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# Synthetic Geometry Group - Alan Yan's Proposal

Note that although these questions are marked as "Synthetic", you are allowed to solve them with any method you want.

1. (Medium Difficulty) Let $$ABC$$ be a triangle and let $$\omega$$ be its incircle. Denote by $$D_1$$ and $$E_1$$ the points where $$\omega$$ is tangent to sides $$BC$$ and $$AC$$, respectively. Denote by $$D_2$$ and $$E_2$$ the points on sides BC and AC, respectively, such that $$CD_2 = BD_1$$ and $$CE_2 = AE_1$$, and denote by $$P$$ the point of intersection of segments $$AD_2$$ and $$BE_2$$. Circle $$\omega$$ intersects segment $$AD_2$$ at two points, the closer of which to the vertex A is denoted by Q. Prove that $$AQ = D_2P$$.

2. (Easy Difficulty) Let triangle $$ABC$$ satisfy $$2BC = AB + AC$$ and have incenter I and circumcircle $$\omega$$. Let D be the intersection of $$AI$$ and $$\omega$$ (with A, D distinct). Prove that I is the midpoint of AD.

3. (Hard Difficulty) Let the incircle $$\omega$$ of triangle $$ABC$$ touch $$BC, CA,$$ and $$AB$$ at $$D , E$$ and $$F$$, respectively. Let $$Y_1 , Y_2, Z_1, Z_2$$ and $$M$$ be the midpoints of $$BF , BD, CE, CD$$ and $$BC$$, respectively. Let the intersection of $$Y_1Y_2$$ and $$Z_1Z_2$$ be $$X$$. Prove that $$MX \perp BC$$.

If you want solutions posted, just comment down below.

Note by Alan Yan
1 year, 1 month ago

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Q2. It is clear that $$ABDC$$ is cyclic. $$\angle DBC = \angle DAC = \angle DAB = \angle DCB$$. So, $$\Delta DCB$$ is isosceles and $$DC = DB$$. But, $$\angle DIB = \angle DBI$$(by some angle chasing). So, $$DI = DB =DC$$. But, since $$ABDC$$ is cyclic, using ptolemy's theorem we get $$AD = 2DB = 2 DI$$, which implies that $$AI = ID$$. So, $$I$$ is mid-point of $$AD$$. · 1 year, 1 month ago

Does anyone want solutions? · 1 year, 1 month ago