Geometry problem - areas of triangles (contest math) The Next CEO of Stack OverflowContest...

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Geometry problem - areas of triangles (contest math)

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Geometry problem - areas of triangles (contest math)



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$begingroup$


This problem is from 2019 Math Kangaroo competition for 9th-10th graders that took place last week, problem #29.
enter image description here



I was able to solve it using coordinate geometry, both triangles have the same area. However, I do not expect 9th graders to know this method. Is there a simpler solution that I am not seeing?






contest-math euclidean-geometry







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    This problem is from 2019 Math Kangaroo competition for 9th-10th graders that took place last week, problem #29.
    enter image description here



    I was able to solve it using coordinate geometry, both triangles have the same area. However, I do not expect 9th graders to know this method. Is there a simpler solution that I am not seeing?






    contest-math euclidean-geometry







    share|cite|improve this question











    $endgroup$















      1












      1








      1


      1



      $begingroup$


      This problem is from 2019 Math Kangaroo competition for 9th-10th graders that took place last week, problem #29.
      enter image description here



      I was able to solve it using coordinate geometry, both triangles have the same area. However, I do not expect 9th graders to know this method. Is there a simpler solution that I am not seeing?






      contest-math euclidean-geometry







      share|cite|improve this question











      $endgroup$




      This problem is from 2019 Math Kangaroo competition for 9th-10th graders that took place last week, problem #29.
      enter image description here



      I was able to solve it using coordinate geometry, both triangles have the same area. However, I do not expect 9th graders to know this method. Is there a simpler solution that I am not seeing?






      contest-math euclidean-geometry




      contest-math euclidean-geometry






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 3 hours ago







      Vasya

















      asked 4 hours ago









      VasyaVasya

      4,1351618




      4,1351618






















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          $begingroup$

          Since $D$ is the midpoint of $BC$, $A_triangle ACD=A_triangle ABD=frac{1}{2}S$.



          Since $AP=2AB$ and $AQ=3AD$, $A_triangle APQ$ is $2times 3=6$ times $A_triangle ABD$.
          Similarly $A_triangle AQR$ and $A_triangle APR$. So $A_triangle PQR = A_triangle APQ+A_triangle AQR - A_triangle APR$, giving the answer.



          All this is just the ratio of areas of triangle with same base and ratio of height (or vice versa), which a year 9 student should already know.






          share|cite|improve this answer









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            $begingroup$

            Since $D$ is the midpoint of $BC$, $A_triangle ACD=A_triangle ABD=frac{1}{2}S$.



            Since $AP=2AB$ and $AQ=3AD$, $A_triangle APQ$ is $2times 3=6$ times $A_triangle ABD$.
            Similarly $A_triangle AQR$ and $A_triangle APR$. So $A_triangle PQR = A_triangle APQ+A_triangle AQR - A_triangle APR$, giving the answer.



            All this is just the ratio of areas of triangle with same base and ratio of height (or vice versa), which a year 9 student should already know.






            share|cite|improve this answer









            $endgroup$


















              4












              $begingroup$

              Since $D$ is the midpoint of $BC$, $A_triangle ACD=A_triangle ABD=frac{1}{2}S$.



              Since $AP=2AB$ and $AQ=3AD$, $A_triangle APQ$ is $2times 3=6$ times $A_triangle ABD$.
              Similarly $A_triangle AQR$ and $A_triangle APR$. So $A_triangle PQR = A_triangle APQ+A_triangle AQR - A_triangle APR$, giving the answer.



              All this is just the ratio of areas of triangle with same base and ratio of height (or vice versa), which a year 9 student should already know.






              share|cite|improve this answer









              $endgroup$
















                4












                4








                4





                $begingroup$

                Since $D$ is the midpoint of $BC$, $A_triangle ACD=A_triangle ABD=frac{1}{2}S$.



                Since $AP=2AB$ and $AQ=3AD$, $A_triangle APQ$ is $2times 3=6$ times $A_triangle ABD$.
                Similarly $A_triangle AQR$ and $A_triangle APR$. So $A_triangle PQR = A_triangle APQ+A_triangle AQR - A_triangle APR$, giving the answer.



                All this is just the ratio of areas of triangle with same base and ratio of height (or vice versa), which a year 9 student should already know.






                share|cite|improve this answer









                $endgroup$



                Since $D$ is the midpoint of $BC$, $A_triangle ACD=A_triangle ABD=frac{1}{2}S$.



                Since $AP=2AB$ and $AQ=3AD$, $A_triangle APQ$ is $2times 3=6$ times $A_triangle ABD$.
                Similarly $A_triangle AQR$ and $A_triangle APR$. So $A_triangle PQR = A_triangle APQ+A_triangle AQR - A_triangle APR$, giving the answer.



                All this is just the ratio of areas of triangle with same base and ratio of height (or vice versa), which a year 9 student should already know.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 4 hours ago









                user10354138user10354138

                7,4722925




                7,4722925






























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