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If in a triangle $ABC$, $A \equiv (1,10)$, circumcenter $\equiv (-\frac13, \frac23)$ and orthocenter $\equiv (\frac{11}3, \frac43)$ then the coordinates of mid-point of side opposite to A is?

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Here clearly point $Q$ is circumcenter and point $P$ is orthocenter. The only thing I see here is $AP ||DQ$. So their slopes are same. So we can get an equation using this. But how am I supposed to get another equation?

Any other way of solving this problem would also be appreciated.

JKnecht
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manshu
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    With orthocenter P and circumcenter Q and with D the mid-point of BC we also have |AP|=2|DQ|. – DanielWainfleet Feb 03 '16 at 19:45
  • @user254665 why is that? – manshu Feb 03 '16 at 19:47
  • From the properties of the Euler line of the triangle and the Feuerbach circle (also called the Euler-Feuerbach circle, the 9-point circle,and other names) which is the circumcircle of the mid-points of the sides. To answer in detail would be an essay. – DanielWainfleet Feb 03 '16 at 20:06
  • @user254665 thanks for the help and the comment at the correct moment. I had actually asked Lucian to explain the reason behind it...:p – manshu Feb 03 '16 at 20:08

1 Answers1

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Hint: $~$ You were given the orthocenter H and the circumcenter O. But what were you not

given ? The centroid G. Now, do you, by any chance, know that the latter is found on the

segment uniting the former two, and, not only that, but it divides said segment in a ratio of

$1:2$ ? $($Remember that the centroid also divides each median into the same ratio as well$).$

In other words, $G\in(OH)$ and $GH=2GO.~$ At the same time, $G=\dfrac{A+B+C}3,~$ and

$M=\dfrac{B+C}2.$

Lucian
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  • @manshu: I am not personally aware of any particularly elegant way of proving it, but, if you are familiar with analytic geometry, then, given enough time and patience, it should be trivial. – Lucian Feb 03 '16 at 20:09