If $|MA|^2=|MB|^2+|MC|^2$. Find the $angle BMC=?$ [closed]












2












$begingroup$


My problem is:




The point $M$ is the internal point of the $ABC$ equilateral triangle. Find the $angle BMC$ if $$|MA|^2=|MB|^2+|MC|^2$$




Is this question related to similar triangles? Or which theorem should I apply?






geometry contest-math euclidean-geometry book-recommendation







share|cite|improve this question











$endgroup$



closed as off-topic by RRL, Lee David Chung Lin, jgon, KReiser, Cesareo Jan 5 at 0:29


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Lee David Chung Lin, jgon, KReiser, Cesareo

If this question can be reworded to fit the rules in the help center, please edit the question.





















    2












    $begingroup$


    My problem is:




    The point $M$ is the internal point of the $ABC$ equilateral triangle. Find the $angle BMC$ if $$|MA|^2=|MB|^2+|MC|^2$$




    Is this question related to similar triangles? Or which theorem should I apply?






    geometry contest-math euclidean-geometry book-recommendation







    share|cite|improve this question











    $endgroup$



    closed as off-topic by RRL, Lee David Chung Lin, jgon, KReiser, Cesareo Jan 5 at 0:29


    This question appears to be off-topic. The users who voted to close gave this specific reason:


    • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Lee David Chung Lin, jgon, KReiser, Cesareo

    If this question can be reworded to fit the rules in the help center, please edit the question.



















      2












      2








      2





      $begingroup$


      My problem is:




      The point $M$ is the internal point of the $ABC$ equilateral triangle. Find the $angle BMC$ if $$|MA|^2=|MB|^2+|MC|^2$$




      Is this question related to similar triangles? Or which theorem should I apply?






      geometry contest-math euclidean-geometry book-recommendation







      share|cite|improve this question











      $endgroup$




      My problem is:




      The point $M$ is the internal point of the $ABC$ equilateral triangle. Find the $angle BMC$ if $$|MA|^2=|MB|^2+|MC|^2$$




      Is this question related to similar triangles? Or which theorem should I apply?






      geometry contest-math euclidean-geometry book-recommendation




      geometry contest-math euclidean-geometry book-recommendation






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 26 '18 at 11:56









      greedoid

      44k1155109




      44k1155109










      asked Dec 26 '18 at 11:39









      ElementaryElementary

      361111




      361111




      closed as off-topic by RRL, Lee David Chung Lin, jgon, KReiser, Cesareo Jan 5 at 0:29


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Lee David Chung Lin, jgon, KReiser, Cesareo

      If this question can be reworded to fit the rules in the help center, please edit the question.







      closed as off-topic by RRL, Lee David Chung Lin, jgon, KReiser, Cesareo Jan 5 at 0:29


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Lee David Chung Lin, jgon, KReiser, Cesareo

      If this question can be reworded to fit the rules in the help center, please edit the question.






















          3 Answers
          3






          active

          oldest

          votes


















          4












          $begingroup$

          Hint:



          Rotate $C$ around $B$ for $60^{circ}$ and let $N$ be a picture of $M$ under this rotation. What can you say for triangles $BMN$ and $AMN$?



          enter image description here



          Edit: Triangle $BMC$ must be equilateral and since $$ AN^2+MN^2 = CM^2+BM^2 = AM^2$$ we see that triangle $ANM$ is rectangular with 90 at $N$...






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I'm practicing the hint you're saying.
            $endgroup$
            – Elementary
            Dec 26 '18 at 11:56










          • $begingroup$
            Is it possible, to add picture, if you have a few time ..I could not do..:(
            $endgroup$
            – Elementary
            Dec 26 '18 at 12:47










          • $begingroup$
            Thank you very much. Your method has exceeded my understanding.I could not. There is no problem in the solution. The problem is with me.
            $endgroup$
            – Elementary
            Dec 26 '18 at 13:00










          • $begingroup$
            I made some edit...
            $endgroup$
            – greedoid
            Dec 26 '18 at 13:03










          • $begingroup$
            Thank you again..
            $endgroup$
            – Elementary
            Dec 26 '18 at 13:26



















          2












          $begingroup$

          Use coordinate geometry. Note that $|MA|^2=|MB|^2+|MC|^2$ is symmetrical in $B & C$. This means $M$ is located symmetrically with respect to $B & C$. Let the coordinates of $A$ be $(0,sqrt{3}a)$, $B$ be $(-a,0)$ and $C$ be $(a,0)$. Since $M$ has to be symmetrical, let the coordinates for $M$ be $(0,y)$. Now,
          $$|MA|^2=|MB|^2+|MC|^2$$
          $$implies y^2+2sqrt3ay-a^2=0$$
          $$implies y=(2-sqrt3)a$$Note that you'll get two values of y but it can't be negative



          Now since you've $M$, you can get the angle $BMC$ to be equal to $150$ degrees. I hope you can get there :)






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            How do you get, $sqrt3 a$
            $endgroup$
            – Elementary
            Dec 26 '18 at 12:05












          • $begingroup$
            Since it's an equilateral triangle, I've taken the length of each side to be $2a$ (calculations will be a little bit easy with 2a instead of just a). You know how I got $sqrt3a$ now right?
            $endgroup$
            – Ankit Kumar
            Dec 26 '18 at 12:06



















          1












          $begingroup$

          That equation is essentially the Pythagorean theorem applied to lines MA, MB, MC. Even though they do not form a triangle in this scenario, the equation shows that they can. Line MA is the hypotenuse in this case, which makes angle BMC 90 degrees.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            No, this is wrong approach.
            $endgroup$
            – Elementary
            Dec 26 '18 at 11:51






          • 1




            $begingroup$
            The answer is incorrect too
            $endgroup$
            – Ankit Kumar
            Dec 26 '18 at 11:52










          • $begingroup$
            But why doesn't that work?
            $endgroup$
            – Viktor Glombik
            Dec 26 '18 at 12:39


















          3 Answers
          3






          active

          oldest

          votes








          3 Answers
          3






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          4












          $begingroup$

          Hint:



          Rotate $C$ around $B$ for $60^{circ}$ and let $N$ be a picture of $M$ under this rotation. What can you say for triangles $BMN$ and $AMN$?



          enter image description here



          Edit: Triangle $BMC$ must be equilateral and since $$ AN^2+MN^2 = CM^2+BM^2 = AM^2$$ we see that triangle $ANM$ is rectangular with 90 at $N$...






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I'm practicing the hint you're saying.
            $endgroup$
            – Elementary
            Dec 26 '18 at 11:56










          • $begingroup$
            Is it possible, to add picture, if you have a few time ..I could not do..:(
            $endgroup$
            – Elementary
            Dec 26 '18 at 12:47










          • $begingroup$
            Thank you very much. Your method has exceeded my understanding.I could not. There is no problem in the solution. The problem is with me.
            $endgroup$
            – Elementary
            Dec 26 '18 at 13:00










          • $begingroup$
            I made some edit...
            $endgroup$
            – greedoid
            Dec 26 '18 at 13:03










          • $begingroup$
            Thank you again..
            $endgroup$
            – Elementary
            Dec 26 '18 at 13:26
















          4












          $begingroup$

          Hint:



          Rotate $C$ around $B$ for $60^{circ}$ and let $N$ be a picture of $M$ under this rotation. What can you say for triangles $BMN$ and $AMN$?



          enter image description here



          Edit: Triangle $BMC$ must be equilateral and since $$ AN^2+MN^2 = CM^2+BM^2 = AM^2$$ we see that triangle $ANM$ is rectangular with 90 at $N$...






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I'm practicing the hint you're saying.
            $endgroup$
            – Elementary
            Dec 26 '18 at 11:56










          • $begingroup$
            Is it possible, to add picture, if you have a few time ..I could not do..:(
            $endgroup$
            – Elementary
            Dec 26 '18 at 12:47










          • $begingroup$
            Thank you very much. Your method has exceeded my understanding.I could not. There is no problem in the solution. The problem is with me.
            $endgroup$
            – Elementary
            Dec 26 '18 at 13:00










          • $begingroup$
            I made some edit...
            $endgroup$
            – greedoid
            Dec 26 '18 at 13:03










          • $begingroup$
            Thank you again..
            $endgroup$
            – Elementary
            Dec 26 '18 at 13:26














          4












          4








          4





          $begingroup$

          Hint:



          Rotate $C$ around $B$ for $60^{circ}$ and let $N$ be a picture of $M$ under this rotation. What can you say for triangles $BMN$ and $AMN$?



          enter image description here



          Edit: Triangle $BMC$ must be equilateral and since $$ AN^2+MN^2 = CM^2+BM^2 = AM^2$$ we see that triangle $ANM$ is rectangular with 90 at $N$...






          share|cite|improve this answer











          $endgroup$



          Hint:



          Rotate $C$ around $B$ for $60^{circ}$ and let $N$ be a picture of $M$ under this rotation. What can you say for triangles $BMN$ and $AMN$?



          enter image description here



          Edit: Triangle $BMC$ must be equilateral and since $$ AN^2+MN^2 = CM^2+BM^2 = AM^2$$ we see that triangle $ANM$ is rectangular with 90 at $N$...







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 26 '18 at 13:03

























          answered Dec 26 '18 at 11:53









          greedoidgreedoid

          44k1155109




          44k1155109












          • $begingroup$
            I'm practicing the hint you're saying.
            $endgroup$
            – Elementary
            Dec 26 '18 at 11:56










          • $begingroup$
            Is it possible, to add picture, if you have a few time ..I could not do..:(
            $endgroup$
            – Elementary
            Dec 26 '18 at 12:47










          • $begingroup$
            Thank you very much. Your method has exceeded my understanding.I could not. There is no problem in the solution. The problem is with me.
            $endgroup$
            – Elementary
            Dec 26 '18 at 13:00










          • $begingroup$
            I made some edit...
            $endgroup$
            – greedoid
            Dec 26 '18 at 13:03










          • $begingroup$
            Thank you again..
            $endgroup$
            – Elementary
            Dec 26 '18 at 13:26


















          • $begingroup$
            I'm practicing the hint you're saying.
            $endgroup$
            – Elementary
            Dec 26 '18 at 11:56










          • $begingroup$
            Is it possible, to add picture, if you have a few time ..I could not do..:(
            $endgroup$
            – Elementary
            Dec 26 '18 at 12:47










          • $begingroup$
            Thank you very much. Your method has exceeded my understanding.I could not. There is no problem in the solution. The problem is with me.
            $endgroup$
            – Elementary
            Dec 26 '18 at 13:00










          • $begingroup$
            I made some edit...
            $endgroup$
            – greedoid
            Dec 26 '18 at 13:03










          • $begingroup$
            Thank you again..
            $endgroup$
            – Elementary
            Dec 26 '18 at 13:26
















          $begingroup$
          I'm practicing the hint you're saying.
          $endgroup$
          – Elementary
          Dec 26 '18 at 11:56




          $begingroup$
          I'm practicing the hint you're saying.
          $endgroup$
          – Elementary
          Dec 26 '18 at 11:56












          $begingroup$
          Is it possible, to add picture, if you have a few time ..I could not do..:(
          $endgroup$
          – Elementary
          Dec 26 '18 at 12:47




          $begingroup$
          Is it possible, to add picture, if you have a few time ..I could not do..:(
          $endgroup$
          – Elementary
          Dec 26 '18 at 12:47












          $begingroup$
          Thank you very much. Your method has exceeded my understanding.I could not. There is no problem in the solution. The problem is with me.
          $endgroup$
          – Elementary
          Dec 26 '18 at 13:00




          $begingroup$
          Thank you very much. Your method has exceeded my understanding.I could not. There is no problem in the solution. The problem is with me.
          $endgroup$
          – Elementary
          Dec 26 '18 at 13:00












          $begingroup$
          I made some edit...
          $endgroup$
          – greedoid
          Dec 26 '18 at 13:03




          $begingroup$
          I made some edit...
          $endgroup$
          – greedoid
          Dec 26 '18 at 13:03












          $begingroup$
          Thank you again..
          $endgroup$
          – Elementary
          Dec 26 '18 at 13:26




          $begingroup$
          Thank you again..
          $endgroup$
          – Elementary
          Dec 26 '18 at 13:26











          2












          $begingroup$

          Use coordinate geometry. Note that $|MA|^2=|MB|^2+|MC|^2$ is symmetrical in $B & C$. This means $M$ is located symmetrically with respect to $B & C$. Let the coordinates of $A$ be $(0,sqrt{3}a)$, $B$ be $(-a,0)$ and $C$ be $(a,0)$. Since $M$ has to be symmetrical, let the coordinates for $M$ be $(0,y)$. Now,
          $$|MA|^2=|MB|^2+|MC|^2$$
          $$implies y^2+2sqrt3ay-a^2=0$$
          $$implies y=(2-sqrt3)a$$Note that you'll get two values of y but it can't be negative



          Now since you've $M$, you can get the angle $BMC$ to be equal to $150$ degrees. I hope you can get there :)






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            How do you get, $sqrt3 a$
            $endgroup$
            – Elementary
            Dec 26 '18 at 12:05












          • $begingroup$
            Since it's an equilateral triangle, I've taken the length of each side to be $2a$ (calculations will be a little bit easy with 2a instead of just a). You know how I got $sqrt3a$ now right?
            $endgroup$
            – Ankit Kumar
            Dec 26 '18 at 12:06
















          2












          $begingroup$

          Use coordinate geometry. Note that $|MA|^2=|MB|^2+|MC|^2$ is symmetrical in $B & C$. This means $M$ is located symmetrically with respect to $B & C$. Let the coordinates of $A$ be $(0,sqrt{3}a)$, $B$ be $(-a,0)$ and $C$ be $(a,0)$. Since $M$ has to be symmetrical, let the coordinates for $M$ be $(0,y)$. Now,
          $$|MA|^2=|MB|^2+|MC|^2$$
          $$implies y^2+2sqrt3ay-a^2=0$$
          $$implies y=(2-sqrt3)a$$Note that you'll get two values of y but it can't be negative



          Now since you've $M$, you can get the angle $BMC$ to be equal to $150$ degrees. I hope you can get there :)






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            How do you get, $sqrt3 a$
            $endgroup$
            – Elementary
            Dec 26 '18 at 12:05












          • $begingroup$
            Since it's an equilateral triangle, I've taken the length of each side to be $2a$ (calculations will be a little bit easy with 2a instead of just a). You know how I got $sqrt3a$ now right?
            $endgroup$
            – Ankit Kumar
            Dec 26 '18 at 12:06














          2












          2








          2





          $begingroup$

          Use coordinate geometry. Note that $|MA|^2=|MB|^2+|MC|^2$ is symmetrical in $B & C$. This means $M$ is located symmetrically with respect to $B & C$. Let the coordinates of $A$ be $(0,sqrt{3}a)$, $B$ be $(-a,0)$ and $C$ be $(a,0)$. Since $M$ has to be symmetrical, let the coordinates for $M$ be $(0,y)$. Now,
          $$|MA|^2=|MB|^2+|MC|^2$$
          $$implies y^2+2sqrt3ay-a^2=0$$
          $$implies y=(2-sqrt3)a$$Note that you'll get two values of y but it can't be negative



          Now since you've $M$, you can get the angle $BMC$ to be equal to $150$ degrees. I hope you can get there :)






          share|cite|improve this answer









          $endgroup$



          Use coordinate geometry. Note that $|MA|^2=|MB|^2+|MC|^2$ is symmetrical in $B & C$. This means $M$ is located symmetrically with respect to $B & C$. Let the coordinates of $A$ be $(0,sqrt{3}a)$, $B$ be $(-a,0)$ and $C$ be $(a,0)$. Since $M$ has to be symmetrical, let the coordinates for $M$ be $(0,y)$. Now,
          $$|MA|^2=|MB|^2+|MC|^2$$
          $$implies y^2+2sqrt3ay-a^2=0$$
          $$implies y=(2-sqrt3)a$$Note that you'll get two values of y but it can't be negative



          Now since you've $M$, you can get the angle $BMC$ to be equal to $150$ degrees. I hope you can get there :)







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 26 '18 at 11:57









          Ankit KumarAnkit Kumar

          1,494221




          1,494221












          • $begingroup$
            How do you get, $sqrt3 a$
            $endgroup$
            – Elementary
            Dec 26 '18 at 12:05












          • $begingroup$
            Since it's an equilateral triangle, I've taken the length of each side to be $2a$ (calculations will be a little bit easy with 2a instead of just a). You know how I got $sqrt3a$ now right?
            $endgroup$
            – Ankit Kumar
            Dec 26 '18 at 12:06


















          • $begingroup$
            How do you get, $sqrt3 a$
            $endgroup$
            – Elementary
            Dec 26 '18 at 12:05












          • $begingroup$
            Since it's an equilateral triangle, I've taken the length of each side to be $2a$ (calculations will be a little bit easy with 2a instead of just a). You know how I got $sqrt3a$ now right?
            $endgroup$
            – Ankit Kumar
            Dec 26 '18 at 12:06
















          $begingroup$
          How do you get, $sqrt3 a$
          $endgroup$
          – Elementary
          Dec 26 '18 at 12:05






          $begingroup$
          How do you get, $sqrt3 a$
          $endgroup$
          – Elementary
          Dec 26 '18 at 12:05














          $begingroup$
          Since it's an equilateral triangle, I've taken the length of each side to be $2a$ (calculations will be a little bit easy with 2a instead of just a). You know how I got $sqrt3a$ now right?
          $endgroup$
          – Ankit Kumar
          Dec 26 '18 at 12:06




          $begingroup$
          Since it's an equilateral triangle, I've taken the length of each side to be $2a$ (calculations will be a little bit easy with 2a instead of just a). You know how I got $sqrt3a$ now right?
          $endgroup$
          – Ankit Kumar
          Dec 26 '18 at 12:06











          1












          $begingroup$

          That equation is essentially the Pythagorean theorem applied to lines MA, MB, MC. Even though they do not form a triangle in this scenario, the equation shows that they can. Line MA is the hypotenuse in this case, which makes angle BMC 90 degrees.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            No, this is wrong approach.
            $endgroup$
            – Elementary
            Dec 26 '18 at 11:51






          • 1




            $begingroup$
            The answer is incorrect too
            $endgroup$
            – Ankit Kumar
            Dec 26 '18 at 11:52










          • $begingroup$
            But why doesn't that work?
            $endgroup$
            – Viktor Glombik
            Dec 26 '18 at 12:39
















          1












          $begingroup$

          That equation is essentially the Pythagorean theorem applied to lines MA, MB, MC. Even though they do not form a triangle in this scenario, the equation shows that they can. Line MA is the hypotenuse in this case, which makes angle BMC 90 degrees.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            No, this is wrong approach.
            $endgroup$
            – Elementary
            Dec 26 '18 at 11:51






          • 1




            $begingroup$
            The answer is incorrect too
            $endgroup$
            – Ankit Kumar
            Dec 26 '18 at 11:52










          • $begingroup$
            But why doesn't that work?
            $endgroup$
            – Viktor Glombik
            Dec 26 '18 at 12:39














          1












          1








          1





          $begingroup$

          That equation is essentially the Pythagorean theorem applied to lines MA, MB, MC. Even though they do not form a triangle in this scenario, the equation shows that they can. Line MA is the hypotenuse in this case, which makes angle BMC 90 degrees.






          share|cite|improve this answer









          $endgroup$



          That equation is essentially the Pythagorean theorem applied to lines MA, MB, MC. Even though they do not form a triangle in this scenario, the equation shows that they can. Line MA is the hypotenuse in this case, which makes angle BMC 90 degrees.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 26 '18 at 11:48









          FrostFrost

          1776




          1776








          • 1




            $begingroup$
            No, this is wrong approach.
            $endgroup$
            – Elementary
            Dec 26 '18 at 11:51






          • 1




            $begingroup$
            The answer is incorrect too
            $endgroup$
            – Ankit Kumar
            Dec 26 '18 at 11:52










          • $begingroup$
            But why doesn't that work?
            $endgroup$
            – Viktor Glombik
            Dec 26 '18 at 12:39














          • 1




            $begingroup$
            No, this is wrong approach.
            $endgroup$
            – Elementary
            Dec 26 '18 at 11:51






          • 1




            $begingroup$
            The answer is incorrect too
            $endgroup$
            – Ankit Kumar
            Dec 26 '18 at 11:52










          • $begingroup$
            But why doesn't that work?
            $endgroup$
            – Viktor Glombik
            Dec 26 '18 at 12:39








          1




          1




          $begingroup$
          No, this is wrong approach.
          $endgroup$
          – Elementary
          Dec 26 '18 at 11:51




          $begingroup$
          No, this is wrong approach.
          $endgroup$
          – Elementary
          Dec 26 '18 at 11:51




          1




          1




          $begingroup$
          The answer is incorrect too
          $endgroup$
          – Ankit Kumar
          Dec 26 '18 at 11:52




          $begingroup$
          The answer is incorrect too
          $endgroup$
          – Ankit Kumar
          Dec 26 '18 at 11:52












          $begingroup$
          But why doesn't that work?
          $endgroup$
          – Viktor Glombik
          Dec 26 '18 at 12:39




          $begingroup$
          But why doesn't that work?
          $endgroup$
          – Viktor Glombik
          Dec 26 '18 at 12:39



          -

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          Province de Neuquén

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          Province de Neuquén Provincia del Neuquén .mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Héraldique Drapeau Localisation de la province de Neuquén Administration Pays   Argentine Capitale Neuquén Gouverneur Omar Gutiérrez ISO 3166-2 AR-Q Démographie Gentilé Neuquino/a Population 550 344  hab. (2010) Densité 5,8  hab./km 2 Géographie Superficie 94 078  km 2 Liens Site web http://www.neuquen.gov.ar modifier   La province de Neuquén est une province d'Argentine, qui se trouve à l'extrémité nord-ouest de la Patagonie argentine. Avec une superficie de 94 378  km 2 , elle est la plus petite province des régions australes du pays. Sa géographie s'abîme sur l'imposante Cordillère des Andes. Sommaire 1 Histoire 1.1 Avant la découverte par les Espagnols 1.2 ...
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          Journaliste

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          Pour les articles homonymes, voir Journaliste (homonymie). Journaliste Journalistes Appellation Journaliste Secteur d'activité Média - Communication Professions voisines Journaliste sportif - Reporter-photographe - Éditorialiste Codes CITP 2451, 2642 IDEO (France) 10197 ROME (France) E1106 modifier - modifier le code - modifier Wikidata Un journaliste est une personne qui recueille, écrit ou distribue des informations. L'activité d'un journaliste est le journalisme. Un journaliste peut travailler sur des sujets généraux ou se spécialiser dans certaines questions. La plupart des journalistes ont tendance à se spécialiser et, en collaborant avec d'autres journalistes, rédigent un article ou mettent en forme un reportage qui couvre l'actualité. Par exemple, un journaliste sportif couvre les nouvelles dans le monde du sport, mais ce journaliste peut aussi faire partie d'un journal qui couvre d'autres su...
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