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I'm looking for commonly used methods in contest geometry to show that 4 point lie on the same circle. Are there any tricks besides using the fact that the angles across add up to 180°?

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    $\begingroup$ Ptolemy’s inequality, Power of a point, inversión and some formula for complex numbers I don’t remember right now are often used too. $\endgroup$ – Dr. Mathva Sep 30 at 11:49
  • $\begingroup$ No three points should be collinear. If you take any two points and draw a line segment, the segment should subtend the same angle at other two points (if points are on the same side of the segment) and should add to 180 degree if they are on opposite sides. $\endgroup$ – Math Lover Sep 30 at 12:14
  • $\begingroup$ It suffices to show that $\angle ABD = \angle ACD$ (when $A,B,C,D$ should be on the circle in this order). $\endgroup$ – Michal Adamaszek Sep 30 at 12:29
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We can prove that quadrilateral $ABCD$ is cyclic by infinitely many methods:

  1. If there is a point $O$ in the plane of the quadrilateral $ABCD$ such that $OA=OB=OC=OD$, so $ABCD$ is cyclic;
  2. If midperpendiculars to sides of quadrilateral $ABCD$ are intersected in the same point, so $ABCD$ is cyclic;
  3. If for quadrilateral $ABCD$ we have $\measuredangle ABD=\measuredangle ACD,$ so $ABCD$ is cyclic;
  4. Let for quadrilateral $ABCD$ we have $AB\cap CD=\{P\},$ $B$ and $C$ be midpoints of $PQ$ and $PR$ respectively, $AM$ and $DM$ be perpendiculars to $AB$ and $CD$ respectively. Now, if $M\in QR$, so $ABCD$ is cyclic; $$\bullet$$ $$\bullet$$ $$\bullet$$
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If we know the coordinates of the four points then they are concyclic if and only if the following determinant is zero. $$0 = \begin{vmatrix} x_1^2+y_1^2&x_1&y_1&1\\ x_2^2+y_2^2&x_2&y_2&1\\ x_3^2+y_3^2&x_3&y_3&1\\ x_4^2+y_4^2&x_4&y_4&1 \end{vmatrix}$$

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Here is a way to tell from the distances between the points without knowing their rotational order.

Leg $d_{ij}$ be the distance between two points $i$ and $j$. By Ptolemy's Theore, if the rotational order of the points is 1-2-3-4 then

$d_{13}d_{24}=d_{12}d_{34}+d_{14}d_{23}; d_{13}d_{24}-d_{12}d_{34}-d_{14}d_{23}=0$

For the other two possible rotational orders we get similar results:

$d_{12}d_{34}-d_{14}d_{23}-d_{13}d_{24}=0$

$d_{14}d_{23}-d_{13}d_{24}-d_{12}d_{44}=0$

Now take all three cases into account with one blow by multiplying the three zero factors together. It looks inelegant, true --- unless we slip in an extra factor:

$d_{13}d_{24}+d_{12}d_{34}+d_{14}d_{23}=0$

The product of a four factor equations is then a simple form that involves only even powers:

$(d_{12}d_{34})^4+(d_{13}d_{24})^4+(d_{14}d_{23})^4-2(d_{12}d_{34})^2(d_{13}d_{24})^2-2(d_{13}d_{24})^2(d_{14}d_{23})^2-2(d_{14}d_{23})^2(d_{12}d_{34})^2=0$

$\color{blue}{(d_{12}d_{34})^4+(d_{13}d_{24})^4+(d_{14}d_{23})^4=2[(d_{12}d_{34})^2(d_{13}d_{24})^2+(d_{13}d_{24})^2(d_{14}d_{23})^2+(d_{14}d_{23})^2(d_{12}d_{34})^2]}$

With this formulation, you do not need square root extractions to get the even powers of the distances in the above expression. You will, however, run into large numbers with seemingly small coordinates. For instance, consider the four points

$1\to(0,5);2\to(5,0);3\to(3,4);4\to(3,-4).$ (They're "rigged" to lie on $x^2+y^2=5^2$).

Then from the Pythagorean formula for squared distance

$d_{12}^2=50,d_{13}^2=10,d_{14}^2=90,d_{23}^2=20,d_{24}^2=20,d_{34}^2=64$

And then

$(d_{12}d_{34})^4+(d_{13}d_{24})^4+(d_{14}d_{23})^4=3200^2+200^2+1800^2=\color{blue}{13520000}$

$2[(d_{12}d_{34})^2(d_{13}d_{24})^2+(d_{13}d_{24})^2(d_{14}d_{23})^2+(d_{14}d_{23})^2(d_{12}d_{34})^2]=2×[(3200×200)+(3200×1800)+(200×1800)]=\color{blue}{13520000}.$

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