Determine the value of $x$ such that the triangle is a right triangle and so on

Question

Given an equilateral triangle $ABC$ with the length of each sides are $a$. On the side of $AB, BC, CA$, each chosen a point $P, Q, R$ such that $AP = x, BQ = 2x, CR = 3x$. Determine the value of $x$ so that: i) the triangle $PQR$ is an isosceles triangle, ii) the triangle $PQR$ is a right triangle, iii) and the area of triangle $PQR$ is minimum.

Really and be honest, I can't solve any numbers of that problem. I have tried to draw and use some approachment, geometrical senses. But, it looks like we need some theorems of a triangle to solve this. And unfortunately, I don't have much clue for this. Please help me. Regards and thank you.

Answer 1

Notice, that $x\in [0,\frac{a}{3}]$.

Denote the lengths: $|PQ|=p, |QR|=q, |RP|=r$

Using the law of cosines you can write: $$p^2=(2x)^2+(a-x)^2-2(2x)(a-x)\cos \frac{\pi}{3}=7x^2-4ax+a^2$$ $$q^2=19x^2-7ax+a^2$$ $$r^2=13x^2-7ax+a^2$$

Ad 1

If we want triangle $PQR$ to be isosceles, we have 3 options:

1. $p=r$ $$6x^2-3ax=0$$ $$x\in \{0, \frac{a}{2}\}$$ Notice, that $\frac{a}{2}>\frac{a}{3}$, thus it is not correct.
2. $p=q$ $$12x^2-3ax=0$$ $$x\in \{0, \frac{a}{4}\}$$
3. $r=q$ $$9x^2=0$$ $$x=0$$

Thus the triangle $PQR$ is isosceles, if $x\in \{0, \frac{a}{4}\}$

Ad 2:

To solve this, we have to use the Pythagoras Theorem. There are three cases, depends on which side would be a hypotenuse:

1. $p$: $$7x^2-4ax+a^2=32x^2-14ax+2a^2$$ $$25x^2-10ax+a^2=0$$ $$a$$ $$x=\frac{a}{5}$$
2. $q$: $$x^2-4ax+a^2=0$$ $$x=a(2\pm\sqrt{3})$$ Notice, that $a(2+\sqrt{3})>\frac{a}{3}$, thus ot is not correct.
3. $r$: $$13x^2-4ax+a^2=0$$ In this case we have $\Delta<0$, thus there are no real solutions.

Thus the triangle $PQR$ is right, if $x\in\{\frac{a}{5},a(2-\sqrt{3})\}$

Ad 3

Minimizing the area of the triangle $PQR$ is also maximizing the area of triangles $APR$, $BQP$ and $CRQ$. Using the formula for the area of the triangle: $$P=\frac{1}{2}st\sin\alpha$$ where $s$ and $t$ are the lengths of sides adjacent to the angle $\alpha$, this area (which is the sum of three triangles) could be described as:

$$P(x)=\frac{1}{2}\sin\frac{\pi}{3}(x(a-3x)+2x(a-x)+3x(a-2x))$$ $$\frac{\sqrt{3}}{4}(6ax-11x^2)$$ The graph of this function is the parabola with the arms pointing downwards, thus there is only one local extremum and it is maximum.

We could compute the derivative: $$P'(x)=\frac{\sqrt{3}}{4}(6a-22x)$$ Of course $P'(x)=0$ for $(6a-22x)=0$, thus the maximum is obtained for $x=\frac{3a}{11}$

Answer 2

I always start by drawing a figure. In the one below, my $D,E,F$ are your $P,Q,R$ The pedestrian approach is to move the figure to a standard position, so let $A=(-1,0),B=(1,0),C=(0,\sqrt 3)$. Then $D=(-1+x,0)$ and you can find the coordinates of $E$ and $F$, the side lengths, and so on. I didn't find an easier way.

Answer 3

Hint:

you can start applying the law of cosines to the triangles $APR$, $BQP$, and $CRQ$:

As an example, find $RP$ as: $$ \overline{RP}^2=x^2+(a-3x)^2-2x(a-3x)\cos (60°)=x^2+(a-3)^2-x(a-3x)=1ex^2-7ax+a^2 $$

In a similar way you can find the other sides of the internal triangle:

$$ \overline{QP}^2=(a-x)^2+4x^2-2x(a-x)=7x^2-4ax+a^2 $$

$$ \overline{RQ}^2=(a-2x)^2+9x^2-3x(a-2x)=19x^2-7ax+a^2 $$

So the internal triangle is isosceles if:

$$ 7x^2-4ax+a^2=19x^2-7ax+a^2 $$

that, for $x\ne 0$ gives $x=\frac{a}{4}$

For the problem $iii)$ note that the area of the internal triangle is minimized when $$ x(a-3x)\sin(60°)+2x(a-x\sin (60°))+3x(a-2x)\sin(60°)=y $$ is maximized.