Calculate the ratio of the sides of a given triangle given the ratio of areas.












3












$begingroup$


Given a triangle $triangle ABC$, points $M$, $N$, $P$ are drawn on the sides of the triangle in a way that $frac{|AM|}{|MB|} = frac{|BN|}{|NC|}= frac{|PC|}{|PA|}=k$, where $k>0$.




Calculate $k$, given that the area of the triangle $triangle MNP$ and the area of the triangle $ triangle ABC$ are in the following ratio: $Area_{triangle MNP} = frac{7}{25} times Area_{triangle ABC}$.




I have tried Heron formula, bot the calculations seem to be too complicated, I have also tried to simplify the problem and assume that k is equal to 1 and then calculate the ratio of the triangles area but it also doesn't help.
I was also looking for similar triangles.



I would appreciate some hint.






algebra-precalculus geometry euclidean-geometry triangle area







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$endgroup$

















    3












    $begingroup$


    Given a triangle $triangle ABC$, points $M$, $N$, $P$ are drawn on the sides of the triangle in a way that $frac{|AM|}{|MB|} = frac{|BN|}{|NC|}= frac{|PC|}{|PA|}=k$, where $k>0$.




    Calculate $k$, given that the area of the triangle $triangle MNP$ and the area of the triangle $ triangle ABC$ are in the following ratio: $Area_{triangle MNP} = frac{7}{25} times Area_{triangle ABC}$.




    I have tried Heron formula, bot the calculations seem to be too complicated, I have also tried to simplify the problem and assume that k is equal to 1 and then calculate the ratio of the triangles area but it also doesn't help.
    I was also looking for similar triangles.



    I would appreciate some hint.






    algebra-precalculus geometry euclidean-geometry triangle area







    share|cite|improve this question











    $endgroup$















      3












      3








      3


      1



      $begingroup$


      Given a triangle $triangle ABC$, points $M$, $N$, $P$ are drawn on the sides of the triangle in a way that $frac{|AM|}{|MB|} = frac{|BN|}{|NC|}= frac{|PC|}{|PA|}=k$, where $k>0$.




      Calculate $k$, given that the area of the triangle $triangle MNP$ and the area of the triangle $ triangle ABC$ are in the following ratio: $Area_{triangle MNP} = frac{7}{25} times Area_{triangle ABC}$.




      I have tried Heron formula, bot the calculations seem to be too complicated, I have also tried to simplify the problem and assume that k is equal to 1 and then calculate the ratio of the triangles area but it also doesn't help.
      I was also looking for similar triangles.



      I would appreciate some hint.






      algebra-precalculus geometry euclidean-geometry triangle area







      share|cite|improve this question











      $endgroup$




      Given a triangle $triangle ABC$, points $M$, $N$, $P$ are drawn on the sides of the triangle in a way that $frac{|AM|}{|MB|} = frac{|BN|}{|NC|}= frac{|PC|}{|PA|}=k$, where $k>0$.




      Calculate $k$, given that the area of the triangle $triangle MNP$ and the area of the triangle $ triangle ABC$ are in the following ratio: $Area_{triangle MNP} = frac{7}{25} times Area_{triangle ABC}$.




      I have tried Heron formula, bot the calculations seem to be too complicated, I have also tried to simplify the problem and assume that k is equal to 1 and then calculate the ratio of the triangles area but it also doesn't help.
      I was also looking for similar triangles.



      I would appreciate some hint.






      algebra-precalculus geometry euclidean-geometry triangle area




      algebra-precalculus geometry euclidean-geometry triangle area






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 17 '18 at 14:49







      user593746

















      asked Dec 17 '18 at 12:44









      Jack BlackwellJack Blackwell

      1249




      1249






















          1 Answer
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          3












          $begingroup$

          Draw $BB1$ is perpendicular to $AC$; $MM1//AC$ and perpendicular to $AC$, then we have $frac{AM}{AB}=frac{MM1}{BB1}$



          So $$frac{AMP}{ABC}=frac{1/2 cdot APcdot MM1}{1/2cdot AC cdot BB1} =frac{APcdot AM}{ACcdot AB}$$



          $$Rightarrow frac{AMP}{ABC} = frac{CPN}{ABC} = frac{BMN}{ABC} = frac{AM}{AB}. frac{AP}{AC} = frac{k}{(k+1)^2}$$



          Or $$frac{MNP}{ABC} = 1 - frac{AMP}{ABC} - frac{CPN}{ABC} - frac{BMN}{ABC} = 1 - frac{3k}{(k+1)^2}=frac{7}{25}$$



          It is not difficult to find $k=frac{2}{3}$ or $k=frac{3}{2}$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            $frac{MNP}{ABC}=frac{Area_{triangle MNP}}{ Area_{triangle ABC}}$
            $endgroup$
            – Word Shallow
            Dec 17 '18 at 13:59












          • $begingroup$
            $triangle AMP$ and $triangle ABC$ are not, in general, similar, because as you pointed out $AM:AB = k:(k+1)$ while $AP:AC = 1:(k+1)$. However, your others arguments still apply.
            $endgroup$
            – Quang Hoang
            Dec 17 '18 at 15:42










          • $begingroup$
            Got it. I edited something good in my solution, see now.
            $endgroup$
            – Word Shallow
            Dec 17 '18 at 15:55











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          1 Answer
          1






          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          3












          $begingroup$

          Draw $BB1$ is perpendicular to $AC$; $MM1//AC$ and perpendicular to $AC$, then we have $frac{AM}{AB}=frac{MM1}{BB1}$



          So $$frac{AMP}{ABC}=frac{1/2 cdot APcdot MM1}{1/2cdot AC cdot BB1} =frac{APcdot AM}{ACcdot AB}$$



          $$Rightarrow frac{AMP}{ABC} = frac{CPN}{ABC} = frac{BMN}{ABC} = frac{AM}{AB}. frac{AP}{AC} = frac{k}{(k+1)^2}$$



          Or $$frac{MNP}{ABC} = 1 - frac{AMP}{ABC} - frac{CPN}{ABC} - frac{BMN}{ABC} = 1 - frac{3k}{(k+1)^2}=frac{7}{25}$$



          It is not difficult to find $k=frac{2}{3}$ or $k=frac{3}{2}$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            $frac{MNP}{ABC}=frac{Area_{triangle MNP}}{ Area_{triangle ABC}}$
            $endgroup$
            – Word Shallow
            Dec 17 '18 at 13:59












          • $begingroup$
            $triangle AMP$ and $triangle ABC$ are not, in general, similar, because as you pointed out $AM:AB = k:(k+1)$ while $AP:AC = 1:(k+1)$. However, your others arguments still apply.
            $endgroup$
            – Quang Hoang
            Dec 17 '18 at 15:42










          • $begingroup$
            Got it. I edited something good in my solution, see now.
            $endgroup$
            – Word Shallow
            Dec 17 '18 at 15:55
















          3












          $begingroup$

          Draw $BB1$ is perpendicular to $AC$; $MM1//AC$ and perpendicular to $AC$, then we have $frac{AM}{AB}=frac{MM1}{BB1}$



          So $$frac{AMP}{ABC}=frac{1/2 cdot APcdot MM1}{1/2cdot AC cdot BB1} =frac{APcdot AM}{ACcdot AB}$$



          $$Rightarrow frac{AMP}{ABC} = frac{CPN}{ABC} = frac{BMN}{ABC} = frac{AM}{AB}. frac{AP}{AC} = frac{k}{(k+1)^2}$$



          Or $$frac{MNP}{ABC} = 1 - frac{AMP}{ABC} - frac{CPN}{ABC} - frac{BMN}{ABC} = 1 - frac{3k}{(k+1)^2}=frac{7}{25}$$



          It is not difficult to find $k=frac{2}{3}$ or $k=frac{3}{2}$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            $frac{MNP}{ABC}=frac{Area_{triangle MNP}}{ Area_{triangle ABC}}$
            $endgroup$
            – Word Shallow
            Dec 17 '18 at 13:59












          • $begingroup$
            $triangle AMP$ and $triangle ABC$ are not, in general, similar, because as you pointed out $AM:AB = k:(k+1)$ while $AP:AC = 1:(k+1)$. However, your others arguments still apply.
            $endgroup$
            – Quang Hoang
            Dec 17 '18 at 15:42










          • $begingroup$
            Got it. I edited something good in my solution, see now.
            $endgroup$
            – Word Shallow
            Dec 17 '18 at 15:55














          3












          3








          3





          $begingroup$

          Draw $BB1$ is perpendicular to $AC$; $MM1//AC$ and perpendicular to $AC$, then we have $frac{AM}{AB}=frac{MM1}{BB1}$



          So $$frac{AMP}{ABC}=frac{1/2 cdot APcdot MM1}{1/2cdot AC cdot BB1} =frac{APcdot AM}{ACcdot AB}$$



          $$Rightarrow frac{AMP}{ABC} = frac{CPN}{ABC} = frac{BMN}{ABC} = frac{AM}{AB}. frac{AP}{AC} = frac{k}{(k+1)^2}$$



          Or $$frac{MNP}{ABC} = 1 - frac{AMP}{ABC} - frac{CPN}{ABC} - frac{BMN}{ABC} = 1 - frac{3k}{(k+1)^2}=frac{7}{25}$$



          It is not difficult to find $k=frac{2}{3}$ or $k=frac{3}{2}$






          share|cite|improve this answer











          $endgroup$



          Draw $BB1$ is perpendicular to $AC$; $MM1//AC$ and perpendicular to $AC$, then we have $frac{AM}{AB}=frac{MM1}{BB1}$



          So $$frac{AMP}{ABC}=frac{1/2 cdot APcdot MM1}{1/2cdot AC cdot BB1} =frac{APcdot AM}{ACcdot AB}$$



          $$Rightarrow frac{AMP}{ABC} = frac{CPN}{ABC} = frac{BMN}{ABC} = frac{AM}{AB}. frac{AP}{AC} = frac{k}{(k+1)^2}$$



          Or $$frac{MNP}{ABC} = 1 - frac{AMP}{ABC} - frac{CPN}{ABC} - frac{BMN}{ABC} = 1 - frac{3k}{(k+1)^2}=frac{7}{25}$$



          It is not difficult to find $k=frac{2}{3}$ or $k=frac{3}{2}$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 25 at 23:08

























          answered Dec 17 '18 at 13:58









          Word ShallowWord Shallow

          8911620




          8911620












          • $begingroup$
            $frac{MNP}{ABC}=frac{Area_{triangle MNP}}{ Area_{triangle ABC}}$
            $endgroup$
            – Word Shallow
            Dec 17 '18 at 13:59












          • $begingroup$
            $triangle AMP$ and $triangle ABC$ are not, in general, similar, because as you pointed out $AM:AB = k:(k+1)$ while $AP:AC = 1:(k+1)$. However, your others arguments still apply.
            $endgroup$
            – Quang Hoang
            Dec 17 '18 at 15:42










          • $begingroup$
            Got it. I edited something good in my solution, see now.
            $endgroup$
            – Word Shallow
            Dec 17 '18 at 15:55


















          • $begingroup$
            $frac{MNP}{ABC}=frac{Area_{triangle MNP}}{ Area_{triangle ABC}}$
            $endgroup$
            – Word Shallow
            Dec 17 '18 at 13:59












          • $begingroup$
            $triangle AMP$ and $triangle ABC$ are not, in general, similar, because as you pointed out $AM:AB = k:(k+1)$ while $AP:AC = 1:(k+1)$. However, your others arguments still apply.
            $endgroup$
            – Quang Hoang
            Dec 17 '18 at 15:42










          • $begingroup$
            Got it. I edited something good in my solution, see now.
            $endgroup$
            – Word Shallow
            Dec 17 '18 at 15:55
















          $begingroup$
          $frac{MNP}{ABC}=frac{Area_{triangle MNP}}{ Area_{triangle ABC}}$
          $endgroup$
          – Word Shallow
          Dec 17 '18 at 13:59






          $begingroup$
          $frac{MNP}{ABC}=frac{Area_{triangle MNP}}{ Area_{triangle ABC}}$
          $endgroup$
          – Word Shallow
          Dec 17 '18 at 13:59














          $begingroup$
          $triangle AMP$ and $triangle ABC$ are not, in general, similar, because as you pointed out $AM:AB = k:(k+1)$ while $AP:AC = 1:(k+1)$. However, your others arguments still apply.
          $endgroup$
          – Quang Hoang
          Dec 17 '18 at 15:42




          $begingroup$
          $triangle AMP$ and $triangle ABC$ are not, in general, similar, because as you pointed out $AM:AB = k:(k+1)$ while $AP:AC = 1:(k+1)$. However, your others arguments still apply.
          $endgroup$
          – Quang Hoang
          Dec 17 '18 at 15:42












          $begingroup$
          Got it. I edited something good in my solution, see now.
          $endgroup$
          – Word Shallow
          Dec 17 '18 at 15:55




          $begingroup$
          Got it. I edited something good in my solution, see now.
          $endgroup$
          – Word Shallow
          Dec 17 '18 at 15:55


















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