A heart shape is constructed using two identical circles with radius $r$. A line is drawn from point $T$ to $C$. There, two tangents are constructed, one to each circle, through $AC$ and $CB$. (The heart is supposed to be symmetrical, I apologize for the rough sketch)

enter image description here

Presuming that the coordinates of points $A,B,C,T$ and length of $r$ as well as the area $A_\text{Heart}$ are known, I would like to calculate the point $S$, the centroid of the heart.

Now, if the heart is "tall" enough, calculating the centroid would be easy. As the heart is perfectly symmetrical, $S_x = C_x = T_x$. The centroid lies in the center of the line that seperates the heart into two areas of equal size ($\frac{1}{2} A_\text{Heart}$).

$$\frac{1}{2} \overline{SR} \overline{SC} = \frac{1}{2} A_\text{Heart} $$

Through basic trigonometry, the knowledge of the angle $\alpha$ gives us

$$ \overline{SR} = 2 \tan(\alpha) \overline{SC} $$

$$ \overline{SC} = \sqrt{ \frac{A_\text{Heart}}{2 \tan(\alpha)}}$$

This leads to the final formula:

$$S(S_x|S_y) = \left(C_x \mid C_y - \sqrt{ \frac{A_\text{Heart}}{2 \tan(\alpha)}}\right)$$

This approach assumes that the area of the triangle $ABC$ is greater than the rest of the heart, meaning that the centroid must lie within the triangle. If the heart were of shorter height, the centroid will inevitably lie outside of the triangle.

I'm open to any ideas.

  • 1
    $\begingroup$ See my answer to this question for remarks on why the centroid / centre of area is not the always the point that divides the figure into two equal areas. $\endgroup$
    – Frentos
    Commented Feb 12, 2016 at 21:38
  • $\begingroup$ Just to clarify, is this a uniform lamina and are you looking for an expression for the distance of the centroid along the line of symmetry from C in terms of $r$ and $\alpha$? $\endgroup$ Commented Feb 12, 2016 at 21:56

1 Answer 1


If the centroid is above the triangle, you want to compute the height that has half the area above it. The centroid is clearly below point $T$ as $T$ splits the area of the circles in half and the triangle will lower the centroid. You want the area of the strip from $T$ down to $S$ to be $\frac 12A_{heart}-\pi r^2$ where the outer edge of the strip follows the circle. Put the origin at $T$ with the $x$ axis directed toward $S$. The area of the strip is $2\int_0^{|TS|} r+\sqrt{r^2-x^2}dr=\frac 12A_{heart}-\pi r^2$ Solve for $x$


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