2001 IMO Shortlist Problems/G8

Problem

Let $ABC$ be a triangle with $\angle BAC = 60^{\circ}$ . Let $AP$ bisect $\angle BAC$ and let $BQ$ bisect $\angle ABC$ , with $P$ on $BC$ and $Q$ on $AC$ . If $AB + BP = AQ + QB$ , what are the angles of the triangle?

Solution

We will have $\angle ABQ = \angle QBC = x$

Let R lie on AC s.t. RAB is equilateral. Let T lie on AB extended past B such that BT = BP, and let U lie on AC such that UAT is equilateral. Since AU = AT = AB + BP = AQ + QB, we have QB = QU. As a result, we calculate $\angle QBU = 90 - \dfrac{60 + \angle ABQ}{2} = 60 - \dfrac{x}{2}$ . Meanwhile, $\angle QBR = 60 - x$ , so we have $\angle RBU = \angle QBU - \angle QBR = \dfrac{x}{2}$ .

Then suppose the bisector of $\angle BTP$ intersects BR at X. Then since $\angle BRT = \angle RBU = \dfrac{x}{2} = \angle BTX$ , we have similar triangles, and by equal ratios $BT^2 = BX \cdot BR$ . Equivalently, $BP^2 = BX \cdot BR$ , so triangles BXP and BPR are similar; in particular, BX = XP.

Since TX is the bisector of $\angle BTP$ , we have then that T, B, X, and P are concyclic. Then $\dfrac{x}{2} = \angle XTP = \angle XBP = \angle ABC - \angle ABR = 2x - 60$ . Solving for x, x = 40. Then $\angle ABC = 80 \implies \angle ACB = 40$ .

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2001 IMO Shortlist Problems/G8

Problem

Solution

Resources