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 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

                    Knooppunt Holsloot

                    Image
                    Knooppunt Holsloot Kenmerken Wegen A37 × N34 Type knooppunt klaverblad Portaal     Verkeer & Vervoer Knooppunt Holsloot is een verkeersknooppunt op de kruising van de autosnelweg A37 en de N34 in Drenthe, tussen Emmen en Coevorden. Oorspronkelijk was deze aansluiting een afritconstructie van het type Haarlemmermeeraansluiting, zoals bij Knooppunt Hooipolder. Sinds 2002 is het knooppunt omgebouwd tot een volledig klaverblad. Als noviteit werd hier voor het eerst gebruikgemaakt van dimbare verlichting in plaats van de normale vaste tijden waarop de verlichting aan en uit gaat, wat een grote kostenbesparing opleverde. [1] Dat Holsloot als volwaardig knooppunt is uitgevoerd, waarbij alle verkeersstromen en afbuigingen ononderbroken kunnen doorgaan, is uitzonderlijk. Het betreft hier namelijk een aansluiting tussen een autosnelweg en niet-snelweg. Een vergelijkbare situatie treft men nabij Hoensbroek bij het knooppunt Ten Esschen.     Aa
                    Read more

                    Altaar (religie)

                    Image
                    Deel van de serie over kerkelijk gerei waaronder paramenten en liturgisch vaatwerk gebruikt in de liturgie Liturgisch vaatwerk Vasa sacra Miskelk · Pateen Kelklepeltje Ciborie · Monstrans Pyxis · Custodiale Vasa non sacra Ampullen · Wijwatervat Olievaatje Wierookvat · Ablutievat Paramenten Amict · Albe · Baarkleed · Cingel Tuniek · Dalmatiek · Fanon Kazuifel · Manipel · Stola Gremiale · Benedictievelum Mijter Koorkledij Rochet · Superplie Koorkap · Cappa magna Kovel Kelkgerei Bursa · Kelkvelum Ciborievelum Kerklinnen Corporale · Kelkdoekje Lavabodoekje · Palla Altaardwaal Kerkinterieur Altaar · Ambo Biechtstoel · Communiebank Doksaal · Doopvont Faldistorium Godslamp · Hoogaltaar Heilig Kruisaltaar Katheder · Preekstoel Sedilia Tabernakel · Volksaltaar Liturgische boeken Altaarmissaal Benedictionale · Brevier Evangeliarium · Evangelistarium Graduale · Kyriale Lectionarium Psalter · Rituaal Sacramentarium · Volksmissaal Overige Flambou
                    Read more

                    Gregoriusmis

                    Image
                    Een afbeelding van de Gregoriusmis uit een getijdenboek gemaakt in de zuidelijke Nederlanden tussen 1500 en 1525 Gregoriusmis, Meester van het Akense Altaar, ca. 1500-1525. Collectie Museum Catharijneconvent De Gregoriusmis of de Mis van Sint Gregorius is een motief in katholieke kunst, dat het eerst verscheen in de late Middeleeuwen en dat nog steeds werd gebruikt tijdens de contrareformatie. Paus Gregorius I (ca. 540-604) is afgebeeld, de H. Mis opdragend, terwijl Christus als de Man van Smarten verschijnt op het altaar dat voor Sint Gregorius staat. Volgens de legende had St. Gregorius gebeden om een teken voor een betwijfelaar van de transsubstantiatieleer, waarop de lijdende Christus tijdens het opdragen van de Mis verscheen om hem te overtuigen en waarbij zijn bloed in de altaarkelk liep. Vaak zijn ook de passiewerktuigen afgebeeld. Zie de categorie Mass of Saint Gregory van Wikimedia Commons voor mediabestanden over dit onderwerp. This page is onl
                    Read more