## BdMO National 2013: Junior 2

BdMO
Posts: 134
Joined: Tue Jan 18, 2011 1:31 pm

### BdMO National 2013: Junior 2

Two isosceles triangles are possible with area of 120 square unit and length of edges integers. One of these two triangles has sides of lengths $17,17$ and $16$. Determine the length of edges of second one.

[Hint: In $\Delta ABC$ if $AB=AC$ and $AD$ is perpendicular to $BC$ then $BD=CD$.]

rubab
Posts: 14
Joined: Wed Feb 19, 2014 3:20 pm

### Re: BdMO National 2013: Junior 2

The height of the isosceles triangle which has three sides of length 17,17,16, is $=\sqrt{17^2-8^2} =15.$ The area of this triangle is 120. Let x,y be the height and base of the new triangle respectively. if $x =\frac{16}{2} =8$ [Half of the base the first triangle] and $y = 15 \times 2 =30$ [double of the height of the first triangle] then the area of the new triangle is 120 and the length of the two equal sides will be $\sqrt {15^2 +8^2} =17$ so the length of the sides of the new triangle are 17,17,30.

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Location: Japan Garden City, Mohammadpur, Dhaka

### Re: BdMO National 2013: Junior 2

Here's one possible approach for this one -

Let $a$ and $b$ denote the equal sides and base of a isosceles triangle of area $120$ square unit .

Then, $\frac b4 \sqrt{4a^2 - b^2} = 120$

$\sqrt{4a^2 - b^2} = \frac{480}{b}$

$4a^2 - b^2 = \frac{480^2}{b^2}$

$4a^2 = \frac{480^2}{b^2} + b^2$

From this we can see that if we replace $b$ with $\frac{480}{b}$ it remains the same . Taking $b = 16$ and $a = 17$ we get our desired triangle having sides $17 , 17$ and $\frac{480}{16} = 30$ .