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| | ~translated into English by Tomas Diaz. ~orders@tomasdiaz.com | | ~translated into English by Tomas Diaz. ~orders@tomasdiaz.com |
| − |
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| − | == Solution 1 ==
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| − | {{Step 1: Let the sides of the triangle be
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| − | a
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| − | a,
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| − | b
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| − | b, and
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| − | c
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| − | c.
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| − | Since the sides are in arithmetic progression, we can represent them as:
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| − |
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| − | b
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| − | =
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| − | a
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| − | +
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| − | c
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| − | 2
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| − | .
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| − | b=
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| − | 2
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| − | a+c
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| − |
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| − | .
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| − | So, the three sides are
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| − | a
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| − | a,
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| − | b
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| − | =
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| − | a
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| − | +
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| − | c
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| − | 2
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| − | b=
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| − | 2
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| − | a+c
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| − |
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| − | , and
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| − | c
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| − | c, where
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| − | a
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| − | a and
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| − | c
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| − | c are the first and third terms of the arithmetic progression.
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| − |
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| − | Step 2: Let the heights corresponding to these sides be
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| − | h
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| − | a
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| − | h
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| − | a
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| − |
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| − | ,
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| − | h
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| − | b
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| − | h
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| − | b
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| − |
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| − | , and
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| − | h
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| − | c
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| − | h
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| − | c
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| − |
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| − | .
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| − | We are also given that the heights are in arithmetic progression. So, we can represent the heights as:
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| − |
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| − | h
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| − | b
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| − | =
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| − | h
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| − | a
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| − | +
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| − | h
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| − | c
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| − | 2
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| − | .
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| − | h
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| − | b
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| − |
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| − | =
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| − | 2
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| − | h
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| − | a
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| − |
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| − | +h
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| − | c
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| − |
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| − |
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| − |
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| − | .
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| − | Step 3: Use the formula for the area of the triangle.
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| − | The area
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| − | A
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| − | A of a triangle can be expressed using any side and the corresponding height:
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| − |
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| − | A
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| − | =
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| − | 1
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| − | 2
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| − | ×
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| − | side
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| − | ×
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| − | corresponding height
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| − | .
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| − | A=
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| − | 2
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| − | 1
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| − |
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| − | ×side×corresponding height.
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| − | Thus, we have the following three expressions for the area of the triangle:
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| − |
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| − | A
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| − | =
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| − | 1
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| − | 2
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| − | a
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| − | h
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| − | a
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| − | =
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| − | 1
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| − | 2
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| − | b
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| − | h
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| − | b
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| − | =
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| − | 1
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| − | 2
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| − | c
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| − | h
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| − | c
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| − | .
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| − | A=
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| − | 2
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| − | 1
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| − |
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| − | ah
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| − | a
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| − |
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| − | =
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| − | 2
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| − | 1
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| − |
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| − | bh
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| − | b
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| − |
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| − | =
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| − | 2
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| − | 1
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| − |
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| − | ch
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| − | c
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| − |
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| − | .
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| − | From these, we can express the heights in terms of the area and the side lengths:
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| − |
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| − | h
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| − | a
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| − | =
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| − | 2
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| − | A
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| − | a
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| − | ,
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| − | h
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| − | b
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| − | =
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| − | 2
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| − | A
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| − | b
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| − | ,
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| − | h
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| − | c
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| − | =
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| − | 2
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| − | A
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| − | c
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| − | .
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| − | h
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| − | a
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| − |
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| − | =
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| − | a
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| − | 2A
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| − |
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| − | ,h
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| − | b
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| − |
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| − | =
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| − | b
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| − | 2A
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| − |
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| − | ,h
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| − | c
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| − |
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| − | =
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| − | c
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| − | 2A
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| − |
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| − | .
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| − | Step 4: Use the fact that the heights are in arithmetic progression.
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| − | Since the heights are in arithmetic progression, we know that:
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| − |
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| − | h
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| − | b
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| − | =
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| − | h
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| − | a
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| − | +
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| − | h
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| − | c
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| − | 2
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| − | .
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| − | h
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| − | b
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| − |
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| − | =
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| − | 2
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| − | h
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| − | a
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| − |
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| − | +h
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| − | c
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| − |
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| − |
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| − |
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| − | .
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| − | Substitute the expressions for
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| − | h
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| − | a
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| − | h
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| − | a
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| − |
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| − | ,
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| − | h
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| − | b
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| − | h
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| − | b
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| − |
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| − | , and
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| − | h
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| − | c
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| − | h
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| − | c
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| − |
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| − | into this equation:
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| − |
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| − | 2
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| − | A
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| − | b
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| − | =
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| − | 2
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| − | A
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| − | a
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| − | +
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| − | 2
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| − | A
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| − | c
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| − | 2
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| − | .
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| − | b
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| − | 2A
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| − |
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| − | =
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| − | 2
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| − | a
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| − | 2A
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| − |
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| − | +
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| − | c
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| − | 2A
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| − |
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| − |
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| − |
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| − | .
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| − | Simplify both sides:
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| − |
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| − | 2
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| − | A
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| − | b
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| − | =
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| − | 2
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| − | A
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| − | 2
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| − | (
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| − | 1
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| − | a
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| − | +
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| − | 1
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| − | c
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| − | )
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| − | =
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| − | A
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| − | (
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| − | 1
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| − | a
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| − | +
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| − | 1
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| − | c
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| − | )
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| − | .
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| − | b
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| − | 2A
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| − |
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| − | =
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| − | 2
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| − | 2A
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| − |
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| − | (
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| − | a
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| − | 1
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| − |
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| − | +
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| − | c
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| − | 1
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| − |
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| − | )=A(
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| − | a
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| − | 1
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| − |
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| − | +
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| − | c
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| − | 1
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| − |
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| − | ).
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| − | Canceling
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| − | A
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| − | A from both sides (assuming
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| − | A
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| − | ≠
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| − | 0
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| − | A
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| − |
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| − | =0):
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| − |
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| − | 2
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| − | b
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| − | =
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| − | 1
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| − | a
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| − | +
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| − | 1
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| − | c
| |
| − | .
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| − | b
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| − | 2
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| − |
| |
| − | =
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| − | a
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| − | 1
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| − |
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| − | +
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| − | c
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| − | 1
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| − |
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| − | .
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| − | Step 5: Substitute the relation
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| − | b
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| − | =
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| − | a
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| − | +
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| − | c
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| − | 2
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| − | b=
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| − | 2
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| − | a+c
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| − |
| |
| − | .
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| − | We know that
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| − | b
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| − | =
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| − | a
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| − | +
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| − | c
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| − | 2
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| − | b=
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| − | 2
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| − | a+c
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| − |
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| − | . Substituting this into the equation above:
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| − |
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| − | 2
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| − | a
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| − | +
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| − | c
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| − | 2
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| − | =
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| − | 1
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| − | a
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| − | +
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| − | 1
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| − | c
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| − | .
| |
| − | 2
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| − | a+c
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| − |
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| − |
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| − | 2
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| − |
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| − | =
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| − | a
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| − | 1
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| − |
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| − | +
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| − | c
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| − | 1
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| − |
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| − | .
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| − | Simplifying the left-hand side:
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| − |
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| − | 4
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| − | a
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| − | +
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| − | c
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| − | =
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| − | 1
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| − | a
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| − | +
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| − | 1
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| − | c
| |
| − | .
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| − | a+c
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| − | 4
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| − |
| |
| − | =
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| − | a
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| − | 1
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| − |
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| − | +
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| − | c
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| − | 1
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| − |
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| − | .
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| − | Now, combine the terms on the right-hand side:
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| − |
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| − | 1
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| − | a
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| − | +
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| − | 1
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| − | c
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| − | =
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| − | a
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| − | +
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| − | c
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| − | a
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| − | c
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| − | .
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| − | a
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| − | 1
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| − |
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| − | +
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| − | c
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| − | 1
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| − |
| |
| − | =
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| − | ac
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| − | a+c
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| − |
| |
| − | .
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| − | Thus, the equation becomes:
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| − |
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| − | 4
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| − | a
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| − | +
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| − | c
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| − | =
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| − | a
| |
| − | +
| |
| − | c
| |
| − | a
| |
| − | c
| |
| − | .
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| − | a+c
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| − | 4
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| − |
| |
| − | =
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| − | ac
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| − | a+c
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| − |
| |
| − | .
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| − | Step 6: Cross-multiply to simplify.
| |
| − | Cross-multiply to eliminate the fractions:
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| − |
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| − | 4
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| − | a
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| − | c
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| − | =
| |
| − | (
| |
| − | a
| |
| − | +
| |
| − | c
| |
| − | )
| |
| − | 2
| |
| − | .
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| − | 4ac=(a+c)
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| − | 2
| |
| − | .
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| − | Expanding both sides:
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| − |
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| − | 4
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| − | a
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| − | c
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| − | =
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| − | a
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| − | 2
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| − | +
| |
| − | 2
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| − | a
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| − | c
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| − | +
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| − | c
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| − | 2
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| − | .
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| − | 4ac=a
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| − | 2
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| − | +2ac+c
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| − | 2
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| − | .
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| − | Rearrange the terms:
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| − |
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| − | 4
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| − | a
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| − | c
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| − | −
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| − | 2
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| − | a
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| − | c
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| − | =
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| − | a
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| − | 2
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| − | +
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| − | c
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| − | 2
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| − | ,
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| − | 4ac−2ac=a
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| − | 2
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| − | +c
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| − | 2
| |
| − | ,
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| − | 2
| |
| − | a
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| − | c
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| − | =
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| − | a
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| − | 2
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| − | +
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| − | c
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| − | 2
| |
| − | .
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| − | 2ac=a
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| − | 2
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| − | +c
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| − | 2
| |
| − | .
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| − | Step 7: Recognize the equation.
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| − | We now have the equation:
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| − |
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| − | 2
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| − | a
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| − | c
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| − | =
| |
| − | a
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| − | 2
| |
| − | +
| |
| − | c
| |
| − | 2
| |
| − | .
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| − | 2ac=a
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| − | 2
| |
| − | +c
| |
| − | 2
| |
| − | .
| |
| − | This is a well-known equation that holds if and only if
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| − | a
| |
| − | =
| |
| − | c
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| − | a=c, meaning the triangle is isosceles with
| |
| − | a
| |
| − | =
| |
| − | c
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| − | a=c.
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| − |
| |
| − | Step 8: Conclude that the triangle is equilateral.
| |
| − | If
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| − | a
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| − | =
| |
| − | c
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| − | a=c, then from the relation
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| − | b
| |
| − | =
| |
| − | a
| |
| − | +
| |
| − | c
| |
| − | 2
| |
| − | b=
| |
| − | 2
| |
| − | a+c
| |
| − |
| |
| − | , we see that
| |
| − | b
| |
| − | =
| |
| − | a
| |
| − | =
| |
| − | c
| |
| − | b=a=c. Therefore, all three sides of the triangle are equal, which means the triangle is equilateral.
| |
| − |
| |
| − | Conclusion:
| |
| − | We have shown that if the sides and the heights of a triangle are both in arithmetic progression, the triangle must be equilateral.}}
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| | | | |
| | == See also == | | == See also == |
| | https://www.oma.org.ar/enunciados/ibe3.htm | | https://www.oma.org.ar/enunciados/ibe3.htm |
