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 == | ||
− | {{ | + | We will have <math>\angle ABQ = \angle QBC = x</math> |
+ | |||
+ | 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>. | ||
+ | |||
+ | 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. | ||
+ | |||
+ | 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 be a triangle with
. Let
bisect
and let
bisect
, with
on
and
on
. If
, what are the angles of the triangle?
Solution
We will have
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 . Meanwhile,
, so we have
.
Then suppose the bisector of intersects BR at X. Then since
, we have similar triangles, and by equal ratios
. Equivalently,
, so triangles BXP and BPR are similar; in particular, BX = XP.
Since TX is the bisector of , we have then that T, B, X, and P are concyclic. Then
. Solving for x, x = 40. Then
.