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Thread: Geometry problem

  1. #1
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    Fast Help!

    I have to complete my homework in 2 days and this problem I found most difficult:

    Let ABC be a right triangle with hypotenuse BC. Suppose that M is the midpoint of BC and H is the feet of the perpendicular dropped from A onto BC. A point P, distinct from A, is chosen on the opposite ray of ray AM. Let the line through H perpendicular to AB intersect PB at Q; and let the line through H perpendicular to AC meet PC at R. Prove that A is the orthocenter of triangle PQR.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by thduong711
    I have to complete my homework in 2 days and this problem I found most difficult:

    Let ABC be a right triangle with hypotenuse BC. Suppose that M is the midpoint of BC and H is the feet of the perpendicular dropped from A onto BC. A point P, distinct from A, is chosen on the opposite ray of ray AM. Let the line through H perpendicular to AB intersect PB at Q; and let the line through H perpendicular to AC meet PC at R. Prove that A is the orthocenter of triangle PQR.
    See diagram. By definition of the Orthocentre to prove that $\displaystyle A$ is the
    orthocentre of triangle $\displaystyle PQR$ it is sufficient to prove that $\displaystyle QA$ extended
    is a normal to $\displaystyle PR$, and that $\displaystyle RA$ extended is a normal to $\displaystyle PQ$.

    Now it is sufficient to prove that in general $\displaystyle QA$ extended is a normal $\displaystyle PR$,
    as by an equivalent argument $\displaystyle RA$ will be a normal to $\displaystyle PQ$.

    Now the best way to proceed as far as I can see is to introduce
    coordinates, and use coordinate geometry to show that angle $\displaystyle QSR$
    in the diagram is a right angle.

    ==================================================

    <<You will need to check the algebra in this carefully>>

    Let $\displaystyle A$ be the origin $\displaystyle (0,0)$, $\displaystyle B$ be $\displaystyle (0,b)$ and $\displaystyle C$ be $\displaystyle (c,0)$.

    Then $\displaystyle M$ is $\displaystyle (c/2,b/2)$.

    The slope of $\displaystyle CB$ is $\displaystyle -b/c$, so the slope of $\displaystyle AH$ is $\displaystyle c/b$, in fact $\displaystyle AH$
    is the line $\displaystyle y=(c/b)x$. The line $\displaystyle CB$ has equation $\displaystyle y=(-b/c).x+b$,
    so $\displaystyle H$ is the point of intersection of:

    $\displaystyle y=(c/b).x$,

    and

    $\displaystyle y=(-b/c).x+b$.

    Which is the point $\displaystyle (b^2.c/(b^2+c^2), b.c^2/(b^2+c^2))$.

    Now $\displaystyle P$ lies on $\displaystyle AM$, and may be written as $\displaystyle -\mu.(c/2,b/2)$ for some
    $\displaystyle \mu>0$.

    To finish find the equation on the line through $\displaystyle P$ and $\displaystyle B$, and from that
    find the coordinates of $\displaystyle Q$. Find the slope of the line through $\displaystyle Q$ and $\displaystyle A$,
    which should be minus the slope of the line through $\displaystyle P$ and $\displaystyle C$.


    RonL
    Attached Thumbnails Attached Thumbnails Geometry problem-orthocentre2.jpg  
    Last edited by CaptainBlack; Dec 20th 2005 at 10:22 AM.
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  3. #3
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    Sorry,but do you have another solution that just using similar/congruent triangles, circles...etc...I haven't learn about your way to solve this, so I don't understand that much, and I think my teacher has another more simple solution
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