How to prove the triangle is isosceles and determine its area












0












$begingroup$


Let $P(a,b)$ be a point on the curve $y = frac{1}{x}$ in first quadrant and let the tangent line at $P$ intersect the $x$-axis at a point $A$. Show that triangle $AOP$ is isosceles and determine its area, where $O$ is the origin.



How do I start to attempt this?






analytic-geometry plane-geometry







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Have you tried to find the tangent line in terms of $P$ yet?
    $endgroup$
    – SmileyCraft
    Dec 13 '18 at 14:18










  • $begingroup$
    Very good. So the slope of $PA$ is $-1/a^2$. Now if you also calculate the slope of $OP$ you should notice something which makes the triangle very symmetric.
    $endgroup$
    – SmileyCraft
    Dec 13 '18 at 15:04










  • $begingroup$
    So essentially, the slope of $OP$ is just the inverse of $PA$? @SmileyCraft
    $endgroup$
    – Valèt Quasar
    Dec 13 '18 at 15:10










  • $begingroup$
    Right, the additive inverse that is. Because $O$ and $A$ lie on the $x$-axis, it follows that $|PA|$=$|OP|$.
    $endgroup$
    – SmileyCraft
    Dec 13 '18 at 15:12










  • $begingroup$
    I see now. Thank you so much for your assistance! @SmileyCraft
    $endgroup$
    – Valèt Quasar
    Dec 13 '18 at 15:13
















0












$begingroup$


Let $P(a,b)$ be a point on the curve $y = frac{1}{x}$ in first quadrant and let the tangent line at $P$ intersect the $x$-axis at a point $A$. Show that triangle $AOP$ is isosceles and determine its area, where $O$ is the origin.



How do I start to attempt this?






analytic-geometry plane-geometry







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Have you tried to find the tangent line in terms of $P$ yet?
    $endgroup$
    – SmileyCraft
    Dec 13 '18 at 14:18










  • $begingroup$
    Very good. So the slope of $PA$ is $-1/a^2$. Now if you also calculate the slope of $OP$ you should notice something which makes the triangle very symmetric.
    $endgroup$
    – SmileyCraft
    Dec 13 '18 at 15:04










  • $begingroup$
    So essentially, the slope of $OP$ is just the inverse of $PA$? @SmileyCraft
    $endgroup$
    – Valèt Quasar
    Dec 13 '18 at 15:10










  • $begingroup$
    Right, the additive inverse that is. Because $O$ and $A$ lie on the $x$-axis, it follows that $|PA|$=$|OP|$.
    $endgroup$
    – SmileyCraft
    Dec 13 '18 at 15:12










  • $begingroup$
    I see now. Thank you so much for your assistance! @SmileyCraft
    $endgroup$
    – Valèt Quasar
    Dec 13 '18 at 15:13














0












0








0





$begingroup$


Let $P(a,b)$ be a point on the curve $y = frac{1}{x}$ in first quadrant and let the tangent line at $P$ intersect the $x$-axis at a point $A$. Show that triangle $AOP$ is isosceles and determine its area, where $O$ is the origin.



How do I start to attempt this?






analytic-geometry plane-geometry







share|cite|improve this question











$endgroup$




Let $P(a,b)$ be a point on the curve $y = frac{1}{x}$ in first quadrant and let the tangent line at $P$ intersect the $x$-axis at a point $A$. Show that triangle $AOP$ is isosceles and determine its area, where $O$ is the origin.



How do I start to attempt this?






analytic-geometry plane-geometry




analytic-geometry plane-geometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 14 '18 at 14:05









amWhy

1




1










asked Dec 13 '18 at 14:14









Valèt QuasarValèt Quasar

32




32








  • 1




    $begingroup$
    Have you tried to find the tangent line in terms of $P$ yet?
    $endgroup$
    – SmileyCraft
    Dec 13 '18 at 14:18










  • $begingroup$
    Very good. So the slope of $PA$ is $-1/a^2$. Now if you also calculate the slope of $OP$ you should notice something which makes the triangle very symmetric.
    $endgroup$
    – SmileyCraft
    Dec 13 '18 at 15:04










  • $begingroup$
    So essentially, the slope of $OP$ is just the inverse of $PA$? @SmileyCraft
    $endgroup$
    – Valèt Quasar
    Dec 13 '18 at 15:10










  • $begingroup$
    Right, the additive inverse that is. Because $O$ and $A$ lie on the $x$-axis, it follows that $|PA|$=$|OP|$.
    $endgroup$
    – SmileyCraft
    Dec 13 '18 at 15:12










  • $begingroup$
    I see now. Thank you so much for your assistance! @SmileyCraft
    $endgroup$
    – Valèt Quasar
    Dec 13 '18 at 15:13














  • 1




    $begingroup$
    Have you tried to find the tangent line in terms of $P$ yet?
    $endgroup$
    – SmileyCraft
    Dec 13 '18 at 14:18










  • $begingroup$
    Very good. So the slope of $PA$ is $-1/a^2$. Now if you also calculate the slope of $OP$ you should notice something which makes the triangle very symmetric.
    $endgroup$
    – SmileyCraft
    Dec 13 '18 at 15:04










  • $begingroup$
    So essentially, the slope of $OP$ is just the inverse of $PA$? @SmileyCraft
    $endgroup$
    – Valèt Quasar
    Dec 13 '18 at 15:10










  • $begingroup$
    Right, the additive inverse that is. Because $O$ and $A$ lie on the $x$-axis, it follows that $|PA|$=$|OP|$.
    $endgroup$
    – SmileyCraft
    Dec 13 '18 at 15:12










  • $begingroup$
    I see now. Thank you so much for your assistance! @SmileyCraft
    $endgroup$
    – Valèt Quasar
    Dec 13 '18 at 15:13








1




1




$begingroup$
Have you tried to find the tangent line in terms of $P$ yet?
$endgroup$
– SmileyCraft
Dec 13 '18 at 14:18




$begingroup$
Have you tried to find the tangent line in terms of $P$ yet?
$endgroup$
– SmileyCraft
Dec 13 '18 at 14:18












$begingroup$
Very good. So the slope of $PA$ is $-1/a^2$. Now if you also calculate the slope of $OP$ you should notice something which makes the triangle very symmetric.
$endgroup$
– SmileyCraft
Dec 13 '18 at 15:04




$begingroup$
Very good. So the slope of $PA$ is $-1/a^2$. Now if you also calculate the slope of $OP$ you should notice something which makes the triangle very symmetric.
$endgroup$
– SmileyCraft
Dec 13 '18 at 15:04












$begingroup$
So essentially, the slope of $OP$ is just the inverse of $PA$? @SmileyCraft
$endgroup$
– Valèt Quasar
Dec 13 '18 at 15:10




$begingroup$
So essentially, the slope of $OP$ is just the inverse of $PA$? @SmileyCraft
$endgroup$
– Valèt Quasar
Dec 13 '18 at 15:10












$begingroup$
Right, the additive inverse that is. Because $O$ and $A$ lie on the $x$-axis, it follows that $|PA|$=$|OP|$.
$endgroup$
– SmileyCraft
Dec 13 '18 at 15:12




$begingroup$
Right, the additive inverse that is. Because $O$ and $A$ lie on the $x$-axis, it follows that $|PA|$=$|OP|$.
$endgroup$
– SmileyCraft
Dec 13 '18 at 15:12












$begingroup$
I see now. Thank you so much for your assistance! @SmileyCraft
$endgroup$
– Valèt Quasar
Dec 13 '18 at 15:13




$begingroup$
I see now. Thank you so much for your assistance! @SmileyCraft
$endgroup$
– Valèt Quasar
Dec 13 '18 at 15:13










3 Answers
3






active

oldest

votes


















0












$begingroup$

To start approaching the problem, first notice that if P lies on the curve then $P (a, b) = P (a, frac{1}{a})$.



Secondly, find the slope of the tangent line at $P$, which is precisely $-frac{1}{a^2}$



Thus equation of the tangent line through $P$ is given by



$$(y - frac{1}{a}) = (-frac{1}{a^2})(x-a) implies a^2y = 2a - x$$



The above line intersects the $x$-axis only at $(2a, 0)$, thus $ A = (2a, 0)$.



Therefore the coordinates of $Delta AOP$ are precisely: $$A (2a, 0), O(0, 0), P (a, frac{1}{a})$$
making $AOP$ isosceles. Moreover, the area of $Delta AOP = frac{1}{2} cdot 2a cdot frac{1}{a} = 1$ unit.






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    Hint:



    WLOG $b=1/a>0$



    $$xy-1implies xdfrac{dy}{dx}+y=0impliesdfrac{dy}{dx}_{text{ at }(a,b)}=-dfrac{1/a}a=-dfrac1{a^2}$$



    The equation of the tangent at $P,$ $$dfrac{y-1/a}{x-a}=-dfrac1{a^2}iff x+a^2y=2a$$



    As $x$ axis, $y=0,$ the intersection will be $A(2a,0)$



    $$|AP|=?, |OP|=?$$






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      The tangent line to $y=frac 1x$ at $P(x_0,y_0)$ is:
      $$y=y_0+y'(x_0)(x-x_0)=frac 1{x_0}-frac 1{x_0^2}(x-x_0).$$
      The point $A(x_1,0)$ has coordinate $x_1$:
      $$0=frac 1{x_0}-frac 1{x_0^2}(x_1-x_0) Rightarrow x_1=2x_0,$$
      which shows the triangle $PAO$ is isosceles, because the altitude $PB$ is the median too.



      Now the area of $Delta PAO$ is:
      $$S=frac 12 cdot AOcdot PB=frac12 cdot 2x_0 cdot frac1{x_0}=1.$$






      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%2f3038053%2fhow-to-prove-the-triangle-is-isosceles-and-determine-its-area%23new-answer', 'question_page');
        }
        );

        Post as a guest















        Required, but never shown

























        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        0












        $begingroup$

        To start approaching the problem, first notice that if P lies on the curve then $P (a, b) = P (a, frac{1}{a})$.



        Secondly, find the slope of the tangent line at $P$, which is precisely $-frac{1}{a^2}$



        Thus equation of the tangent line through $P$ is given by



        $$(y - frac{1}{a}) = (-frac{1}{a^2})(x-a) implies a^2y = 2a - x$$



        The above line intersects the $x$-axis only at $(2a, 0)$, thus $ A = (2a, 0)$.



        Therefore the coordinates of $Delta AOP$ are precisely: $$A (2a, 0), O(0, 0), P (a, frac{1}{a})$$
        making $AOP$ isosceles. Moreover, the area of $Delta AOP = frac{1}{2} cdot 2a cdot frac{1}{a} = 1$ unit.






        share|cite|improve this answer











        $endgroup$


















          0












          $begingroup$

          To start approaching the problem, first notice that if P lies on the curve then $P (a, b) = P (a, frac{1}{a})$.



          Secondly, find the slope of the tangent line at $P$, which is precisely $-frac{1}{a^2}$



          Thus equation of the tangent line through $P$ is given by



          $$(y - frac{1}{a}) = (-frac{1}{a^2})(x-a) implies a^2y = 2a - x$$



          The above line intersects the $x$-axis only at $(2a, 0)$, thus $ A = (2a, 0)$.



          Therefore the coordinates of $Delta AOP$ are precisely: $$A (2a, 0), O(0, 0), P (a, frac{1}{a})$$
          making $AOP$ isosceles. Moreover, the area of $Delta AOP = frac{1}{2} cdot 2a cdot frac{1}{a} = 1$ unit.






          share|cite|improve this answer











          $endgroup$
















            0












            0








            0





            $begingroup$

            To start approaching the problem, first notice that if P lies on the curve then $P (a, b) = P (a, frac{1}{a})$.



            Secondly, find the slope of the tangent line at $P$, which is precisely $-frac{1}{a^2}$



            Thus equation of the tangent line through $P$ is given by



            $$(y - frac{1}{a}) = (-frac{1}{a^2})(x-a) implies a^2y = 2a - x$$



            The above line intersects the $x$-axis only at $(2a, 0)$, thus $ A = (2a, 0)$.



            Therefore the coordinates of $Delta AOP$ are precisely: $$A (2a, 0), O(0, 0), P (a, frac{1}{a})$$
            making $AOP$ isosceles. Moreover, the area of $Delta AOP = frac{1}{2} cdot 2a cdot frac{1}{a} = 1$ unit.






            share|cite|improve this answer











            $endgroup$



            To start approaching the problem, first notice that if P lies on the curve then $P (a, b) = P (a, frac{1}{a})$.



            Secondly, find the slope of the tangent line at $P$, which is precisely $-frac{1}{a^2}$



            Thus equation of the tangent line through $P$ is given by



            $$(y - frac{1}{a}) = (-frac{1}{a^2})(x-a) implies a^2y = 2a - x$$



            The above line intersects the $x$-axis only at $(2a, 0)$, thus $ A = (2a, 0)$.



            Therefore the coordinates of $Delta AOP$ are precisely: $$A (2a, 0), O(0, 0), P (a, frac{1}{a})$$
            making $AOP$ isosceles. Moreover, the area of $Delta AOP = frac{1}{2} cdot 2a cdot frac{1}{a} = 1$ unit.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 14 '18 at 13:54

























            answered Dec 13 '18 at 20:45









            thesagniksahathesagniksaha

            164115




            164115























                1












                $begingroup$

                Hint:



                WLOG $b=1/a>0$



                $$xy-1implies xdfrac{dy}{dx}+y=0impliesdfrac{dy}{dx}_{text{ at }(a,b)}=-dfrac{1/a}a=-dfrac1{a^2}$$



                The equation of the tangent at $P,$ $$dfrac{y-1/a}{x-a}=-dfrac1{a^2}iff x+a^2y=2a$$



                As $x$ axis, $y=0,$ the intersection will be $A(2a,0)$



                $$|AP|=?, |OP|=?$$






                share|cite|improve this answer









                $endgroup$


















                  1












                  $begingroup$

                  Hint:



                  WLOG $b=1/a>0$



                  $$xy-1implies xdfrac{dy}{dx}+y=0impliesdfrac{dy}{dx}_{text{ at }(a,b)}=-dfrac{1/a}a=-dfrac1{a^2}$$



                  The equation of the tangent at $P,$ $$dfrac{y-1/a}{x-a}=-dfrac1{a^2}iff x+a^2y=2a$$



                  As $x$ axis, $y=0,$ the intersection will be $A(2a,0)$



                  $$|AP|=?, |OP|=?$$






                  share|cite|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    Hint:



                    WLOG $b=1/a>0$



                    $$xy-1implies xdfrac{dy}{dx}+y=0impliesdfrac{dy}{dx}_{text{ at }(a,b)}=-dfrac{1/a}a=-dfrac1{a^2}$$



                    The equation of the tangent at $P,$ $$dfrac{y-1/a}{x-a}=-dfrac1{a^2}iff x+a^2y=2a$$



                    As $x$ axis, $y=0,$ the intersection will be $A(2a,0)$



                    $$|AP|=?, |OP|=?$$






                    share|cite|improve this answer









                    $endgroup$



                    Hint:



                    WLOG $b=1/a>0$



                    $$xy-1implies xdfrac{dy}{dx}+y=0impliesdfrac{dy}{dx}_{text{ at }(a,b)}=-dfrac{1/a}a=-dfrac1{a^2}$$



                    The equation of the tangent at $P,$ $$dfrac{y-1/a}{x-a}=-dfrac1{a^2}iff x+a^2y=2a$$



                    As $x$ axis, $y=0,$ the intersection will be $A(2a,0)$



                    $$|AP|=?, |OP|=?$$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 13 '18 at 14:22









                    lab bhattacharjeelab bhattacharjee

                    225k15157275




                    225k15157275























                        0












                        $begingroup$

                        The tangent line to $y=frac 1x$ at $P(x_0,y_0)$ is:
                        $$y=y_0+y'(x_0)(x-x_0)=frac 1{x_0}-frac 1{x_0^2}(x-x_0).$$
                        The point $A(x_1,0)$ has coordinate $x_1$:
                        $$0=frac 1{x_0}-frac 1{x_0^2}(x_1-x_0) Rightarrow x_1=2x_0,$$
                        which shows the triangle $PAO$ is isosceles, because the altitude $PB$ is the median too.



                        Now the area of $Delta PAO$ is:
                        $$S=frac 12 cdot AOcdot PB=frac12 cdot 2x_0 cdot frac1{x_0}=1.$$






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          The tangent line to $y=frac 1x$ at $P(x_0,y_0)$ is:
                          $$y=y_0+y'(x_0)(x-x_0)=frac 1{x_0}-frac 1{x_0^2}(x-x_0).$$
                          The point $A(x_1,0)$ has coordinate $x_1$:
                          $$0=frac 1{x_0}-frac 1{x_0^2}(x_1-x_0) Rightarrow x_1=2x_0,$$
                          which shows the triangle $PAO$ is isosceles, because the altitude $PB$ is the median too.



                          Now the area of $Delta PAO$ is:
                          $$S=frac 12 cdot AOcdot PB=frac12 cdot 2x_0 cdot frac1{x_0}=1.$$






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            The tangent line to $y=frac 1x$ at $P(x_0,y_0)$ is:
                            $$y=y_0+y'(x_0)(x-x_0)=frac 1{x_0}-frac 1{x_0^2}(x-x_0).$$
                            The point $A(x_1,0)$ has coordinate $x_1$:
                            $$0=frac 1{x_0}-frac 1{x_0^2}(x_1-x_0) Rightarrow x_1=2x_0,$$
                            which shows the triangle $PAO$ is isosceles, because the altitude $PB$ is the median too.



                            Now the area of $Delta PAO$ is:
                            $$S=frac 12 cdot AOcdot PB=frac12 cdot 2x_0 cdot frac1{x_0}=1.$$






                            share|cite|improve this answer









                            $endgroup$



                            The tangent line to $y=frac 1x$ at $P(x_0,y_0)$ is:
                            $$y=y_0+y'(x_0)(x-x_0)=frac 1{x_0}-frac 1{x_0^2}(x-x_0).$$
                            The point $A(x_1,0)$ has coordinate $x_1$:
                            $$0=frac 1{x_0}-frac 1{x_0^2}(x_1-x_0) Rightarrow x_1=2x_0,$$
                            which shows the triangle $PAO$ is isosceles, because the altitude $PB$ is the median too.



                            Now the area of $Delta PAO$ is:
                            $$S=frac 12 cdot AOcdot PB=frac12 cdot 2x_0 cdot frac1{x_0}=1.$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 13 '18 at 16:24









                            farruhotafarruhota

                            20.2k2738




                            20.2k2738






























                                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%2f3038053%2fhow-to-prove-the-triangle-is-isosceles-and-determine-its-area%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

                                Ellipse (mathématiques)

                                Image
                                Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
                                Read more

                                Quarter-circle Tiles

                                Image
                                up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
                                Read more

                                Mont Emei

                                Image
                                Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
                                Read more