Challenging Geometry Problem 1607. Share your solution by posting it in the comment box provided.
Audience: Mathematics Education - K-12 Schools, Honors Geometry, and College Level.
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Let BE =a, CE = b, AE = c, DE = d and BD = CD = t
ReplyDelete<CBD = <BCD = 60 hence Triangle BCD is equilateral and
BC^2 = t^2 = 3^2 + 5^2 - 2.3.5.cos60 = 19.............(1)
Now Triangle BED // Triangle ABD hence
a = 3d/V19 ..............(2)
Similarly b = 5d/V19 ........(3) since Triangles CED // ACD
Using Ptolemy in cyclic quadrilateral ABEC,
5a + 3b = V19.c and substituting from (1) and (2) and simplifying
30d/V19 = V19. c from which,
c = 30d/19
Now d (c+d) = 19 so
d (30d/19 + d) = 19 hence
d^2 = 19^2/49 and therefore
d = 19/7 = DE
Sumith Peiris
Moratuwa
Sri Lanka
Alternate Solution using the same letters as before. Let < BAD = @
DeleteConstruct Equilateral Triangle AFD, F and C on the same side of AD
Triangles FCD and ABD are congruent SAS and so
< DFC = < DAB = @ = < DBE so < EBC = < EAC = 60 - @ and so < CAF = @
Since DFC = @, < CFA = 60 - @ and so < ACF = 120
Hence AF^2 = 5^2 + 3^2 - 2.3.5.cos 120 and so
AF = 7 = AD = c + d
Hence DE d = d = BD^2 / (c + d) = 19 /7
Sumith Peiris
Moratuwa
Sri Lanka
Let ∠ABC=θ.
ReplyDelete△Using the cosine law on ABC:
cosθ=1/(2√19), sinθ=5√3/(2√19)
△Using the cosine law on ABD:
AD^2=3^2+(√19)^2-1*3*√19*cos(60°+θ)＝49
Therefore, AD=7
Since DE*DA=19, DE=19/7
Rotate △ABD 60° about point D.
ReplyDelete(Point B coincides with point C, and point A moves to the new point A')
Then, △AA'D becomes an equilateral triangle.
Let ∠CAD = θ. Then, ∠AA'C = θ.
Applying the cosine rule to △ACD and △AA'C:
cosθ = (25 + AD² - 19) / (2 × 5 × AD) = (9 + AD² - 25) / (2 × 3 × AD)
Solving this gives
AD = 7.