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Let $A_5$ be arbitrary point on the diagonal $A_1A_3$ of trapezoid $A_1A_2A_3A_4$, where $A_1A_2 ||A_3A_4$. If $J_1$ and $J_2$ are the centers of the circumscribed circles of $A_1A_2A_5$ and $A_3A_4A_5$, prove that $J_1J_2$ is independent of the choice of $A_5$. If it is correct, suggest proofs. If it is not, change it so that it becomes true.

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If you stare at the figure long enough the statement becomes obvious. Assume $A_1 A_2$ and $A_4 A_3$ horizontal from left to right. Then $J_1$ is on the vertical $m_1$ through the midpoint of $A_1 A_2$, and $J_2$ is on the vertical $m_2$ through the midpoint of $A_4 A_3$. These two verticals do not depend on the choice of $A_5$.

Given $A_5$ on $d:=A_1 A_3$ the point $J_1$ is on the perpendicular $n_1$ to $d$ through the midpoint of $A_1 A_5$, and $J_2$ is on the perpendicular $n_2$ to $d$ through the midpoint of $A_5 A_3$. The distance between these two perpendiculars is equal to ${1\over2}|A_1 A_3|$, for all choices of $A_5$. It follows that the distance between $J_1=m_1\wedge n_1$ and $J_2=m_2\wedge n_2$ does not depend on the choice of $A_5$ either.

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