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$l,k,2$ are length of three hights of triangle$ABC$.Where $dfrack+2k-2$ is not integer ,but rational number, Integer $l$'s maximum value is $f(k)$, find $f(k)$




Image discription of problem Image discription of problem(I'm sorry about I can't draw a diagram in MathJax.....



This question is Have posted by me. But I can't understand Xinye Bao's answer.




First, we set the Area of the triangle be “$lkt$”. Then the three sides of the triangle are $2kt, 2lt,$ and $lkt$. So, we have $2k+2l>lk, 2k+lk>2l,$ and$ 2l+lk>2k$. To these, we have $2k>l(k-2)$, $2k>l(2-k)$, and $(2+k)l>2k$. if $k>2$, $l<dfrac2kk-2$. If $k<2$, $l<dfrac2k2-k$. if $k=2$, l just needs to be more than $1$.




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  • I can't understand


"we set the Area of the triangle be “$lkt$"."




Plz help me......




geometry




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  • Possible duplicate of What is minimum value of $l$ where $l$ is one of the prependiculer line of triangle
    – Hans Lundmark
    Aug 25 at 9:05










  • Please don't repeat the same question again. Ask for clarifications in comments instead.
    – Hans Lundmark
    Aug 25 at 9:06










  • @HansLundmark Yes I wrote that
    – user366725
    Aug 25 at 9:06










  • @HansLundmark I'm sorry about that......
    – user366725
    Aug 25 at 9:06














up vote
-2
down vote

favorite
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$l,k,2$ are length of three hights of triangle$ABC$.Where $dfrack+2k-2$ is not integer ,but rational number, Integer $l$'s maximum value is $f(k)$, find $f(k)$




Image discription of problem Image discription of problem(I'm sorry about I can't draw a diagram in MathJax.....



This question is Have posted by me. But I can't understand Xinye Bao's answer.




First, we set the Area of the triangle be “$lkt$”. Then the three sides of the triangle are $2kt, 2lt,$ and $lkt$. So, we have $2k+2l>lk, 2k+lk>2l,$ and$ 2l+lk>2k$. To these, we have $2k>l(k-2)$, $2k>l(2-k)$, and $(2+k)l>2k$. if $k>2$, $l<dfrac2kk-2$. If $k<2$, $l<dfrac2k2-k$. if $k=2$, l just needs to be more than $1$.




This is his answer



  • I can't understand


"we set the Area of the triangle be “$lkt$"."




Plz help me......




geometry




share|cite|improve this question






















  • Possible duplicate of What is minimum value of $l$ where $l$ is one of the prependiculer line of triangle
    – Hans Lundmark
    Aug 25 at 9:05










  • Please don't repeat the same question again. Ask for clarifications in comments instead.
    – Hans Lundmark
    Aug 25 at 9:06










  • @HansLundmark Yes I wrote that
    – user366725
    Aug 25 at 9:06










  • @HansLundmark I'm sorry about that......
    – user366725
    Aug 25 at 9:06












up vote
-2
down vote

favorite
1









up vote
-2
down vote

favorite
1






1






$l,k,2$ are length of three hights of triangle$ABC$.Where $dfrack+2k-2$ is not integer ,but rational number, Integer $l$'s maximum value is $f(k)$, find $f(k)$




Image discription of problem Image discription of problem(I'm sorry about I can't draw a diagram in MathJax.....



This question is Have posted by me. But I can't understand Xinye Bao's answer.




First, we set the Area of the triangle be “$lkt$”. Then the three sides of the triangle are $2kt, 2lt,$ and $lkt$. So, we have $2k+2l>lk, 2k+lk>2l,$ and$ 2l+lk>2k$. To these, we have $2k>l(k-2)$, $2k>l(2-k)$, and $(2+k)l>2k$. if $k>2$, $l<dfrac2kk-2$. If $k<2$, $l<dfrac2k2-k$. if $k=2$, l just needs to be more than $1$.




This is his answer



  • I can't understand


"we set the Area of the triangle be “$lkt$"."




Plz help me......




geometry




share|cite|improve this question















$l,k,2$ are length of three hights of triangle$ABC$.Where $dfrack+2k-2$ is not integer ,but rational number, Integer $l$'s maximum value is $f(k)$, find $f(k)$




Image discription of problem Image discription of problem(I'm sorry about I can't draw a diagram in MathJax.....



This question is Have posted by me. But I can't understand Xinye Bao's answer.




First, we set the Area of the triangle be “$lkt$”. Then the three sides of the triangle are $2kt, 2lt,$ and $lkt$. So, we have $2k+2l>lk, 2k+lk>2l,$ and$ 2l+lk>2k$. To these, we have $2k>l(k-2)$, $2k>l(2-k)$, and $(2+k)l>2k$. if $k>2$, $l<dfrac2kk-2$. If $k<2$, $l<dfrac2k2-k$. if $k=2$, l just needs to be more than $1$.




This is his answer



  • I can't understand


"we set the Area of the triangle be “$lkt$"."




Plz help me......





geometry





share|cite|improve this question













share|cite|improve this question




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edited Aug 25 at 6:20

























asked Aug 25 at 5:51









user366725

637




637











  • Possible duplicate of What is minimum value of $l$ where $l$ is one of the prependiculer line of triangle
    – Hans Lundmark
    Aug 25 at 9:05










  • Please don't repeat the same question again. Ask for clarifications in comments instead.
    – Hans Lundmark
    Aug 25 at 9:06










  • @HansLundmark Yes I wrote that
    – user366725
    Aug 25 at 9:06










  • @HansLundmark I'm sorry about that......
    – user366725
    Aug 25 at 9:06
















  • Possible duplicate of What is minimum value of $l$ where $l$ is one of the prependiculer line of triangle
    – Hans Lundmark
    Aug 25 at 9:05










  • Please don't repeat the same question again. Ask for clarifications in comments instead.
    – Hans Lundmark
    Aug 25 at 9:06










  • @HansLundmark Yes I wrote that
    – user366725
    Aug 25 at 9:06










  • @HansLundmark I'm sorry about that......
    – user366725
    Aug 25 at 9:06















Possible duplicate of What is minimum value of $l$ where $l$ is one of the prependiculer line of triangle
– Hans Lundmark
Aug 25 at 9:05




Possible duplicate of What is minimum value of $l$ where $l$ is one of the prependiculer line of triangle
– Hans Lundmark
Aug 25 at 9:05












Please don't repeat the same question again. Ask for clarifications in comments instead.
– Hans Lundmark
Aug 25 at 9:06




Please don't repeat the same question again. Ask for clarifications in comments instead.
– Hans Lundmark
Aug 25 at 9:06












@HansLundmark Yes I wrote that
– user366725
Aug 25 at 9:06




@HansLundmark Yes I wrote that
– user366725
Aug 25 at 9:06












@HansLundmark I'm sorry about that......
– user366725
Aug 25 at 9:06




@HansLundmark I'm sorry about that......
– user366725
Aug 25 at 9:06















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