Convex quadrilateral ABCD is inscribed in a circle, ∠A=60o, BC=CD=1, rays AB and DC intersect at point E, rays BC and AD intersect each other at point F. It is given that the perimeters of triangle BCE and triangle CDF are both integers. Find the perimeter of quadrilateral ABCD.
Problem
Source: China TST 2004 Quiz
Tags: geometry, perimeter, inequalities, trigonometry, trig identities, Law of Sines, geometry unsolved
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11.04.2009 00:58
Certain lengths and angles are marked in the diagram.
First, notice that a+b=120∘, and ∠E=a−30>0. Similarly, b−30>0. Thus, 0<a−30<60. Let us consider the triangle that has the smaller perimeter. WLOG let it be △CDF. We shall consider the case where the two triangles are congruent. Since ∠BCE=∠DCF=60∘, congruency is achieved when a=b=60∘. In this case, △CDF is 30-60-90 and has perimeter 1+2+√3<5. Notice that this case results in the maximum possible perimeter for△CDF. Since the perimeter must be an integer, we conclude that the perimeter is less than 5, i.e. 1+x+y≤4.\
Now we consider the case where the perimeter of △CDF is the smallest. This occurs when b−30 is maximized, or when b−30=60, making △CDF an equilateral triangle with perimeter 3. This can never be achieved because it would imply ∠E=a−30=0, so we conclude that the perimeter of △CDF must be greater than 3, i.e. 1+x+y≥4. Combined with the previous inequality 1+x+y≤4 we discover that the perimeter of △CDF must be 4, or x+y=3. Now we find x+y.\
From the law of sines,
1sin(b−30)=ysin60⟹y=sin60sin(b−30)1sin(b−30)=xsin∠CDF=xsin(150−b)⟹x=sin(150−b)sin(b−30)x+y=sin60+sin(150−b)sin(b−30)=√3+cosb+√3sinb√3sinb−cosb=3⟹√3+cosb+√3sinb=3√3sinb−3cosb⟹√3+4cosb=2√3sinb=2√3√1−cos2b⟹3+8√3cosb+16cos2b=12−12cos2b⟹28cos2b+8√3cosb−9=0⟹cosb=−8√3+√192+100856=3√314⟹sinb=√1−cos2b=1314
With one last application of the law of sines on triangle ABD, we have
ADsina=BDsin60=2⟹AD=2sina
Similarly AB=2sinb=137. Since a+b=120, then sina=sin(120−b)=√3cosb+sinb2=928+1328=1114. Thus, AD=2sina=117, so the perimeter of quadrilateral ABCD is 137+117+2=
387
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