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Fall Mathematical Competition

Geometry Difficulty 5.4 AIME, harder Prove it Bulgaria

Let MM and NN be the midpoints of the sides ACAC and BCBC of △ABC\triangle ABC (AC>BCAC > BC) and let the bisector of ∠B\angle B intersect the segment MNMN at a point PP. The incircle of △ABC\triangle ABC has center II and is tangent to BCBC at a point QQ. Denote by RR the intersection point of the perpendiculars from PP and QQ to MNMN and BCBC, respectively, and by SS the intersection point of the lines ABAB and RNRN.

a) Prove that the quadrilateral PCQI is cyclic.
b) Express the length of the segment BS by the lengths a, b, c of the sides of △ABC\triangle ABC.

Solution

a) Obviously
⊨ABP=⊨BPN=⊨PBN=β/2. \vDash ABP = \vDash BPN = \vDash PBN = \beta/2.
Therefore BN=CN=PNBN = CN = PN, whence ⊨BPC=90∘\vDash BPC = 90^\circ. Since ⊨CQI=90∘\vDash CQI = 90^\circ, the quadrilateral PCQI is cyclic.

Figure 1

b) Answer. BS=b+c2BS = \frac{b+c}{2}.

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