Olympiad Maths Prep

Track / Stage 7 / 249 of 300 #1649 of 2000

Problem 1649

National olympiad second round; IMO P1/P4
Geometry Difficulty 7.6 Prove it

Let ABCDABCD be a convex quadrilateral whose diagonals intersect at O.O. Given that AB→+AD→+AO→=BC→+DC→+OC→,\overrightarrow{AB}+\overrightarrow{AD}+\overrightarrow{AO}=\overrightarrow{BC}+\overrightarrow{DC}+\overrightarrow{OC},prove that ABCDABCD is a parallelogram.

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

1. Given Condition: We start with the given vector equation:
AB→+AD→+AO→=BC→+DC→+OC→ \overrightarrow{AB} + \overrightarrow{AD} + \overrightarrow{AO} = \overrightarrow{BC} + \overrightarrow{DC} + \overrightarrow{OC}
We need to prove that quadrilateral ABCDABCD is a parallelogram.

2. Vector Decomposition: Let's decompose the vectors in terms of the points A,B,C,D,OA, B, C, D, O:
AO→=O→−A→,BO→=O→−B→,CO→=O→−C→,DO→=O→−D→ \overrightarrow{AO} = \overrightarrow{O} - \overrightarrow{A}, \quad \overrightarrow{BO} = \overrightarrow{O} - \overrightarrow{B}, \quad \overrightarrow{CO} = \overrightarrow{O} - \overrightarrow{C}, \quad \overrightarrow{DO} = \overrightarrow{O} - \overrightarrow{D}
Substituting these into the given equation, we get:
AB→+AD→+(O→−A→)=BC→+DC→+(O→−C→) \overrightarrow{AB} + \overrightarrow{AD} + (\overrightarrow{O} - \overrightarrow{A}) = \overrightarrow{BC} + \overrightarrow{DC} + (\overrightarrow{O} - \overrightarrow{C})

3. Simplifying the Equation: Simplify the equation by combining like terms:
AB→+AD→+O→−A→=BC→+DC→+O→−C→ \overrightarrow{AB} + \overrightarrow{AD} + \overrightarrow{O} - \overrightarrow{A} = \overrightarrow{BC} + \overrightarrow{DC} + \overrightarrow{O} - \overrightarrow{C}
AB→+AD→−A→=BC→+DC→−C→ \overrightarrow{AB} + \overrightarrow{AD} - \overrightarrow{A} = \overrightarrow{BC} + \overrightarrow{DC} - \overrightarrow{C}

4. Rewriting in Terms of Points: Rewrite the vectors in terms of the points:
(B→−A→)+(D→−A→)−A→=(C→−B→)+(C→−D→)−C→ (\overrightarrow{B} - \overrightarrow{A}) + (\overrightarrow{D} - \overrightarrow{A}) - \overrightarrow{A} = (\overrightarrow{C} - \overrightarrow{B}) + (\overrightarrow{C} - \overrightarrow{D}) - \overrightarrow{C}
Simplify further:
B→+D→−2A→=C→−B→+C→−D→−C→ \overrightarrow{B} + \overrightarrow{D} - 2\overrightarrow{A} = \overrightarrow{C} - \overrightarrow{B} + \overrightarrow{C} - \overrightarrow{D} - \overrightarrow{C}
B→+D→−2A→=C→−B→+C→−D→ \overrightarrow{B} + \overrightarrow{D} - 2\overrightarrow{A} = \overrightarrow{C} - \overrightarrow{B} + \overrightarrow{C} - \overrightarrow{D}
B→+D→−2A→=2C→−B→−D→ \overrightarrow{B} + \overrightarrow{D} - 2\overrightarrow{A} = 2\overrightarrow{C} - \overrightarrow{B} - \overrightarrow{D}

5. Equating the Vectors: Equate the vectors on both sides:
B→+D→−2A→=2C→−B→−D→ \overrightarrow{B} + \overrightarrow{D} - 2\overrightarrow{A} = 2\overrightarrow{C} - \overrightarrow{B} - \overrightarrow{D}
Combine like terms:
B→+D→+B→+D→=2A→+2C→ \overrightarrow{B} + \overrightarrow{D} + \overrightarrow{B} + \overrightarrow{D} = 2\overrightarrow{A} + 2\overrightarrow{C}
2B→+2D→=2A→+2C→ 2\overrightarrow{B} + 2\overrightarrow{D} = 2\overrightarrow{A} + 2\overrightarrow{C}
Divide by 2:
B→+D→=A→+C→ \overrightarrow{B} + \overrightarrow{D} = \overrightarrow{A} + \overrightarrow{C}

6. Conclusion: The equation B→+D→=A→+C→\overrightarrow{B} + \overrightarrow{D} = \overrightarrow{A} + \overrightarrow{C} implies that the diagonals of quadrilateral ABCDABCD bisect each other. This is a defining property of a parallelogram.

■\blacksquare

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.