Maths Olympiad Prep

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Geometry Difficulty 5.8 AIME, harder Prove it Japan

For a triangle ABCABC the line tangent to its circum-circle at AA and the line BCBC intersects at a point PP. Let QQ and RR be the points which are symmetrically located from PP with respect to the lines ABAB and ACAC, respectively. Prove that the lines BCBC and QRQR intersect perpendicularly.

Solution

Without loss of generality, we may assume that the points BB, CC, PP lie on the same straight line in this order. Let us denote by Q′Q' the point of intersection of the lines ABAB and PQPQ and by R′R' the point of intersection of the lines ACAC and PRPR.
If ∠ACB≠90∘\angle ACB \neq 90^\circ, we see from PQ′⊥AQ′PQ' \perp AQ' and PR′⊥AR′PR' \perp AR' that the 4 points AA, Q′Q', PP, R′R' lie on the circumference of a circle. Therefore, if we let SS be the point of intersection of the lines BCBC and Q′R′Q'R', then we get
∠BSQ′=∠BPQ′+∠PQ′S=∠BPQ′+∠PAR′=∠BPQ′+∠PBQ′=∠BQ′Q=90∘, \begin{align*} \angle BSQ' &= \angle BPQ' + \angle PQ'S \\ &= \angle BPQ' + \angle PAR' \\ &= \angle BPQ' + \angle PBQ' \\ &= \angle BQ'Q = 90^\circ, \end{align*}
which means that BC⊥Q′R′BC \perp Q'R'.
If ∠ACB=90∘\angle ACB = 90^\circ, we see that the points Q′Q' and R′R' coincide with the points AA and CC, respectively, and therefore, BC⊥Q′R′BC \perp Q'R' holds in this case as well.
Q′Q', R′R' are the mid-points of the line segments PQPQ and PRPR, respectively, and therefore, we must have QR∥Q′R′QR \parallel Q'R', and thus we have shown that BC⊥QRBC \perp QR must hold.

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