If the lengths of two altitudes drawn from two vertices of a triangle on their opposite sides are $2014$ and $1$ unit, then what will be the length of the altitude drawn from the third vertex of the triangle on its opposite side?

Source: BdMO National 2014

Re: How two altitudes determine the third

Posted: Wed Feb 22, 2017 2:21 pm

by aritra barua

Let,ABC denote a right angled triangle.Let,the sides be 2014,1 and y units.Then the area of the triangle will be 1007 sq units.If we construct another altitude x in length,then the area will be xy/2 units.Then,we apply an equation,xy/2=1007;then x results in 2014/√(2014^2+1) units which denotes the length of the 3rd altitude.

Re: How two altitudes determine the third

Posted: Fri Feb 02, 2018 2:17 am

by samiul_samin

You can use the triangle inequality to get the answer.

Re: How two altitudes determine the third

Posted: Mon Feb 19, 2018 7:26 pm

by samiul_samin

This is the problem of BdMO National Secondary 2014/8 Answer

If the third altitudes is $r$ then $\dfrac {2014}{2013}>r>\dfrac{2014}{2015}$

Solution source

বিজ্ঞানচিন্তা January 2017.Solution is written by Tushar Chakraborty

Solution

Suppose,three sides of the triangle is $a,b,c$ and altitudes are $p,q,r$ in same order.
So,the area of the triangle $=ap=bq=cr$
So, $ap=bq\Rightarrow {\dfrac ab=\dfrac pq}$
And $bq=cr\Rightarrow {\dfrac bc=\dfrac rq}=\dfrac {p}{\dfrac{pq}{r}}$
So, $ a/b/c=p/q/ {pq/r}$
Using Triangular inequality ,we get
$a<b+c$
$p>q+pq/r$
$1/r>(p-q)/pq$
And
$b<a+c$
$q<p+ (pq/r)$
$1/r<(p+q/pq)$
And
$c<a+b$
$pq/r <p+q$
$1/r< (p+q)/pq$
We can write these $3$ inequalities as,
$(p-q)/pq <1/r<(p+q)/pq$
$pq/ (p-q)>r>pq/(p+q)$
If $p-q<0$ we will take only the differenece of $p-q$
According to the question ,
$p=2014$ & $q=1$
So, the answer is $\dfrac {2014}{2013}>r>\dfrac {2014}{2015}$