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Two non-congruent integer sided isosceles triangles have the same area and perimeter. The ratio of the lengths of the bases of the two triangles is 8:7. Find the minimum possible value of their common perimeter.

I have tried substituting the area for $a$ and the bases as $8x$ and $7x$. I then did Pythagoras and got the equation $15a=169x$. However, I still don't know how to find the minimum value. Does anyone know how?

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Let's call the base of the first triangle $x$, the base of the second $\frac{8}{7}x$. The other side of the first and of the second will be $y$ and $z$. We'll have:

$$ \left\{\begin{matrix} x+2y=\frac{8}{7}x+2z\\ x\sqrt{y^2-\frac{x^2}{4}}=\frac{8}{7}x\sqrt{z^2-\frac{16x^2}{49}} \end{matrix}\right.\Rightarrow \left\{\begin{matrix} x+2y=\frac{8}{7}x+2z\\ \left(y-\frac{x}{2}\right)\left(y+\frac{x}{2}\right)=\frac{64}{49}\left(z-\frac{4x}{7}\right)\left(z+\frac{4x}{7}\right) \end{matrix}\right.$$

[The first equation is the equality of the two perimeters, the second the one of areas(in which I calculated the height with the Pythagora Theorem). Then we squared the second equation and scomposed the two differences of squares]

Now there is a little trick! Multiply both sides of the second equation by $2$: $$ \left\{\begin{matrix} x+2y=\frac{8}{7}x+2z\\ \left(y-\frac{x}{2}\right)\left(x+2y\right)=\frac{64}{49}\left(z-\frac{4x}{7}\right)\left(\frac{8x}{7}+2z\right) \end{matrix}\right.$$

Since $\left(x+2y\right)$ and $\left(\frac{8x}{7}+2z\right)$ are equal for the first equation we can simplify them and obtain a linear system: $$ \left\{\begin{matrix} x+2y=\frac{8}{7}x+2z\\ \left(y-\frac{x}{2}\right)=\frac{64}{49}\left(z-\frac{4x}{7}\right) \end{matrix}\right.$$

Now we can easily solve this for $(y,z)$ and obtain:

$$ \left\{\begin{matrix} x=x\\ y=\frac{233}{210}x \\ z=\frac{109}{105}x \end{matrix}\right.$$ Notice that since the sides are integer, at least $x=210$ Since the perimeter is:

$$P(x)=x+2y=x+2\frac{233}{210}x=\frac{338}{105}x$$

And the minimum is:

$$P(210)=676$$

:)

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  • $\begingroup$ This really helped! Thanks! $\endgroup$ – Matthew Tan Mar 31 at 13:57
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In isosceles $\triangle ABC$ let the base $|AB|=c$, $|AC|=|BC|=a$, $S$ is the area and semiperimeter $\rho=\tfrac12(a+a+c)=a+\tfrac c2$.

We have \begin{align} S^2&=\rho(\rho-a)^2(\rho-c) \\ S^2&=\tfrac14 \rho c^2(\rho-c) . \end{align}

Given two bases $7x$ and $8x$ and common $S$ and $\rho$, we have \begin{align} \left(\frac{7x}2\right)^2(\rho-7x) &=\left(\frac{8x}2\right)^2(\rho-8x) ,\\ \rho&=\frac{169}{15}\,x , \end{align}

corresponding perimeter is \begin{align} p&=2\rho=\frac{338}{15}\,x . \end{align}

Thus $x=15$ provides the minimal integer perimeter $p=338$, but for the triangle with the base $c=7x=105$ the other side $a=(338-105)/2=\frac{233}2$ is not integer, so we have to double it and get $x=30$ with $p=676$ as the answer.

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