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?






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 4 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

                    Ponta tanko

                    Image
                    Usona ponta tanko M60A1 Ponta tanko estas, simile kiel kirasita savveturilo, inter specialaj kirasitaj transportiloj. Kiel pontaj tankoj eblas konsideri ĉiujn memmovigaj rimedoj ebligantaj superi obstaklojn ĝis larĝeco 24 metroj rekte en spaco de bataloperacoj. Ties ĉefa tasko estas ebligi al la propraj taĉmentoj daŭrigi en rapida antaŭenirado sen senbezona reteniĝo. El tio pli-malpli rezultas taktikaj-teknikaj postuloj, kiujn ili devas plenumi. Precipe temas pri kapablo operacii komune kun unue-sinsekvaj taĉmentoj kaj transpontigi obstaklojn dum kelke da minutoj. This page is only for reference, If you need detailed information, please check here
                    Read more

                    Tantalo (mitologio)

                    Image
                    Ĉi tiu artikolo temas pri mitologio. Por aliaj signifoj vidu la paĝon Tantalo (apartigilo) . Tantalo , laŭ Gioacchino Assereto. Laŭ la helena mitologio, Tantalo (el la greka Τάνταλος , ŝancelanta aŭ tre malfeliĉa ) estis filo de Zeŭso kaj la oceanidino Pluto, laŭ la plej divastigata versio. Li estis reĝo de Frigio [1] aŭ de la monto Sipilo, en Lidio (Malgrandazio) [2] kaj post sia morto li suferis eternan torturadon ĉe la Tartaro, la plej profunda parto de Hadeso, rezervita por la plej perversaj homoj. Tamen, kelkaj rakontoj asertas ke li estis reĝo de Argo aŭ de Korinto [3] , aŭ ke li migris norden el la monto Sipilo por akiri la tronon de Paflagonio, el kie Ilo forpelis lin pro tio, ke Tantalo forkaptis kaj seksperfortis lian fraton Ganimedon [4] , kaj tiel ofendis la diojn. Verŝajne li ankaŭ regis urbon en okcidenta Anatolio, kiun oni nomis Tantalis [5] pro li, aŭ Sipilo , pro la monto sur kies flanko ankoraŭ staras ĝiaj ruinaĵoj [6] . Laŭ la geografiisto Strabo,...
                    Read more

                    Erzsébet Schaár

                    Image
                    Erzsébet Schaár Memportreto de Erzsébet Schaár en Pécs Personaj informoj Naskiĝo 27-an de julio 1908 ( 1908-07-27 ) en Budafok Morto 29-an de aŭgusto 1975 ( 1975-08-29 ) (67-jara) en Budapeŝto Ŝtataneco Hungario Profesio skulptisto Familianoj Edz(in)o Tibor Vilt v   •   d   •   r Erzsébet Schaár [erĵEbet ŝAr], laŭ hungarlingve kutima sinsekvo Schaár Erzsébet estis hungara skulptisto. Ŝia edzo estis Tibor Vilt. Déryné Széppataki Róza en Miskolc fare de Erzsébet Schaár Erzsébet Schaár [1] naskiĝis la 27-an de julio 1908 en Budafok, ŝi mortis la 29-an de aŭgusto 1975 en Budapeŝto. Enhavo 1 Biografio 2 Elektitaj solaj ekspozicioj 3 Elektitaj kolektivaj ekspozicioj 4 Elektitaj skulptaĵoj 5 Elektitaj disĉiploj 6 Fontoj 7 Referencoj Biografio | Erzsébet Schaár studis en Industriarta Lernejo, poste en Reĝa Belarta Altlernejo (1924-1926). Ŝia ĉefinstruisto estis Zsig...
                    Read more