Geometry problem - areas of triangles (contest math)












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?










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?










    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?










      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






      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






















          1 Answer
          1






          active

          oldest

          votes


















          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









          $endgroup$














            Your Answer





            StackExchange.ifUsing("editor", function () {
            return StackExchange.using("mathjaxEditing", function () {
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            });
            });
            }, "mathjax-editing");

            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "69"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3167832%2fgeometry-problem-areas-of-triangles-contest-math%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            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












              $begingroup$

              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






























                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Mathematics Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3167832%2fgeometry-problem-areas-of-triangles-contest-math%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    Popular posts from this blog

                    Liste der Baudenkmale in Friedland (Mecklenburg)

                    Single-Malt-Whisky

                    Czorneboh