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Difference between revisions of "2001 IMO Shortlist Problems/G8"

(New page: == Problem == Let <math>ABC</math> be a triangle with <math>\angle BAC = 60^{\circ}</math>. Let <math>AP</math> bisect <math>\angle BAC</math> and let <math>BQ</math> bisect <math>\angle ...)
 
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== Solution ==
 
== Solution ==
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{{solution}}
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We will have <math>\angle ABQ = \angle QBC = x</math>
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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 <math>\angle QBU = 90 - \dfrac{60 + \angle ABQ}{2} = 60 - \dfrac{x}{2}</math>.  Meanwhile, <math>\angle QBR = 60 - x</math>, so we have <math>\angle RBU = \angle QBU - \angle QBR = \dfrac{x}{2}</math>.
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Then suppose the bisector of <math>\angle BTP</math> intersects BR at X.  Then since <math>\angle BRT = \angle RBU = \dfrac{x}{2} = \angle BTX</math>, we have similar triangles, and by equal ratios <math>BT^2 = BX \cdot BR</math>.  Equivalently, <math>BP^2 = BX \cdot BR</math>, so triangles BXP and BPR are similar; in particular, BX = XP. 
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Since TX is the bisector of <math>\angle BTP</math>, we have then that T, B, X, and P are concyclic.  Then <math>\dfrac{x}{2} = \angle XTP = \angle XBP = \angle ABC - \angle ABR = 2x - 60</math>.  Solving for x, x = 40.  Then <math>\angle ABC = 80 \implies \angle ACB = 40</math>.
  
 
== Resources ==
 
== Resources ==

Revision as of 02:42, 5 August 2017

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 $\angle ABC = 80 \implies \angle ACB = 40$.

Resources

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