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Is my idea correct? 3 concentric circles of radius 1, 2 and 3 are given. An equilateral triangle is formed having its vertices lie on the side of the three concentric circles. What is the length of tbe equilateral triangle?

My idea is to set a point at the middle of the triangle, then use the distance of it to the vertices given that the three concentric circles are set as $$x^2 + y^2 = 1$$$$x^2 + y^2 = 4$$ and $$x^2 + y^2 = 9$$ i will manipulate the formula afterwards,,,

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While the posted geometric solutions are much simpler, it is possible to do this with algebra and coordinate geometry.

Centering the circles at the origin, we get the equations that you provided: $$x^2+y^2=1$$ $$x^2+y^2=4$$ $$x^2+y^2=9$$ Let's choose an arbitrary point on the smallest circle, say $(0, 1)$ for simplicity. Let $l$ be the length of each side of the equilateral triangle. So the vertices on the other two circles must be a distance of $l$ from our chosen point $(0, 1)$. Equivalently, the two vertices must be on the circle with radius $l$ centered at $(0, 1)$ We can set up an equation to represent this: $$x^2+(y-1)^2=l^2$$

Graph of the 4 circles Red is the circle of radius 1, Blue is the circle of radius 2, Green is the circle of radius 3, Dotted Black is the circle centered at $(0, 1)$ with radius $l$.

Finding the intersection of this circle with the other two circles, we get the following two equations to represent the vertices: $$x^2+y^2-4=x^2+(y-1)^2-l^2$$ $$x^2+y^2-9=x^2+(y-1)^2-l^2$$ Solving the equations for $y$, we get the following. $y_1$ is the y-coordinate of the vertex on the circle of radius 2, and $y_2$ is the y-coordinate of the vertex on the circle of radius 3: $$y_1=\frac{5-l^2}{2}$$ $$y_2=\frac{10-l^2}{2}$$ We can plug this into their respective equations to find the x-coordinates: $$x_1=\sqrt{4-\left(\frac{5-l^2}{2}\right)^2}$$ $$x_2=\sqrt{9-\left(\frac{10-l^2}{2}\right)^2}$$ These coordinates are a distance of $l$ from the point on the smallest circle. It now remains to make these two points a distance of $l$ from each other: $$l=\sqrt{\left(\sqrt{9-\left(\frac{10-l^2}{2}\right)^2}-\sqrt{4-\left(\frac{5-l^2}{2}\right)^2}\right)^2+\left(\frac{10-l^2}{2}-\frac{5-l^2}{2}\right)^2}$$ Solving this equation for $l$ yields the answer of $l=\sqrt{7}$

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Using the construction that @Michael Rozenberg suggested

enter image description here

I will leave the following exercise for you (which isn't that hard)

Prove that the quadrilateral $ABCD$ is cyclic.

Thus $\angle BDC=180°-\angle CAB=120°$. In virtue of the law of Cosines $$\begin{array}a [CB]^2&=[CD]^2+[DB]^2-2·[CD]·[DB]·\cos(\angle BDC)\\ &=1+4-2·1·2·(-0.5)\\ &=5+2=7 \end{array}$$

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  • $\begingroup$ Nice! I thought about algebraic solution only. $\endgroup$ – Michael Rozenberg Mar 15 at 17:51
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The hint.

Take $A$ on the biggest circle and rotate the smallest circle by $60^{\circ}$ around $A$.

Now, take an intersection point $B$ with the middle circle.

Thus, $AB$ is a side of the needed triangle.

I took $A(-3,0)$ and got $AB=\sqrt7.$

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Here's the image

I'll use the process of Dr. Mathva in a different way.

We'll first prove $ABCD$ is cyclic.

Let $AB=AC=BC=s$

$DC=1, DB=2, DA=3$

We see that $AB×DC+BD×AC=1s+2s=3s=AD×BC$

By Converse of Ptolemy's theorem, we conclude that $ABCD$ is cyclic.

After this you can find out a through pure trigonometric means. I got $s = \sqrt{7}$

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