Question 1209061: Verify that the points (0,0), (a,0) and (c,d) are the vertices of an equilateral triangle.
Note: c = (a/2) and d = (a•sqrt{3})/2
Then show that the midpoints of the 3 sides are the vertices of a second equilateral triangle.
Answer by textot(100)   (Show Source):
You can put this solution on YOUR website! **1. Verify Equilateral Triangle**
* **Points:**
* A: (0, 0)
* B: (a, 0)
* C: (c, d) = (a/2, (a√3)/2)
* **Distances:**
* AB = √[(a - 0)² + (0 - 0)²] = √a² = a
* BC = √[(a/2 - a)² + ((a√3)/2 - 0)²] = √[(-a/2)² + (a√3/2)²] = √(a²/4 + 3a²/4) = √(4a²/4) = a
* AC = √[(a/2 - 0)² + ((a√3)/2 - 0)²] = √[(a/2)² + (a√3/2)²] = √(a²/4 + 3a²/4) = √(4a²/4) = a
Since AB = BC = AC = a, the points A, B, and C form an equilateral triangle.
**2. Find Midpoints**
* **Midpoint of AB (M1):**
* [(0 + a)/2, (0 + 0)/2] = (a/2, 0)
* **Midpoint of BC (M2):**
* [(a + a/2)/2, (0 + (a√3)/2)/2] = (3a/4, (a√3)/4)
* **Midpoint of AC (M3):**
* [(0 + a/2)/2, (0 + (a√3)/2)/2] = (a/4, (a√3)/4)
**3. Verify Equilateral Triangle for Midpoints**
* **M1M2:**
* √[((3a/4) - (a/2))² + ((a√3)/4 - 0)²] = √[(a/4)² + (a√3/4)²] = √(a²/16 + 3a²/16) = √(4a²/16) = a/2
* **M2M3:**
* √[((a/4) - (3a/4))² + ((a√3)/4 - (a√3)/4)²] = √[(-a/2)² + 0²] = a/2
* **M3M1:**
* √[((a/2) - (a/4))² + (0 - (a√3)/4)²] = √[(a/4)² + (a√3/4)²] = √(a²/16 + 3a²/16) = a/2
Since M1M2 = M2M3 = M3M1 = a/2, the midpoints M1, M2, and M3 also form an equilateral triangle.
**Therefore, it is proven that if the points (0, 0), (a, 0), and (c, d) (where c = a/2 and d = (a√3)/2) are the vertices of an equilateral triangle, then the midpoints of the sides of this triangle also form an equilateral triangle.**
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