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I would like to know how I shrink a triangle by $30\%$ using dilation not changing its center?

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    Shrinking in... total area or perimeter? – 2017-02-05
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    scale it by 0.7? – 2017-02-05
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    If it helps, see here: http://math.stackexchange.com/questions/17561/how-to-shrink-a-triangle – 2017-02-05

4 Answers 4

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Affine solution:

enter image description here

The task is to shrink the triangle $ABC$ "around the center" $O$.

To create a dilation trisect $BC$. Let the length of the red segment be $\frac23$ $BC$.

Then from $O$ draw lines through the end points of the red segment. Draw a line (white) through $C$ so that it is parallel to $OB$...

So, you can construct the thick black segment which is parallel to the red segment and its length is the same as that of $BC$.

The red segment and the thick black segment determine a dilation with center $O$.

The yellow lines show how to construct the image of the thick black point on $BC.$ Moving this black point, the resulting images will trace the shrunk image of $BC$. Actually, you don't have to trace. It is enough to construct one point and then to construct a parallel with $BC$.

Do the same with the other sides of the triangle. Then the vertices will shop up as well.

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Multiply every coordinate of the original triangle by the scale factor $\left(\frac 7{10}\right)$

Also see this for reference.

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Analytically: translate the center to the origin, shrink everything, translate back.

Geometrically: move the points on each line through the center toward the center by decreasing length by a factor $0.7$.

This changes all lengths to 70% of what they were. It will fix whatever point you have decided is the "center" - probably the intersection of the medians. For areas, use factor $\sqrt{0.7}$.

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I presume you want to shrink the Area by $30\%$. Assume the center is at $p$. Then you start by translating the triangle so that its center is in $0$, and then note that the area scales like the square of the scaling factor, so you have to scale the coordinates by $\sqrt{0.7}$, and then translate it back to its original position.

Composing these maps gives: $$f(\vec{x})=\sqrt{0.7}\ (\vec{x}-\vec{p})+\vec{p}$$