If an equilateral triangle can be placed on three concentric circles with radii
1
,
2
, and
3
such that each vertex is on a different circle, then the side length of that equilateral triangle is
a
. Find
a
.
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If you simple check this whether it is cyclic or not by Ptolemy , then it's very easy . Ptolemy satisfies here so this Quad PQOR is cyclic. Now apply Cosine Rule on angle O of triangle ORQ
There is a formula which I posted some time ago. It works only for lengths which can be used to construct a triangle including degenerate case of a line segment, as is in this case. It would be impossible to construct equilateral triangle with circles of radii: 1,2,4 for example. The formula for this case works as below:
x = ( a 2 + b 2 + c 2 − 4 3 A ) / 2 = 7
Let the coordinates of the vertices of the equilateral triangle be ( 0 , 2 3 a ) , ( 2 a , 0 ) , and ( − 2 a , 0 ) , and the coordinates of the center of the three concentric circles be ( x , y ) . Then we can assume:
⎩ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎧ ( x − 0 ) 2 + ( y − 2 3 a ) 2 = 1 2 ( x − 2 a ) 2 + ( y − 0 ) 2 = 2 2 ( x + 2 a ) 2 + ( y − 0 ) 2 = 3 2 . . . ( 1 ) . . . ( 2 ) . . . ( 3 )
⎩ ⎪ ⎨ ⎪ ⎧ ( 3 ) − ( 2 ) : ( 3 ) + ( 2 ) − 2 ( 1 ) 2 a x = 5 2 3 a y − a = 1 1 ⟹ x = 2 a 5 ⟹ y = 2 3 a 1 1 + a
Therefore ( 2 ) becomes:
( 2 a 5 − 2 a ) 2 + ( 2 3 a 1 1 + a ) 2 ( 2 a 5 − a ) 2 + ( 2 3 a 1 1 + a ) 2 3 a 2 − 3 0 a + 7 5 + a 2 + 2 2 a + 1 2 1 a 2 − 1 4 a + 4 9 ( a − 7 ) 2 ⟹ a = 4 = 4 = 4 8 a = 0 = 0 = 7
Same start as in
David Vreken
's solution (rotation by
6
0
∘
).
Then I used Stewart's theorem on
△
P
O
′
O
:
P Q 2 = O ′ O P O 2 ⋅ O ′ Q + P O ′ 2 ⋅ Q O − O ′ Q ⋅ Q O = 3 3 2 ⋅ 2 + 3 2 ⋅ 1 − 2 ⋅ 1 = 7
Hence, P Q = 7 ⇒ a = 7
Ps. Nice variation of A Triangle Problem .
Consider the three roots of the polynomial z 3 − 1 = 0 in the complex plane (as usual, we'll call these 1 , ω , ω 2 ). These lie at the vertices of an equilateral triangle centred at the origin with sidelength 3 .
We can transform these three points into the equilateral triangle in the question by applying a stretch and a translation; ie we apply the map z ↦ t z + u + v i to { 1 , ω , ω 2 } where t , u , v are real, and t > 0 . In this transformation, t represents the scale factor of the stretch, and u + v i represents the translation. Note that by choosing t to be real and positive, no rotation is applied to the triangle; we can do this without loss of generality because we can just rotate the original diagram so that the triangle has the right orientation (although, in this case, it already does).
We now want to make sure these points lie on the three given circles; that is we get the following equations:
∣ t + u + v i ∣ ∣ t ω + u + v i ∣ ∣ t ω 2 + u + v i ∣ = 2 = 3 = 1
Solving these, we find t = 3 7 . Applying this scale-factor to the original sidelength, we find the triangle in the question has sidelength 7 , giving the answer a = 7 .
The solution of these equations isn't too bad - substituting in for ω and expanding, they become
( t + u ) 2 + v 2 ( − 2 t + u ) 2 + ( 2 t 3 + v ) 2 ( − 2 t + u ) 2 + ( − 2 t 3 + v ) 2 = 4 = 9 = 1
Subtracting the third of these from the second gives t v 3 = 4 .
Doubling the first equation and subtracting both the others gives 3 t u = − 1 .
It's then easy to substitute into the first equation and find t 2 = 3 7 ; as mentioned above, we disregard the negative root.
Two LaTeX questions - 1) could anyone tell me how to include equation numbers in an align environment? 2) How do you insert a horizontal section break line?
Thanks!
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Nice solution! Not sure about the answer to LaTeX question #1, but for #2 you just put an underscore symbol _ all by itself on its own line and that will put a horizontal section break line in your solution.
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Label the triangle △ P Q R and let O be the center of the circles, and then rotate △ P Q R 6 0 ° clockwise about P to make △ P Q ′ Q :
From the radii of the circles and the rotation, O Q = O ′ Q ′ = 1 , O R = O ′ Q = 2 , O P = O ′ P = 3 , and O P O ′ = 6 0 ° .
Since O P = O ′ P = 3 and O P O ′ = 6 0 ° , △ O P O ′ is an equilateral triangle, so O O ′ = 3 and ∠ O ′ O P = ∠ O O ′ P = 6 0 ° .
Since O Q = 1 and O ′ Q = 2 , O Q + O ′ Q = 3 = O O ′ , so Q is on O O ′ . Therefore, ∠ Q O P = ∠ O ′ O P = 6 0 ° .
Using the law of cosines on △ Q O P , P Q = O P 2 + O Q 2 − 2 ⋅ O P ⋅ O Q ⋅ cos 6 0 ° = 3 2 + 1 2 − 3 ⋅ 1 = 7 , which makes a = 7 .