d + e)(4d^{2} + 5de + 4e^{2})}.\]

Since \(h - t = \frac{2a + 4d + 4e}{5}\) , this gives us

\[\frac{h - s}{h - t} = \frac{5}{4}\cdot \frac{(a^{2} + de)(2d^{2} + 3de + 2e^{2}) + a(d + e)(4d^{2} + 5de + 4e^{2})}{(d + e)(2ad + 2ae + de)(a + 2d + 2e)}.\]

It is easy to see that this is real by using the symmetry of the expressions (both the numerator and denominator satisfy \(\mathcal{E} = a^{2}d^{3}e^{3}\overline{\mathcal{E}}\) . We conclude that \(H\) , \(S\) , and \(T\) are collinear, as desired.


Let \(A B C\) be an acute triangle. Points \(B\) , \(D\) , \(E\) , and \(C\) lie on a line in this order and satisfy \(B D =\) \(D E = E C\) . Let \(M\) and \(N\) be the midpoints of \(A D\) and \(A E\) , respectively. Let \(H\) be the orthocentre of triangle \(A D E\) . Let \(P\) and \(Q\) be points on lines \(B M\) and \(C N\) , respectively, such that \(D\) , \(H\) , \(M\) , and \(P\) are concyclic and \(E\) , \(H\) , \(N\) , and \(Q\) are concyclic. Prove that \(P\) , \(Q\) , \(N\) , and \(M\) are concyclic.

The orthocentre of a triangle is the point of intersection of its altitudes.

Solution:
As usual, we reduce to proving \(T = (3A + D + E) / 5\) is on the radical axis and compute; this time in Cartesian coordinates.

Let \(A(0,h)\) , \(D(b,0)\) , \(E(c,0)\) be coordinates for \(ADE\) . Then \(H(0, - \frac{b c}{h})\) , \(M(\frac{b}{2},\frac{h}{2})\) , \(N(\frac{c}{2},\frac{h}{2})\) and \(T(\frac{b + c}{5},\frac{3h}{5})\) . We compute \(O_{D}(x_{D},y_{D})\) the circumcentre of \(DMH\) and obtain \(O_{E}\) by symmetry. We then have to verify that \(O_{D}O_{E} \perp HT\) , which can be done by comparing slopes.

The centre \(O_{D}\) can be given by the intersection of perpendicular bisectors of \(DM\) and \(DH\) . This gives the following system of equations on \(x_{D},y_{D}\) :

\[h\cdot x_{D} + c\cdot y_{D} = \frac{b h}{2} -\frac{b c^{2}}{2 h}\] \[-b\cdot x_{D} + h\cdot y_{D} = -\frac{3 b^{2}}{4} +\frac{h^{2}}{4}\]

Solving the system gives

\[(h^{2} + b c)x_{D} = \frac{3 b^{2}c}{4} -\frac{b c^{2}}{2} +\frac{h^{2}b}{2} -\frac{h^{2}c}{4}\] \[(h^{2} + b c)y_{D} = -\frac{h b^{2}}{4} +\frac{h^{3}}{4} -\frac{b^{2}c^{2}}{2 h}\]

The formulas for \(x_{E}, y_{E}\) will be the same, swapping \(b \leftrightarrow c\) by symmetry; thus \(y_{D} - y_{E}\) and \(x_{D} - x_{E}\) will be antisymmetric in \(b, c\) and divisible by \(b - c\) , and explicitly:

\[(h^{2} + b c)(x_{D} - x_{E}) = \frac{b - c}{4} (5b c + 3h^{2})\] \[(h^{2} + b c)(y_{D} - y_{E}) = -\frac{b - c}{4} h(b + c)\]

So \(\frac{y_{D} - y_{E}}{x_{D} - x_{E}} = -\frac{h(b + c)}{5b c + 3h^{2}}\) .

The other slope is more immediate:

\[\frac{y_{T} - y_{H}}{x_{T} - x_{H}} = \frac{3h / 5 + b c / h}{(b + c) / 5} = \frac{5b c + 3h^{2}}{h(b + c)} = -\frac{x_{D} - x_{E}}{y_{D} - y_{E}}\]

so indeed the two slopes correspond to perpendicular lines.

## Day 2


Let \(A B C\) be an acute triangle with incentre \(I\) and \(A B \neq A C\) . Let lines \(B I\) and \(C I\) intersect the circumcircle of \(A B C\) at \(P \neq B\) and \(Q \neq C\) , respectively. Consider points \(R\) and \(S\) such that \(A Q R B\) and \(A C S P\) are parallelograms (with \(A Q \parallel R B\) , \(A B \parallel Q R\) , \(A C \parallel S P\) , and \(A P \parallel C S\) ). Let \(T\) be the point of intersection of lines \(R B\) and \(S C\) . Prove that points \(R\) , \(S\) , \(T\) , and \(I\) are concyclic.

Solution:
We will prove that \(\triangle B I R\sim \triangle C I S\) , since the statement then follows from \(\angle T R I = \angle B R I = \angle C S I =\) \(\angle T S I\) .

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Step 1. Let us prove \(\angle R B I = \angle S C I\) . We will use directed angles:

\[(B R,B I) = (B R,A B) + (A B,B I) = (A Q,A B) + (B I,B C)\] \[\qquad = (C Q,B C) + (B I,B C) = (C I,C B) + (B I,B C),\]

which is symmetric in \(B,C\) . Therefore, analogously we would obtain the same expression for \((C S,C I)\)

Step 2. Let us prove \(B R / B I = C S / C I\) . Clearly \(B R = A Q\) and \(C S = A P\) . Angle chasing gives \(\angle I C B = \angle Q C B = \angle A P Q\) , and similarly \(\angle P Q A = \angle C B I\) , and so \(\triangle I B C\sim \triangle A Q P\) , from which the desired \(A Q / B I = A P / C I\) follows. This finishes the solution.

Remark. In the alternative solutions below, the notation \((A B C)\) refers to the circle passing through points \(A\) , \(B\) , and \(C\) . \(D\) will refer to the midpoint of arc \(B C\) not containing point \(A\) , unless stated otherwise.

Let \(I_{A}\) be the \(A\) - excenter of \(\triangle A B C\) . It is well- known that points \(B\) , \(C\) , \(I\) , and \(I_{A}\) lie on a circle with diameter \(I I_{A}\) . The center of this circle is point \(D\) . It is clear that \(A\) , \(I\) and \(D\) are collinear.

In some of the solutions, some of the facts below may be used:

\(\cdot B,T,I,C\) are concyclic. Indeed, \(\angle B T C = \angle Q A P = A + \frac{B + C}{2} = 90^{\circ} + \frac{A}{2} = \angle B I C\) , so \(B,T,I,C\) are concyclic.

\(\angle Q A P = \angle B T C = \angle R T S\) . Indeed, since \(Q A\parallel B T\) and \(P A\parallel C T\) , then \(\angle Q A P = \angle B T C =\) \(\angle R T S\)

\(\cdot R Q\) and \(P S\) are tangents to \((A B C)\) . Indeed, \(\angle R Q B = \angle Q B A = \angle Q A B\) and similarly for \(P S\)

\(\cdot \triangle A Q P\sim I B C\) (see Solution 1 for the proof).

\(\cdot Q P\) is the perpendicular bisector of \(A I\) . Indeed, \(Q A = Q I\) and \(P A = P I\) so \(P Q\perp A I\) and \(P Q\) bisects \(A I\) .


Solution:
We use complex numbers, with \((A B C)\) as the unit circle. Set \(D = d\) , \(P = p\) , \(Q = q\) , so that \(A = a = - \frac{p q}{d}\) , \(B = b = - \frac{d q}{p}\) , and \(C = c = - \frac{d q}{q}\) . Write \(z\sim w\) if \(z / w\) is a nonzero real number. We observe that

\[\frac{R - T}{S - T}\sim \frac{R - B}{S - C} = \frac{Q - A}{P - A} = \frac{q + \frac{p q}{d}}{p + \frac{p q}{d}} = \frac{(d + p)q}{(d + q)p}.\]

So, it suffices to show that \(\frac{I - R}{I - S}\sim \frac{(d + p)q}{(d + q)p}\) . Indeed,

\[I - R = (d + p + q) - (Q + B - A) = d + p + \frac{d q}{p} - \frac{p q}{d} = (d + p)\left(1 + \frac{(d - p)q}{d p}\right) = \frac{(d + p)(d p + d q - p q)}{d p},\]

so

\[\frac{I - R}{I - S} = \frac{\frac{d + p}{d p}}{\frac{d + q}{d q}} = \frac{(d + p)q}{(d + q)p}.\]


Solution:
In the following, all segment notations denote vectors.

As mentioned above, we find \(\triangle A Q P\sim \triangle I B C\) , and by definitions of the parallelograms we have \(B R = A Q\) and \(C S = A P\) as well as \(\angle R T S = \angle Q A P\) , so it suffices to show \(\angle R I S = \angle Q A P\) . From the similarity \(\triangle A Q P\sim \triangle I B C\) , we have a spiral map \(\lambda\) such that \(I B = \lambda A Q\) and \(I C = \lambda A P\) . It follows that \(I R = I B + B R = (\lambda +1)A Q\) and \(I S = I C + C S = (\lambda +1)A P\) . Because \(\lambda +1\) is also a spiral map, we have \(\triangle I R S\sim \triangle A Q P\) and in particular \(\angle R I S = \angle Q A P\) , as we wanted to show.

Remark. This solution is deeply related to the complex numbers solution; indeed, the vectors can be interpreted as complex numbers and the spiral map as a complex scalar multiplication. But it only relies on the additive structure of the complex numbers as a real plane and the linear map acting on them (rather than, e.g., multiplying two points together), making vectors a slightly more natural language for the claims.

Remark. The number 1 in the solution above represents the identity map.


Solution:
Let \(E\) , \(F\) , \(G\) be the midpoints of \(A I\) , \(B Q\) , \(C P\) . As in Solution 1, angle chase shows that \(\triangle A Q P\sim\) \(\triangle I B C\) .

Note that by the Mean Geometry Theorem we have that \(\frac{1}{2} A Q P + \frac{1}{2} I B C = E F G\) is similar to \(\triangle I B C\) . Homothety with center \(A\) and scale- factor 2 maps \(E F G\) to \(I R S\) . Hence \(\angle R I S = \angle F E G = \angle Q A P =\) \(\angle B T C = \angle R T S\) , so \(R,T,I,S\) are concyclic.

Remark. As shown above, \(E\) lies on \(Q P\) and \(A I\perp P Q\) . One can prove that \(\angle F E G = \angle B I C\) in another way. Let \(J\) be the midpoint of \(P Q\) . Then \(\angle B I C = \angle F J G\) by midlines and \(\angle F J G = \angle F E G\) by the lemma below applied in \(B C P Q\) .

Lemma. Let \(A B C D\) is a cyclic quadrilateral and \(E\) is the intersection of its diagonals. Then the midpoints of \(A B\) , \(B C\) , \(C D\) and the foot of the perpendicular from \(E\) to \(B C\) are concyclic.

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i6vp0wmXU9O4vYhhk7TL/dPv6VtUUAU9O1GHUrRZ4Tjs6HqjdwauVgajay6Zetq1ihZf+Xq3H8a/wB4e4rYtLuG9tY7iBw8bjIIoAnooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACsHVf8AS/EOl2XVYy1y4+nA/Wt6sHTP9L8S6nd9VhC2yH6cn9aAF1v9zqmjXPpcmM/Rxj+eKZFfPf3N5azX1pGqzvD5BT52Qcf3u49ql8Uo39j+ev3reVJvyYVHeabu8O6sILsRPeJLKk+MeUXXjn29aAONvvFlgtlYabE/2qSCVg1rEwZ3ZWIRDj9fYVsaTpUkVy+u6/NG2pSDC7mAS2T+4v8AU1wvhDwZdaxrcbXSTabFaxFt8MgWRtwwApH8PWvRYfhz4eR989vNdP3a4nZ8/rSabZ1Up04wtdp+n/BH3PijQrPPn6rarjsJAf5Vzlp400t9ZvruFLy8LbY4hb27N8oH4DrXY/2BoOkWkk8Ok2UYiQuW8lS3Az1PNP8ADFr5GhW7MoEkuZXwO7HNFmL2lFdG/n/wDnh4l1i6H+g+Fr9gejXDLEPxyagj1HxZeyBIRoNoWOAst35rE+wWvQMCsC5vbbSdU1a+uflht7SORsDJ4LdPftRYX1hLSMF+Zljw14ruuLvxMkKnqtrbKMfieaX/AIV9BMc3+t6vdnuDcbFP4Af1qTw38QbPxBqTWD2ctnMULxeY4YOB16dDz0rrSwz04xRYHiqvR29Ejmbf4f8Ahi2O4aVHK396ZmkP/jxNU/EOsaH4RiS3s9Ot5NSl4gtoYxuJ7E45AqDxZ46FgW07R1FxfsdhdeVjJ7e7VQ8M+GWspG1TVHNxqk3zFnOfLz2HvWigo6vfsKE5zXNOTsvx8l+pjx+Ar7VpJr3WNSMUt65kuIIYx8mR91W7HHsa0V0vxX4eAOh6wby2QYFteYbAHYE/4iuvoprQSm03b/gHMWvxOns5Rb6/pE1vIOC8XQ++D/jXW6b4t0TVcC21CPef+WcnyN+RqldWdtexGK6gjlQ9nXNcxf8AgaAkyadL5Z6+TMu9PwPUU+SEtnZico/aj93+T/Rnp4YEZzS7hXkMU+r+H2CyT3unoOjoftFsfwPK/nXS2Hi3VvJEsljb6nAOs2nyfMPqh5zUShKGrWnfoCpwm7U5Jvs9H9z/AEO6zSZrn7Hxnot44ia5+zT9PKuQY2z6c8VurIjqGVgQehBzUkzpzg7TViSqmo2EOpWj20wyrdCOqnsR7irdFBBiaVfzwXJ0nUW/0pBmKXtMnqPf1rbrP1XS49StwNxjnjO6GVeqNUOj6m90ZLS8Xyr6DiROzD+8PY0Aa1FFFAGHrfyarosvpclP++lNbgrD8SfJHp83aO9iJ/PFblADZHEcbOeigk1zuh6bb6h4ajW6iDiZ3kz0IJY8g9jWnrtx9l0S8l7iIgfU8D+dSaTbm00m0gPVIVU/XHNAGX59/oB23Re804dJgMyRD/aHce9blvcw3UCTQSLJGwyGU5FSEZGD0rDuNHnsp3vNHcRuTmS2b/Vy/wDxJ96AN2is3TtZhvnaCRGt7tP9ZBJww9x6j3FaQOaADtXOXKP4dvGvIUJ02ZszxjnymP8AGPb1ro6a6K6FHAKsMEEcGgBI5EljV0YMjDII6EU+uciZ/Dd4IJCTpUzfunP/ACwY/wAJ9q6IEYGOlAC0UUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUARzyrBbyTOcKilj9AM1k+FomXRUnf/WXLtO3/AAI/4Yo8UTNHokkUf+suGWBR6ljj+VatvCtvbxwp92NAo/AUANvYFurKe3bpJGyH8RiuTW8kvtA0/SUci4ncwSkdVRD8x/ICuyJwMmuL8PNA/jHUplQhJgTbE9Dg4cj6mgDVMaWHiy1EahYri0MQA6ZTkfpW+AKw/EeYG06+H/LC6UN/utwf6VuDpQBi+KHP9kfZkOHupUgX8Tz+gNbEUaxRJGowqKFA+lYt/wD6X4p0626pbo9ww9+grdzigArmvEjLqYk0G3jWSa7j23DHpFH6n39Kv6zqpskS3tl8y+n+WGP3/vH2FQW8dn4b0ua7v7keY3zzzv1ZvQf0FNJvRCbSVzF0z4d6Jo7SXN1NNdsqECSdwvljuRtx6da5rWNUYtJFod7exWBJi3+YW89/7kanr9a0Na1WfWYBc36y22kM2LayXia8btn0WrujaO8ci6hfogudu2KFPuW6f3VHr6mtdKfnL8v+CVTp+0XPPSH4vy9PMxtC8NalpTLevHaz3BXiNiQY8+h9fU1v/wBr3EP/AB96ZcR/7UeHH6VrUVN23dmspuXy+75GfDrmnTHaLlUb0k+U/rV9XV1DIysD0KnIqOa1guBiaGOT/eUGqDaBY53QCW3b1hkK/wD1qCDUorJ+wapB/wAe+peYB/DOgP6il+2avAP3+nxzKP4oJP6GgZqFQwIIBB7GsW78L2M0v2izMlhcjkS2zbfzHSphr9op23KT2zf9NYyB+Yq9De2tyAYbiN8+jCqjOUXdMmUYyVmjnrkavaoY9V0621y0HG9EAlA+n+FN01dLuZMaDrN3pVyD/wAe0jcZ9NjcGuqqhqGi6fqi4u7ZHbs44YfiOaf7uXxKz7r/ACHCdamrQlddnqhy6p4p00f6TaW2qQj+O3by5Mf7p4P4VbtfG+kyyeTdtNp8/wDzzu49n69KwF0/XdHwdK1AXcA6W15ycegapV8T6fcH7Hr9i1nIeNtwm6Mn2apdCW8Hf+u25X1ijLSrHlflp/wPyO6injnjEkUiSIRkMhyDWbq+myXIjurRxFfwcxv2b/ZPsa59fDGnt/pOj3c9izch7SXKH/gPQ1Klx4s0zjNrq0I9f3UuP5GsdVoy/YRlrCX36f8AA/E6HSdTXUbcllMdxGds0J6o3+FaNee3via2gvFv3trnTdQQYkjuIz5c6+m4cZ9Ca7PStXstYsUurKZZEYcgHlT6Gi6ZnOjOHxIp+Kh/xInf/nnJG/5MK2lO5QfXmsrxIA/h2+A6iPd+Rz/Sr1nIJLKB89Y1P6UzMy/EP+kPYaeP+Xi4UsP9lfmP8hW4AB0rCtT9v8TXVz1is0EEZ/2zyx/kK3aACjFFFAFDUtJttRQGQFJk5jmjOHQ+xrPj1S70mRbfWMNCThL1B8p9nHY1v0ySNJYzHIoZGGCCMg0AKjK6BlIKkZBHenVz76feaKxl0vM1p1ezc9PdD/StLTtTttSiLwOQy8PGwwyH0IoAsXVtFd27286B43GGBrFsLmXR7xNJvXLQv/x6Tt3H9w+4roAc1V1Cwh1K0a3nX5TyGHVT2I96ALVFYmk388Vy2l6iR9qjGY5O0ydiPf1rboAKKKKACiiigAooooAKKKKACiiigAooooAwdU/0vxHpdmOViLXLj6cD9TW9WDpn+leJdUu+qxbbZD9OW/Wt40AZPiG8e10xkh/4+LgiGIDrubjNUNQ07+ytMsbm1XL6dywH8aH7/wDjUyn+1PFDP1t9OXaPQyt1/IVusiyIUcZVhgg9xQBla2qal4auTEdweLzEI9uR/Krum3Iu9Ntrgf8ALSJW/HFZGjE2U9xodxkhAXtyf4oj2/Ck0K6+yeHbhZDzYvKjfRSTQBJpB+1a5q171UOtuh9lHP6mtDVdSi0yzM0nzMTtjjHV27AVn6K8emeGI7q6bZuBnkJ9WOf8BXO3etu+pJcG2NzqcgxZWI6Qr/ff0Jqoxc3ZClJRV2aL3kOgQPqmrt5up3XEcCcsPRFH5ZNY98ZG2ax4lUySFsWWlIcgMemR3b+VWWhi8PqdX1mU3+sTnbEijOGPRIx2HvT9N064luzqurEPfOMIg5WBf7q+/qa0bUPdhv3KpUuf95V+Hou4mm6bPLdf2rqpD3zDEcY+7br/AHV9/U1s0UVCVjWc3J3CiiimSFFFFABRRRQAhVWGGUEehFUZ9G0+4JL2sYb+8g2n9Kv0UAZP9ivD/wAemo3MOOisd6j8DRjW7fvbXaj1yjH+la1FAjK/tiaH/j7025j9WQeYv5ilOp6RqEZhmlhdT1jmGP0NalQzWltcDE0Eb/7yg0bBuYh8OC2cz6FqEti552Kd8TfVakTxHqmlnZreml4h/wAvVp8y/Ur2qw2gWQObcy2zesUhH6Un2HVYP9RqKyqP4biPP6jmr9pzK01f+u5n7Pl1g7f12NWz1PS9ag/0eeG4Uj5kOMj6qa53WfCttZXP9qaas1rj/XrZuUOP7wA4/Cqt9o0ksnnS6V5c45FxYS7W/LvUMXiTWdGYJcBr63/u3Eflyj8ehpeyjL4Xr2ZpDEVKe/4fqjWnHiJNGna01G31S0lhbi4TZJjB/iHBP1p1r46j07QIW1HTbu2kEIEchTfG5xx8wrFj8aaRp5llWSSKzn4ltXQ74nPdQOoPoKk8La1aeJprPT13qthEZZIpVwXOcAgdwMis5Upxu2noarEUp6SSv5aP+vkdt4Yns20iIW95BcSuTJKY3BO88nNbgOa5S88L6Tdv5v2b7PP2mt2MbD8uv41Etl4j0wf8S/V0u4h0hvkycf745qLjdGEvgl952Qork08X3NlhdZ0e5twOs0A82P68citnT9f0zVRmyvYZj/dDfN+XWi5nKhUirtaGnRSZoyaZkGKzNR0aO6lF1byG1vV+7NGOvsw7itPNLQBiWmsyQXAstXjFvcE4SUH93L9D2PtW1uzUN1Z297btBcxLJG3VWFYuL/w+cjzL3TR26ywj2/vCgDR1bS01G2ADmK4jO+GZeqNUWkam9yHtLtRHfwcSJ/eH94exq7aXlvfwLNbyrJG3cfyNUdX0xrlo7u0YRX8HMb9mH90+xoA1qKz9J1RNStydpjnjOyaI9Ub/AArQoAKKKKACiiigAooooAKKKKACoriVbe3lmb7qIWP4CpaxvFErJoskKH95cssC4/2jg/pmgBPC8TJokczj95cM07Z/2jkfpir2qXy6dps9038C/KPU9APzqe3hWC3iiXgIgUfgKxdT/wCJnrtppo5hg/0mf8Pug/jzQBc0Cyay0qMS8zykyzHuXbk1pngUgpJP9W30NAMxNXWK+RLiwuInv7U+ZGFcEsP4lPsRXMT6vBKNRtYpFT+0HicKxwVJ4f8ALFWtJjljt9BlzaspkKhYo9sgyrck5OQO/ArxfxAWk1/VGuzm7N1ICV653Hbs9sYxiuunhvaScea2hzSxHJHmtc9Vv9VvPE2qJp2irmztsL5zD5FI43H1PoK2AmneDtOaQ77i+nb7x5luHPb6fyrAhGv+G/DdntudLjMqKIoIYmeSVyB+Z9TVi28L+Iru7XVNQ1mKO9K4VRbhxCPRcnipm18EWkv68jejSv8AvaqbX5+n6s0dN024kuzq2qkPfOMIg+7br/dX39TWxWQPDustxJ4lmx/sWyLS/wDCKXLf6zxDqZP+y4X+VQowX2vzNZ1ak3dx/I1qKyf+EOif/Xatqkv1uDS/8IRpR+9JeP8A71w1P93/ADfgR+8/l/E0mmiT70iD6sKja+tEGWuYR9XFVV8FaCv3rNn/AN+Zz/WpV8IaAnTS4T/vZP8AM0r0u7C1Xshra1pafe1C1H1lFQP4l0WP72p23/AXz/KtFPDuip93SrP/AL8qf6VPHpOnRf6uxtk/3YlH9KOal5j5avkYJ8XaEOl+p/3Y3P8AIU0+LtLP+rNzL/uW7f4V04toB0hjH0UVIqKv3VUfQUc9Ls/v/wCAHJV7r7v+Ccn/AMJVbN/q7DUn+lsaP+EhuWH7rQNUcevk4rraKPaw/l/EPZz/AJvw/wCHOTGs6s/+q8N3mf8AbdV/nTvt3iRvu+HQP967T+ldVRS9qv5V+P8AmHsn/M/w/wAjlvM8VSfd0yxi/wCulwW/lR5Xi5uPL0lPxc/1rqaKPbdoofsv7zOX/s7xU/J1Gwj9lhJ/nS/2L4jk+/rsKf7lsK6eij20uy+4PYx7v7zmf+Ea1Zvv+Jbj6JAo/rQfCMsoxca9qMgPUKyrn9K6aoLu7hsbZ7idwsaDJP8ASl7ef9JB7GH9NnnXivwFaJZRGwuLmbUjKPJjmcEP65wOOO9ReFfBGsaZqf8AaF9dxac6o0cKh1cuTjOe2OOnvXc6TbS3MzareLiWQYhQ/wDLJP8AE1MFtm1TUhfCEgKmzzccRbecZ7bt1axxFVwcL6GE6FOM1NLUh/4n9t1FpeL+Mbf4Un9uSwcXul3UPqyjzF/MVqWzQvZ27W4YQmNfLD5zt7ZzzUtc2x1rVXM6DXNMuTtS7jDf3X+U/kajvNA0fVP3ktpE0naWP5W/Mc1dnsLS5GJ7aKT/AHkBrPbw5Zq261kuLVvWGUj9KVioylF3i7EC6RrGmnOla1KyDpBer5q/TdwRUq+JtWsDjVdEkKDrPZN5i/8AfPUUv2HWbcfuNTSZR/DcR/8Aswo/tDV7b/j50sSj+9byZ/Q0WLdVS+OKf5/galh4n0fUyEt76Pze8b/I4/A4NbG4YzmuGu7rw/qR2anZeVJ6zwFWH/AhRbaZJCN2g+IZY1HSGVxPH9MHkUXZPs6UvhlZ+f8Amv8AI7nIoyK5JNd1/TuNR0hbuMdZrF8n/vg81fsfF+jX0nlfahbz/wDPK5Hlt+RoTTIlh6iV0rry1JbvRnjna90qUW10eXT/AJZy/wC8P61Lp2spcSm0u4za3y9YXP3vdT3FaQdWGVII9RVW+0y11KHy7hOVOUdeGQ+oPamYlHVbGeG5Gq6cv+kxjEsQ6Tp6fX0rR0+/h1G0S4hb5W6qeqnuD71kre3uiOItSLXNn0W8VeV/3x/WmXKNp051nTj5tpKN1zFGchh/fX3oA6Oiora4iu7dJ4XDxuMqw71LQAUUUUAFFFFABRRRQAVg6n/pfiTS7TqsO65cfThf1reNYOlf6X4i1W86rEVtkP05b9cUAbU8yW8Ek0hwiKWY+wrI8OQvJbzalOMTXrmTB7J0UflTfELNdG10mJsPdv8AvMdoxy3+FbcaLGioowqjAA7CgB1Iehpc4rI1jVHtyllZASX1xxGvZB3Y+woAz79YXv8A+zNJt4Ybl1/0i4jjAMSHtkdzUmpy6T4c0qONrZJXK+VBCFDPKx7D69zTJ7i08J6UPMLXF5M3CjmS4kP+fwqrpelXEl22r6uQ+oOMIg5W3X+6vv6mhyN6VJW557fn/wAAw9KsJ9D1GO/1hAyTDbEQSy2eT9z6e9duCGUEEEHkEU2WJJomilQOjDDKRwaxVE2gPg75tMJ4PVoP8VpFTm5O7N2imRSxzxLLE6ujDIZTkGn0EhRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUVBd3kFlAZbiQIo/Mn0A7mgCSaaO3haWVwkaDLMewrEt4ZNcu1vrpCtlGc28LD7x/vn+lSRWs+sSrc38ZjtVOYbY9/8Aaf8AwraxgYHSgApkkUU2zzoYpNhyu9A236Zp9FAWTAnJyaKKKACiiigAooooAa6JIpV1VlPZhms+fQNMnO5rRFb+9GSp/StKigNDG/sS4g/48tUuYv8AZkxIP1qteWGpzxlL2z0/Uo/9pdjfrxXRUUAm07o4cWz6a260/tbSSP4UPnw/kc1etPGGq25xMtnqUY6tA/kyf98N/SuqqtcafZ3X+vtopD6sgz+dFuxp7Vv40n/Xff8AEr23jPRrsiG5ke0lfjyruMpn8eh/OnGxmss3ehyJJA/zPalso/up7H9Ko3HhPTbhGVVliU9Qr5U/gcisj/hC77T5DLo+sS27f3RlQfqOh/KjUlxpS2vH8V/n+ZrWWqQ6XcvJAHSxdv8ASLZxh7Zz3x/dNdgjh0DKQVIyCO9eVapN4jt0zqVraXhQYFxBIscoH8iPbFZuj/E2801ZLGPThdJCQf3s3llAf4RgHP8AIU7pkToShHm3R7TRVDRtTi1jSLTUYAyx3MYkCt1XPUH6VfoMQooooAKKKKAIriZbe2kmf7salj+FZXhiJotDjmk/1lwzTuT3LHOfyo8UysuiPAh/eXLrAv8AwI8/pmjWpmsdFSztuJ5gttCB2zxn8BQBHo2dQ1S81Zh8hPkW+f7ink/ia3qrWNoljYw20Y+WJAo96S+vobC0e5nbCIM+5PoKAINW1JNOtt+0yTOdkMS9XbsKxJLmPw3ZtqGoZuNVuztWNOWZuyL7CoZ79bADW9URnvZvksbNeWAPQAep7ntS6VpU8l4dX1ciTUZB8qD7tuv91ff1NJvob0qStzz2/P8A4HcXS9LuJLs6tqzCTUZB8ijlbdf7q+/qa3KKKCpzc3dhSEBgQQCD2NLRQSY02n3Omytc6Vgxk5ktGPyn3X0NW7HVbe+JjXdFOv34ZBhh/jV6qd7pltfAGVSJV+7Khwy/Q0AXKKx/M1XTf9Ygv7cfxJ8soHuO9W7PVrK94imHmd43+Vh+BoC5dooooAKKKKACiiigAooooAKKKKACis661uzt38pGM856RQjc3/1qg8jVNSH+kuLK3P8AyyjOZGHu3b8KAJbzWI4pfs1pGbu7PSOPov8AvHtTbTSnecXmpSCe56ov8EXso/rV20srexi8q3iWNe+Op+p71YoAKKKKACiiigAoozVSfU7G2BM95BHj+9IBQNJvYt0VhS+MNCjbat8JW9IUaT+QqMeKHuP+PLRNUuPQ+TsH5k0ro19hU/l+/T8zoaKwhdeKbn/UaFDbg9GubkfyXNOXTPFtx/rdR0+1HpDCXP8A48aLi9mlvJL5/wCVzbzTWdUGWYKPUnFZA8IXk/8Ax++ItQkz1EO2MfyNTR+A9DJ3XEU903rcTs/9aNRfulvK/ov87DrjX9Itc+dqNspHbzAT+Qqg3jLSGJW3a4um9LeBn/pW/beG9FtMeRpdqpHQ+WCfzNaaRpGu1FCj0AxT1F7SiujfzsccNf1G4/48/DmoP7zbYh+pNPU+Lro4Sw0+0HrNOzkf98iuv2jFAUDtRbzJ9ul8MF+L/U5IaF4kuP8Aj41+GAd1trUfzbNPHgpJeb3WtUufUedsX8hXVbRS4osg+tVOmnokc5D4I8PREE6ekzDvM7SfzNQap8P/AA/qkscjWrWroNpNq3l719Gx1/nz1rqdorM1iG4mSBICrYk3PCZDH5i46Aj86dkZTq1J/FJv5luytILC0htLWMRwQoI40HQADgVZrJ0OVWS4i8q4gkikw8Ezh/LJAOFYZyOc9T1rWoICiiigAooooAwNS/0vxLplp1WENcv+HA/WiL/iZ+J5JusGnr5aehkbqfwHFUzfCC+1vVj8whC20XuR2/76Na+i2J0/S4o5D++bMkpPdjyaANCWaOGF5JHCIi5Zm6AVxeoavFI6ardo726ttsLID5p3/vYo8Q67FO2whpLJJNixx/eu5eyL/sjuam0fSJvtP9q6qVfUGXEaL9y3Tsqj+ZpN9EdFKkmvaT2/P/gdx2laVcPdtq+rsJNQk+4nVbdf7q/1NblFFA5zc3dhRRRQSFFFFABRRRQAVTu9Lsr7m4t0ZuzgYYfiKuUUAZH9mahaf8eOolkHSK6G8fg3Wk/tLU7b/j70tpFHV7Z93/jp5rYooCxkL4k07OJjNAfSWJh/SrSaxpsgyt/b/QyAH9auMqsMMoI9xmqz6ZYynL2kDE/7AoDUet7aN926hP0kFI1/Zp966gH1kFV20PSm62EH/fApP7H0mIZNlbKB3KigNWOk1vS4h81/B9FcMf0qsfElm3FvHc3DdhFETTZL/wAOWBJaewjI7KVJ/Sof+Ex0g/LatPckdFt4Gb+mKLo0VGpLaJY+2azdf8e+nx26H+O4fJ/IUf2PPdc6lfyzDvFF8ifjjk1ANf1G4/48vDmoOOzTbYh+ppw/4S+5+5p+nWg9Zpi5H4LSuP2LXxNL5o1raytbNNttBHGO+1ev41PWKNB8TXH/AB8a9DAD1W2th/Ns1IPBSS/8fus6ncnuPO2D8gKeouWmt5/cmaMt1bwAmWeKMDqWcCs2fxTodvkPqduSO0bbz+masQ+B/D0J3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