Art of Problem Solving
Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Van Aubel's Theorem: Difference between revisions

New page: = Theorem = Construct squares <math>ABA'B'</math>, <math>BCB'C'</math>, <math>CDC'D'</math>, and <math>DAD'A'</math> externally on the sides of quadrilateral <math>ABCD</math>, and let the...
 
Ap246 (talk | contribs)
editor
50 edits
 
(16 intermediate revisions by 5 users not shown)
Line 1: Line 1:
= Theorem =
= Theorem =
Construct squares <math>ABA'B'</math>, <math>BCB'C'</math>, <math>CDC'D'</math>, and <math>DAD'A'</math> externally on the sides of quadrilateral <math>ABCD</math>, and let the centroids of the four squares be <math>P, Q, R,</math> and <math>S</math>, respectively.  Then <math>PR = QS</math> and <math>PR \perp QS</math>.
On each side of quadrilateral <math>ABCD</math>, construct an external square and its center: <math>ABA'B'</math>, <math>BCB'C'</math>, <math>CDC'D'</math>, <math>DAD'A'</math>; yielding centers <math>P_{AB}, P_{BC}, P_{CD}, P_{DA}</math>.  Van Aubel's Theorem states that the two line segments connecting opposite centers are perpendicular and equal length:
<geogebra>21cd94f930257bcbd188d1ed7139a9336b3eb9bc</geogebra>
<math>P_{AB}P_{CD} = P_{BC}P_{DA}</math>, and <math>\overline{P_{AB}P_{CD}} \perp \overline{P_{BC}P_{DA}}</math>.


= Proofs =
= Proofs =


== Proof 1: Complex Numbers==
== Proof 1: Complex Numbers==
<asy>
size(220);
import TrigMacros;
rr_cartesian_axes(-3,8,-2,8,complexplane=true,usegrid = false);
pair A, B, C, D, O, P, Q, R, SS;
O = (0,0) ;
A = (2,1.5);
B= (4,1.8);
C = (5.3,3);
D= (3,5.3);
draw(A--B--C--D--cycle);
draw(A--(A + rotate(-90)*(B-A))--(B + rotate(90)*(A-B))--B);
draw(B--(B + rotate(-90)*(C-B))--(C + rotate(90)*(B-C))--C);
draw(C--(C + rotate(-90)*(D-C))--(D + rotate(90)*(C-D))--D);
draw(D--(D + rotate(-90)*(A-D))--(A + rotate(90)*(D-A))--A);
P = (B + (A + rotate(-90)*(B-A)))/2;
Q = (C + (B + rotate(-90)*(C-B)))/2;
R = (D + (C + rotate(-90)*(D-C)))/2;
SS = (A + (D + rotate(-90)*(A-D)))/2;
//draw(WW--Y,red);
//draw(X--Z,blue);
dot("$a$",A,SW);
dot("$b$",B,2*E);
dot("$c$",C,E);
dot("$d$",D,NNW);
dot("$p$",P,E);
dot("$q$",Q,S);
dot("$r$",R,N);
dot("$s$",SS,S);
</asy>
Putting the diagram on the complex plane, let any point <math>X</math> be represented by the complex number <math>x</math>.  Note that <math>\angle PAB = \frac{\pi}{4}</math> and that <math>PA = \frac{\sqrt{2}}{2}AB</math>, and similarly for the other sides of the quadrilateral.  Then we have
Putting the diagram on the complex plane, let any point <math>X</math> be represented by the complex number <math>x</math>.  Note that <math>\angle PAB = \frac{\pi}{4}</math> and that <math>PA = \frac{\sqrt{2}}{2}AB</math>, and similarly for the other sides of the quadrilateral.  Then we have


Line 27: Line 61:
\end{eqnarray*}</cmath>
\end{eqnarray*}</cmath>


Finally, we have <math>(p-r) = i(q-s) = e^{i \pi/2}(q-r)</math>, which implies <math>PR = QS</math> and <math>PR \perp QS</math>, as desired.
Finally, we have <math>(p-r) = i(q-s) = e^{i \pi/2}(q-s)</math>, which implies <math>PR = QS</math> and <math>PR \perp QS</math>, as desired.


== Proof 2: Mean Geometry Theorem ==
==See Also==
[[Category:Theorems]]

Latest revision as of 13:01, 4 March 2023

Theorem

On each side of quadrilateral $ABCD$, construct an external square and its center: $ABA'B'$, $BCB'C'$, $CDC'D'$, $DAD'A'$; yielding centers $P_{AB}, P_{BC}, P_{CD}, P_{DA}$. Van Aubel's Theorem states that the two line segments connecting opposite centers are perpendicular and equal length: $P_{AB}P_{CD} = P_{BC}P_{DA}$, and $\overline{P_{AB}P_{CD}} \perp \overline{P_{BC}P_{DA}}$.

Proofs

Proof 1: Complex Numbers

[asy] size(220); import TrigMacros; rr_cartesian_axes(-3,8,-2,8,complexplane=true,usegrid = false); pair A, B, C, D, O, P, Q, R, SS; O = (0,0) ; A = (2,1.5); B= (4,1.8); C = (5.3,3); D= (3,5.3); draw(A--B--C--D--cycle); draw(A--(A + rotate(-90)*(B-A))--(B + rotate(90)*(A-B))--B); draw(B--(B + rotate(-90)*(C-B))--(C + rotate(90)*(B-C))--C); draw(C--(C + rotate(-90)*(D-C))--(D + rotate(90)*(C-D))--D); draw(D--(D + rotate(-90)*(A-D))--(A + rotate(90)*(D-A))--A); P = (B + (A + rotate(-90)*(B-A)))/2; Q = (C + (B + rotate(-90)*(C-B)))/2; R = (D + (C + rotate(-90)*(D-C)))/2; SS = (A + (D + rotate(-90)*(A-D)))/2; //draw(WW--Y,red); //draw(X--Z,blue); dot("$a$",A,SW); dot("$b$",B,2*E); dot("$c$",C,E); dot("$d$",D,NNW); dot("$p$",P,E); dot("$q$",Q,S); dot("$r$",R,N); dot("$s$",SS,S); [/asy] Putting the diagram on the complex plane, let any point $X$ be represented by the complex number $x$. Note that $\angle PAB = \frac{\pi}{4}$ and that $PA = \frac{\sqrt{2}}{2}AB$, and similarly for the other sides of the quadrilateral. Then we have


\begin{eqnarray*} p &=& \frac{\sqrt{2}}{2}(b-a)e^{i \frac{\pi}{4}}+a \\ q &=& \frac{\sqrt{2}}{2}(c-b)e^{i \frac{\pi}{4}}+b \\ r &=& \frac{\sqrt{2}}{2}(d-c)e^{i \frac{\pi}{4}}+c \\ s &=& \frac{\sqrt{2}}{2}(a-d)e^{i \frac{\pi}{4}}+d \end{eqnarray*}

From this, we find that \begin{eqnarray*} p-r &=& \frac{\sqrt{2}}{2}(b-a)e^{i \frac{\pi}{4}}+a - \frac{\sqrt{2}}{2}(d-c)e^{i \frac{\pi}{4}}-c \\ &=& \frac{1+i}{2}(b-d) + \frac{1-i}{2}(a-c). \end{eqnarray*} Similarly, \begin{eqnarray*} q-s &=& \frac{\sqrt{2}}{2}(c-b)e^{i \frac{\pi}{4}}+a - \frac{\sqrt{2}}{2}(a-d)e^{i \frac{\pi}{4}}-c \\ &=& \frac{1+i}{2}(c-a) + \frac{1-i}{2}(b-d). \end{eqnarray*}

Finally, we have $(p-r) = i(q-s) = e^{i \pi/2}(q-s)$, which implies $PR = QS$ and $PR \perp QS$, as desired.

See Also

Retrieved from "https://wiki.aopstest.com/wiki/index.php?title=Van_Aubel%27s_Theorem&oldid=190617"