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  1. #1
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    triangle question

    Hi, here is my problem:

    Inside of a triangle ABC, a random point M, in which three straight lines parallel to the respective sides of the triangle intersect, was chosen. Those straight lines have divided the triangle into six parts, three of which are also triangles. Let r_1, r_2, r_3 be the radii of circles inscribed in those three triangles, and let r be the radius of a circle inscribed in triangle ABC. Show that r=r_1+r_2+r_3.

    Help appreciated. Thank you.
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  2. #2
    Moo
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    I found it !

    Okay look at the sketch. It is not necessary to draw the inscribed circles.


    triangle question-inscribed.jpg


    Consider triangles DEM and ABC. All their sides are parallel to each other. Hence they're congruent (or similar). Let k_1 be the ratio between the two. This means that k_1=\frac{DE}{AB}=\frac{DM}{AC}={\color{red}\frac{  EM}{BC}}={\color{blue}\frac{r_1}{r}}=\dots
    This always holds true for two congruent triangles.

    It'll be the same for triangle FGM : k_2={\color{blue}\frac{r_2}{r}}={\color{red}\frac{  MG}{BC}}=\dots

    and for triangle HMI : k_3={\color{blue}\frac{r_3}{r}}={\color{red}\frac{  HI}{BC}}

    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
    Now consider the blue parts of the equalities.
    You can see that if you sum them, it should give 1, according to the result you're given.

    So we'll prove that the sum of the red parts is equal to 1 !

    {\color{red}\frac{EM}{BC}+\frac{MG}{BC}+\frac{HI}{  BC}}

    But since EM//BH and EB//MH (parallel), then EMHB is a parallelogram. Hence \bold{EM=BH}

    Similarly, MGCI is a parallelogram. And hence \bold{MG=IC}


    So the red sum becomes :
    =\frac{BH}{BC}+\frac{IC}{BC}+\frac{HI}{BC}=\frac{B  H+IC+HI}{BC}

    you can see that the points B, H, I and C are aligned. Hence BH+IC+HI=BC

    So the red part is equal to 1.

    Therefore for the blue part :
    1=\frac{EM}{BC}+\frac{MG}{BC}+\frac{HI}{BC}=\frac{  r_1}{r}+\frac{r_2}{r}+\frac{r_3}{r}=\frac{r_1+r_2+  r_3}{r}

    And you're done
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