# Find opposite vertices of a rhombus, given the other 2

I am stuck with this problem. I posted an earlier problem with a square, where rotation with i of 90 degrees was possible. This one is a rhombus, how should I proceed?

Given ABCD is a rhombus with AC = 2BD, and b = 3 + i, d = 1 - 3i. Find a and c.

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This is not a question involving complex numbers, the tag is a little misleading –  soandos May 26 '11 at 5:10
A,B,C and D are complex numbers on an Argand diagram. Sorry I should have been clearer. –  mathguy80 May 26 '11 at 5:14
You're both right - there's nothing lost in setting the problem in ${\bf R}^2$, with $B=(3,1)$ and $D=(1,-3)$, getting the answer there (using, say, Isaac's ideas), then interpreting the answer back in $\bf C$. –  Gerry Myerson May 26 '11 at 5:38
There shouldn't be any functional difference to doing this problem with $b=3+i$ and $d=1-3i$ versus $B=(3,1)$ and $D=(1,-3)$—the pictures and relationships are the same, just different notation. –  Isaac May 26 '11 at 5:40
Why do you use capitals for $AC$ and $BD$ and lower case for $b$ and $d$? They are the same points, I think. Doing this just gives cause for worry. –  Ross Millikan May 27 '11 at 13:11

Since you're given opposite vertices and information about the lengths of the diagonals, it is probably useful to know some things about specific quadrilaterals. Kites—quadrilaterals where two consecutive sides are congruent and the other two (consecutive) sides are congruent—have perpendicular diagonals. Parallelograms have diagonals that bisect each other (that intersect at their midpoints). Rhombi (rhombuses) are simultaneously kites and parallelograms, which gives you some information about the diagonals of the rhombus in your particular problem.

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Thanks. I get the properties of Rhombi but I am little to too dense, not able to connect the dots. Can you tell me how to work out one of these points? –  mathguy80 May 26 '11 at 6:14
@mathguy80: knowing those properties and the two points, you can find the location of the midpoint of the diagonals, the slope of the diagonal between the known points and thus the slope of the diagonal between the unknown points, the distance between the known points and thus the distance between the unknown points and the distance from the common midpoint to the unknown points... Does that give you more of a start? –  Isaac May 26 '11 at 6:16
Almost got it. Here's what I have so far. Equation of BD is y = 2x - 5, and then got equation of AC is y = -2x + 3. Also got Midpoint at M(2,-1). The distance BD = $$\sqrt 20$$ units and hence AC = $$2\sqrt 20$$ units. I tried to use distance formula with AM but that gives a quadratic equation with 2 variables. Can you clarify the last step? Thanks. –  mathguy80 May 26 '11 at 7:07
That looks odd. Where did the line breaks go? –  mathguy80 May 26 '11 at 7:10
@mathguy80: The slope for AC should be $-\frac{1}{2}$ (the product of the slopes of perpendicular lines is $-1$), which changes that equation a bit. Using the distance formula with AM should give you an equation of a circle—the set of points a fixed distance ($\sqrt{20}$) from a fixed point (M). If you substitute in your equation for AC, you should find 2 solutions—A and C. –  Isaac May 26 '11 at 7:13
Hint: The points $A,C$ are the points of intersection between the diagonal $AC$ and the circle centered at the mid point of $BD$ with radius $|\frac{AC}{2}|$. How do you express this information in terms of Cartesian equations?