Problem: β³ A B C β³ A B C 55 5 5 D , E D , E F F B C βΎ , C A βΎ B C , C A A B βΎ A B B D = 7 , C E = 30 B D = 7 , C E = 3 0 A F = 40 A F = 4 0 P P β³ A B C β³ A B C
β A E P = β B F P = β C D P β A E P = β B F P = β C D P
Find 2 ( β A E P ) tan 2 ( β A E P )
Solution:
Let's denote the angle as β A E P = ΞΈ β A E P = ΞΈ P D = x P D = x P E = y P E = y P F = z . P F = z .
Then the area of the triangle can be written in two ways, namely
[ A B C ] = 5 5 2 3 4 = [ A B F ] + [ B P C ] + [ A P C ] = 1 2 β
55 ( x + y + z ) [ A B C ] = 4 5 5 2 3 β β = [ A B F ] + [ B P C ] + [ A P C ] = 2 1 β β
5 5 ( x + y + z )
So if we find x + y + z x + y + z
Notice how [ A F P ] = 180 β [ A E P ] [ A F P ] = 1 8 0 β [ A E P ] A F P E , B F P D , C E P D A F P E , B F P D , C E P D
P is also the Miguel point of the triangle.
Using Law of cosines, we can find the lengths of D F , F E D F , F E E F E F
D F = 13 , E F = 35 , D E = 42 D F = 1 3 , E F = 3 5 , D E = 4 2
Thus, by heron's the area of [ D E F ] = 120 3 [ D E F ] = 1 2 0 3 β
Notice that β F P D = β D P E = β D P F = 120 β F P D = β D P E = β D P F = 1 2 0 180 β 60 1 8 0 β 6 0
From then, the area of [ D E F ] [ D E F ]
1 2 sin β‘ 12 0 β ( x y + y z + x z ) = 120 3 2 1 β sin 1 2 0 β ( x y + y z + x z ) = 1 2 0 3 β
which would yield x y + y z + x z = 480 x y + y z + x z = 4 8 0
Now, we just use law of sines 3 times off P and triangle DEF to get
x 2 + y 2 + 2 x y cos β‘ ( 120 ) = 4 2 2 x 2 + y 2 + 2 x y cos ( 1 2 0 ) = 4 2 2
x 2 + z 2 + 2 x y cos β‘ ( 120 ) = 1 3 2 x 2 + z 2 + 2 x y cos ( 1 2 0 ) = 1 3 2
y 2 + z 2 + 2 y z cos β‘ ( 120 ) = 3 5 2 y 2 + z 2 + 2 y z cos ( 1 2 0 ) = 3 5 2
and solving these and manipulating yields
x + y + z = 2299 x + y + z = 2 2 9 9 β
which yields
sin β‘ ΞΈ = 55 3 2 ( x + y + z ) sin ΞΈ = 2 ( x + y + z ) 5 5 3 β β
from the original expression of the area of the big triangle in terms of the sum of x, y, and z.
Thus, we get
sin β‘ ΞΈ = 55 3 2299 β cos β‘ ΞΈ = 11 2299 β 2 ΞΈ = ( 5 3 ) 2 = 75 sin ΞΈ = 2 2 9 9 β 5 5 3 β β β cos ΞΈ = 2 2 9 9 β 1 1 β β tan 2 ΞΈ = ( 5 3 β ) 2 = 7 5 β
The problems on this page are the property of the MAA's American Mathematics Competitions
