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| | == Solution == | | == Solution == |
| − | 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>.
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| | == Resources == | | == Resources == |
Latest revision as of 02:42, 5 August 2017
be a triangle with 
. Let 
bisect 
and let 
bisect 
, with 
on 
and 
on 
. If 
, what are the angles of the triangle?
