Inscribed Equilateral
Equilateral Δ A B C with side length 1 2 is inscribed in a circle. Points D , E , F are on minor arcs BC ⌢ , A C ⌢ , and A B ⌢ , respectively, such that A D + B E + C F = 4 0 . Find the perimeter of hexagon A E C D B F .
The answer is 40.
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7 solutions
Sorry for the silly question but how can we show that 40 is a possible value of AD+BE+CF?
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Its given in the question.
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I think I was not clear. Assume that we don't know that AD+BE+CF=40, is it possible to determine the range of values AD+BE+CF can have?
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@Pranav Arora – The lengths of A D , B E , C F are maximised when they are all diameters of the circle. It can be found that d = 8 3 which implies that the maximum of A D + B E + C F is 2 4 3 The lengths of A D , B E , C F are minimised when the points D , E , F lie on the triangle. In this case A D = B E = C F = 1 2 hence A D + B E + C F = 3 6 . However, this minimum cannot be achieved since A E C D B F will be a triangle, so, 3 6 < A D + B E + C F ≤ 2 4 3 4 0 is in this range and so is an acceptable value to use.
@Pranav Arora – If you are looking for the highest value of AD+BE+CF,then you can simply compute the highest value that each of AD,BE,CF can take and add them together.In this case,the highest value occurs when AD,BE and CF are all diameters of the circle.Diameter of the circumcircle is given by
s i n ( 6 0 ) 1 2 = 8 3
Therefore,the highest value is 2 4 3 .
wow, so smart
better than the best solution
Thats a good one.....
Good solution
MAKE A PROPER FIG. FIRST. WE WILL APPLY PTOLEMY'S THEOREM .
ALL THE QUADRILATERAL ABOUT WHICH WE TALKING ARE CYCLIC , SO IN A B D C , A D . B C = A B . C D + B D . A C ( P T O L E M Y ′ S T H E O R E M ) 1 2 . A D = 1 2 C D + 1 2 . A C ie A D = C D + A C
SIMILARLY REPEATING STEPS IN QUADRILATERALS BAFC & AEBC , WE GET CE=AE+BE & BF=AF+FC AND PERIMETR F HEXAGON = CD+AC+AE+BE+AF+FC= AD+CE+BE= 40 (GIVEN)
By Pompeiu's theorem D A = D B + D C E B = E C + E A F C = F A + F B Hence the perimeter is 4 0 .
By Ptolemy, AD 12 = CD 12 + BD*12 or AD = BD + DC. Likewise, BE = CE + AE & CF = AF + BF. In other words, 40 = AD + BE + CF = Reqd. Perimeter
Its kinda becomes easier i solving the problem by seeing the tags attached to it! :P Its clear we have to use Ptolemy's theorem as the tag suggests! Just kidding! :P Its clear to sue the Ptolemy's theorem when we have cyclic quadrilateral and we have to find the length of 2 sides of it . Well, applying Ptolemy's theorem in cyclic quadrilateral A B D C , we see that
B D × A C + D C × A B = A D × B C
Putting the values of the sides we get B D + D C = A D . Similarly, we will have, C E + E A = B E and A F + F B = C F
Hence, adding the three equations, we have B D + D C + C E + E A + A F + F D = A D + B E + C F ,
or, B D + D C + C E + E A + A F + F D = 4 0
lol, I solved this problem by coordinate bashing because I didn't know ptolmey's theorem...
Applying Ptolemy's Theorem and dividing by 1 2 , A D = B D + C D . Similarly, B E = A E + C E and C F = A F + B F . Therefore the perimeter is simply A D + B E + C F = 4 0 .
By Ptolemy's Theorem, we have A D ⋅ B C = A B ⋅ C D + A C ⋅ B D ⇒ A D = B D + C D . We then see by the analogous relations that A E + E C + C D + D B + B F + F A = A D + B E + C F = 4 0 , so the answer is 4 0 .
Consider the cyclic quadrilateral A B D C . By Ptolemy's theorem we have, B C ⋅ A D \ 1 2 A D \ = A B ⋅ C D + A C ⋅ B D = 1 2 C D + 1 2 B D ⇒ A D = C D + B D ( 1 ) Following the same process, we find that, \begin{aligned} {BE} &= {CE} + {AE} \tag{2}\\\ {CF} &= {BF} + {AF} \tag{3} \end{aligned} Thus, from ( 1 ) , ( 2 ) , ( 3 ) , we have the total perimeter of the hexagon A E C D B F is, P \ \ = A E + C E + C D + B D + B F + A F = A D + B E + C F = 4 0