Regular dodecahedron A B C D E F G H I J K L with side length 1 is drawn. Segments A F , B G , and C H are then drawn.
The area of the tiny little triangle formed from the intersections of these segments can be expressed as c a − b for positive integers a , b , c . Find the value of a + b + c .
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how did you get FN as 3^(1/2)
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△ G F N is a 3 0 − 6 0 − 9 0 triangle and since G F = 1 then F N = 3 .
!!!.....The answer is correct ( I am talking about a + b + c ), but you have typed the value of the area of the 'tiny triangle' WRONG!....There is ′ + ′ in the solution, but I should be ′ − ′ .....Right?
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@Daniel Liu Minor typo. Note that in your final line, it should be a − .
Okay, fixed. Thanks.
M N = O N − O M = ( 2 − 3 ) 2 M N = 2 2 − 3 O S = 2 M N t a n ( 3 0 0 ) = 2 M N × 3 1 [ M N O ] = 2 M N ∗ O S = 2 1 × M N ( M N / 2 ) × ( 1 / 3 ) = 4 3 M N 2 = 4 3 7 − 4 3 [ M N O ] = 1 2 7 3 − 1 2 = 1 2 4 9 ∗ 3 − 1 2 ( a + b + c ) = 1 7 1
I always like to solve most of your problems by the hard way :D, this time I used analytic geometry and complex numbers. Rotate the dodecahedron 3 0 ° to the left, with the objective to make that D and J be in the x-axis. Now, we know that the side length is 1 , so the radius, with a little of trigonometry, will be r = 2 cos ( 1 5 ° ) . Notice that the vertices of the figure, in the Argand diagram are: V = 2 cos ( 1 5 ° ) × c i s ( 3 0 ° k ) from k = 0 to 1 1 .
So, F = 2 cos ( 1 5 ° ) × c i s ( 6 0 ° ) , G = 2 cos ( 1 5 ° ) × c i s ( 9 0 ° ) , H = 2 cos ( 1 5 ° ) × c i s ( 1 2 0 ° ) , A = 2 cos ( 1 5 ° ) × c i s ( 2 7 0 ° ) , B = 2 cos ( 1 5 ° ) × c i s ( 3 0 0 ° ) and C = 2 cos ( 1 5 ° ) × c i s ( 3 3 0 ° )
In the cartesian plane, they are: F = ( 4 6 + 2 , 4 6 + 3 2 ) G = ( 0 , 2 6 + 2 ) H = ( − 4 6 + 2 , 4 6 + 3 2 ) A = ( 0 , − 2 6 + 2 ) B = ( 4 6 + 2 , − 4 6 + 3 2 ) C = ( 4 6 + 3 2 , − 4 6 + 2 )
Let's find the equation of the line that passes through: A F ⇒ y = ( 2 + 3 ) x − 2 6 + 2 B G ⇒ y = − ( 2 + 3 ) x + 2 6 + 2 C H ⇒ y = − x + 2 2
Their intersection points are the vertices of the triangle, and they are: O = ( 2 6 − 2 , 0 ) N = ( 1 2 6 + 3 2 , 1 2 3 2 − 6 ) M = ( 4 3 2 − 6 , 4 6 − 2 )
And finally, find the area of the triangle using determinants: A T = 1 2 7 3 − 1 2
Hence, a = 1 4 7 , b = 1 2 , c = 1 2 and a + b + c = 1 7 1
In a previous problem involving a square and a small triangle, people realized coordinate bashing was easy. However, in this problem, coordinate bashing is clearly not the easiest way! However, it still works, as shown by you :P
It can be shown that ∠ A B G = 9 0 ∘ and ∠ F A B = 6 0 ∘ , therefore, the tiny △ M N O is a 3 0 ∘ − 1 2 0 ∘ − 3 0 ∘ isosceles triangle. If ∣ M N ∣ = d then the height of the tiny △ M N O is 2 d t a n 3 0 ∘ and its area is given by:
A = 2 1 d ( 2 d t a n 3 0 ∘ ) = 4 3 d 2
To get d , we use ∣ B G ∣ − ∣ B N ∣ − ∣ M G ∣ = ∣ M N ∣ = d . From △ A B N ( 3 0 ∘ − 6 0 ∘ − 9 0 ∘ ), we find that ∣ B N ∣ = 3 ; similarly, ∣ M G ∣ = 3 .
Let the center of the dodecahedron be O ′ , then △ O ′ A B is a 7 5 ∘ − ( 3 0 ∘ − ( 7 5 ∘ isosceles triangle. If ∣ B G ∣ = H , then:
2 H = 2 1 t a n 7 5 ∘ ⇒ H = t a n 7 5 ∘ = 2 + 3
⇒ d = ∣ B G ∣ − ∣ B N ∣ − ∣ M G ∣ = H − 2 3 = 2 + 3 − 2 3 = 2 − 3
⇒ A = 4 3 d 2 = 4 3 ( 2 − 3 ) 2 = 4 3 7 − 4 3 = 1 2 7 3 − 1 2 = 1 2 1 4 7 − 1 2
Therefore, a = 1 4 7 and b = c = 1 2 , and a + b + c = 1 7 1
Let st. line POQR through O meet CF at P, NM at Q and AH at R.
COF,.NOM,.AOH are (isosceles) SIMILAR triangles with respective altitudes ,OP, OQ, OP.
RQ = 1, RO = RQ + QO = 1 + QO...>>> QO = RO - 1....RP = RO + OP.
Ratio of area of similar triangles = square of ratio of respective sides, or altitude......ect.
This is a dodecahedron hence external angles are 30, its sides are 1.
By simple trigo we find out the base lengths, PQ, and PR, as under.
CF = 1+sqrt(3).....AH = 2+sqrt(3)....PQ = sqrt(3)/2....RP = 1+sqrt(3)/2 =(1/2)(2+sqrt(3))
By ratios of similar triangles and ratio rules we find altitudes OR and OQ.
RO/AH=OP/CF=(RO+OP)/(AH+CF)=RP/(AH+CF)...>RO=AH
RP/(AH+CF).
RO =( 2+sqrt(3))
(1/2)(2+sqrt(3))./.( 2 + sqrt(3) + 1 + sqrt(3) )
RO=(1/2)
( (2+sqrt(3)) )^2./.( 3+2
sqrt(3) ) ...................................................(1)
..... = 1 + 1/(2* ( 3+2
sqrt(3) ) = 1 + QO......QO = (1/2)/( 3+2
sqrt(3) ).....(2)
Areas NOM./.AOH = ( QO./.RO )^2 = ( (2)/( 1) ) ^2 = 1./.( (2+sqrt(3)) )^4
Areas AOH = (1/2)AH
RO =(1/2) * ( 2+sqrt(3 ) ) * (1/2) * ( (2+sqrt(3)) )^2./.( 3+2
sqrt(3) ) )
..= (1/4) * ( 2 + sqrt(3 ) )^3./..( 3+2
sqrt(3) ) )
Areas NOM = (1/4) * ( 2+sqrt(3) )^3./..( 3+2
sqrt(3) ) ) * 1./.( (2+sqrt(3))^4
..= (1/4)./.( ( 2+sqrt(3) ) * ( 3+2 * sqrt(3)) ) = (- 12 + 7 * sqrt(3))/12 =
= (- 12+7 * sqrt(3) )./.12
= (- 12+sqrt(147) )./.12 = ( - b + sqrt(a) ) / c...>>>a+b+c= 171.
I have not given details of multiplication and division of surds.
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Refer to the diagram.
To find [ M N O ] , I will try to find the area of other shapes and subtract them from each other to get M N O . In this case, I notice: [ M N O ] = [ N G F ] − [ O M G F ]
Extending G F and C H to meet at P :
[ O M G F ] = [ O F P ] − [ P G M ]
Thus [ M N O ] = [ N G F ] − [ O F P ] + [ P G M ]
We see that: [ N G F ] = 2 1 ⋅ ( 1 ) ⋅ ( 3 ) = 2 3
F P = 1 + 2 1 ⋅ 3 ⋅ 2 = 1 + 3
∴ O F 3 1 + 3 = 1 + 3 3
∴ [ O F P ] = 2 1 ( 1 + 3 ) ( 1 + 3 3 ) = 3 2 3 + 1
P G = 3 ⟹ [ P G M ] = 4 3 3
Thus [ M N O ] = 2 3 − ( 3 2 3 + 1 ) + 4 3 3 = 1 2 7 3 − 1 2 = 1 2 1 4 7 − 1 2
and so our answer is 1 4 7 + 1 2 + 1 2 = 1 7 1