Đầu năm đi đó đi đây

Đầu năm nước Việt thì đầy bài thi…

(Đúng! Năm nào cũng thi chọn đội tuyển Việt Nam vào ngày mồng hai DL. Đây chỉ là bài duy nhất chưa có bài giải sau ba ngày thi).

This is the only problem without solution three days after completion of competition.

Problem 3 of Vietnam Team Selection Test 2014

ABC is a triangle with AB < AC and is inscribed in circle Γ with center O. Let I be the midpoint of arc BC that does not contain A. K is a point on AC that is different from C such that IK = IC. The line BK cuts the circle Γ and AI at D and E, respectively, with D different from B.  The line DI cuts the line AC at F.

a) Prove that EF = ½BC.

b) On DI pick point M such that CM || AD. The line KM cuts BC at N. The circumcircle of triangle BKN cuts Γ at P, with P different from B. Prove that the line PK passes through the midpoint of AD.

Solution by BBT

a) Extend IK to meet circle Γ at Q. Denote (Ω) the length of arc Ω. Since I is the midpoint of the minor arc BC, BI = CI = KI and both BIK and CIK are isosceles triangles with bases BK and CK, respectively. As a result, ∠IBK = ∠IKB and ∠ICK = ∠IKC. Furthermore, because angle ∠IBK subtends (DI) whereas angle ∠IKB subtends (BI) and DQ, and with BI = CI, we have CD = DQ.

Similarly, with ∠ICK = ∠IKC and angle ∠ICK subtends (AI) whereas angle ∠IKC subtends (CI) and AQ, and with BI = CI, we have AB = AQ. Now with (BI) = (CI), (AB) = (AQ), and (CD) = (DQ), we have (AB) + (DI) = (AD) + (BI) = ½(Γ), or ∠AEB = 90° and E is the midpoint of BK. Similarly, (AD) + (CI) = (AI) + (CD) = ½(Γ), or ∠AFD = 90° and F is the midpoint of CK. Therefore, EF = ½BC.

b) From ∠AEB = 90° and ∠AFD = 90°, AI ⊥BD, AC ⊥DI and K is the orthocenter of triangle ADI, or AD ⊥IK. However, given CM || AD, we also have CM ⊥IK. Combining with AC ⊥DI (or CK ⊥IF), M is the orthocenter of triangle CKI. As a result ∠CKN = ∠KCM.

Denote the circumcircle of triangle BKN Φ, r and Rthe radii of Γ and Φ, respectively. We will prove that the two triangles BNK and ACD are similar. Indeed, ∠NBK = ∠CBK = ∠CAD (both subtend (CD) on circle Γ), and ∠BNK = ∠BCK + ∠CKN = ∠BCK + ∠MCK (because M is the orthocenter of isosceles triangle CKI) = ∠BCA + ∠CAD (because CM || AD). Now note that ∠BCA subtends (AB) and ∠CAD subtends (CD) = ∠CKD = ∠KCD (because CDK is isosceles with base CK). The two triangles BNK and ACD have their respective angles being congruent and are similar. This result gives us AD/BK = r/R.

Furthermore, ∠BNK = ∠BPK subtends (BK) on Φ and ∠BPK also subtends (BJ) on Γ, and we also have BJ/BK = r/R, or AD/BK = BJ/BK  and AD = BJ, or AB = DJ. This implies that AJ || BD, or AJ || DK and because AI ⊥BD, AI ⊥AJ, or ∠IAJ = 90°, and I, O, and J are collinear and IJ⊥BC. This gives us ∠AIJ = ∠CBD. Therefore, AJ = CD = DK. Let line PK pass through the mid-point of AD at point L. Since AJ || DK and AJ = DK, AJDK is a parallelogram and L is the midpoint of AD.