A simple one on cyclic quads

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A simple one on cyclic quads

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cyclic quadrilateral

power of a point

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Source: India RMO 1992 Problem 4

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1592 posts

Rushil

#1 h Oct 15, 2005, 4:42 AM • 2 Y

Y by Glucosa, Adventure10

$ABCD$ is a cyclic quadrilateral with $AC \perp BD$ ; $AC$ meets $BD$ at $E$ . Prove that $EA^2 + EB^2 + EC^2 + ED^2 = 4 R^2$
where $R$ is the radius of the circumscribing circle.

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2643 posts

yetti

#2 h Oct 16, 2005, 12:23 AM • 4 Y

Y by aayush-srivastava, Mathcollege, Adventure10, Mango247

Let O be the quadrilateral circumcenter. The power of the point E to the circumcircle (O) is

$EA \cdot EC = EB \cdot ED = (R - EO)(R + EO) = R^2 - EO^2$

Let M, N be the midpoints of AC, BD. WLOG, assume $EA \le EC, EB \le ED$ , otherwise relabel the quadrilateral.

$EO^2 = EN^2 + EM^2 = (MA - EA)^2 + (NB - EB)^2 =$

$= \left(\frac{AC}{2} - EA\right)^2 + \left(\frac{BD}{2} - EB\right)^2 = \left(\frac{EA + EC}{2} - EC\right)^2 + \left(\frac{EB + ED}{2} - EB\right)^2 =$

$= \frac{(EC - EA)^2}{4} + \frac{(ED - EB)^2}{4} = \frac{EA^2 + EB^2 + EC^2 + ED^2}{4} - \frac{EA \cdot EC + EB \cdot ED}{2} =$

$= \frac{EA^2 + EB^2 + EC^2 + ED^2}{4} - (R^2 - EO^2)$

$EA^2 + EB^2 + EC^2 + ED^2 = 4R^2$

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749 posts

#3 h Oct 16, 2005, 12:52 AM • 4 Y

Y by tantheta67, Adventure10, abvk1718, Mango247

Another soln

Let a,b,c,d = EA,EB,EC,ED

we can calculate the circumradius of ABD as $\frac{\sqrt{(a^2 + b^2)(a^2 + d^2)(b+d)^2}}{2(b+d)a}$ using the classic $R = \frac{abc}{4\Delta}$ formula.

Then $4R^2 = \frac{(a^2 + b^2)(a^2 + d^2)}{a^2} = a^2 + b^2 + d^2 + \frac{b^2d^2}{a^2}$ . From power of a point, bd = ac so this becomes $4R^2 = a^2 + b^2 + c^2 + d^2$ .

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777 posts

#4 h Oct 20, 2005, 4:23 PM • 4 Y

Y by Rushil_india, Adventure10, Mango247, and 1 other user

$SUM = (EA^2+EB^2)+(EC^2+ED^2) = AB^2+CD^2$

Let $w$ the circumcircle of $ABCD$

$L \in w, AL \parallel BD\Rightarrow$

$AB = LD$

$\widehat{CAL}=90^0 \Rightarrow CL$ is a diameter $\Rightarrow$

$\widehat{CDL}=90^0$

So the triangle $CDL$ is right

Finally,

$SUM = AB^2+CD^2 = LD^2+CD^2 = CL^2 = (2R)^2$

By the way, if the line $AL$ is tangent, then all above is correct if we replace $L$ by $A$ .

( $OA \perp BD , AC \perp BD \Rightarrow AC$ is diameter
Now we have $AB = AD \Rightarrow$ $SUM = AB^2+CD^2 = AD^2+CD^2 =AC^2$ )

Attachments:

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4 posts

ruza

#5 h May 29, 2006, 6:48 PM • 3 Y

Y by Adventure10, Mango247, Math_DM

This is just idea.
From pithagora theorem we obtain AE^2+BE^2+CE^2+DE^2=(AB^2+BC^2+CD^2+AD^2)/2
Then if we apply cosine rule to triangles AOB,BOC,COD,AOD and then sum it we get:
AB^2+BC^2+CD^2+AD^2=2R^2(4-cosAOB-cosBOC-cosCOD-cosDOA)
I think that its not hard to prove that cosAOB+cosBOC+cosCOD+cosDOA=0 where AOB+BOC+COD+DOA=360(these are angles)

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1012 posts

#6 h Aug 10, 2011, 4:28 AM • 4 Y

Y by tantheta67, abvk1718, Adventure10, Mango247

Here is another simple solution:

We have: $EA^2+EB^2+EC^2+ED^2=\frac{a^2+b^2+c^2+d^2}{2}$ so we want to prove that: $a^2+b^2+c^2+d^2=8R^2$ . We have from the Law of Sines: $\{\begin{array}{ccc }a=2R\sin{\angle{ACB}}\\\\b=2R\sin{\angle{BCD}}\\\\c=2R\sin{\angle{DBC}}=2R\cos{\angle{ACB}}\\\\d=2R\sin{\angle{ACD}}=2R\cos{\angle{BCD}} \end{array}$ thus: $a^2+b^2+c^2+d^2=4R^2(\sin^2{\angle{ACB}}+\cos^2{\angle{ACB}}+\sin^2{\angle{BCD}}+\cos^2{\angle{BCD}})=4R^2*2=8R^2$ as desired.

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7892 posts

AIME15

#7 h Aug 10, 2011, 2:14 PM • 2 Y

Y by Adventure10, Mango247

EDIT: never mind, this post is bogus

This post has been edited 1 time. Last edited by AIME15, Aug 13, 2011, 10:45 PM

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7054 posts

#8 h Aug 10, 2011, 3:32 PM • 3 Y

Y by basketball9, Adventure10, Mango247

Indeed, the shortest proof of this simple problem is basketball9's. For $x=m(\angle ADB)$ obtain $AB=2R\sin x$ $CD=2R\cos x$ .

Thus, $EA^2+EB^2+EC^2+ED^2=$ $\left(EA^2+EB^2\right)+\left(EC^2+ED^2\right)=$ $AB^2+CD^2=$ $4R^2\left(\sin^2x+\cos^2x\right)=4R^2$ .

However, I"ll present here some remarkable properties of the cyclical ortodiagonal quadrilaterals. Denote :

$\left|\begin{array}{cccc} AB=a & BC=b & CD=c\\\ DA=d & AC=e & BD=f\end{array}\right|\ ;$ the midpoints $\{M,N,P,Q,U,V\}$ of the sides and the diagonals

$\left\{[AB],[BC],[CD],[DA],[AC],[BD]\right\}$ respectively ; the projections $\{X,Y,Z,T\}$ of the point $E$ on the sides

$\left\{[AB],[BC],[CD],[DA]\right\}$ respectively ; $G\in MP\cap NQ$ ; the circumcircle $w=C(O,R)$

of the given cyclical quadrilateral $ABCD$ for which $AC\perp BD$ . Prove easily that the following remarkable properties :

$\blacktriangleright\ EA^2+EB^2+EC^2+ED^2=AB^2+CD^2=AD^2+BC^2=4R^2\ .$

$\blacktriangleright\ AB^2+BC^2+CD^2+DA^2=AC^2+BD^2+4\cdot OE^2=8R^2\ .$

$\blacktriangleright\ \left\{\begin{array}{ccc} X\in PE & OM=PE=\frac c2\\\\ Y\in QE & ON=QE=\frac d2\\\\ Z\in ME & OP=ME=\frac a2\\\\ T\in NE & OQ=NE=\frac b2\end{array}\right|\implies$ the points $\{M,N,P,Q,X,Y,Z,T\}$ are concyclically (circle of eight points - $\beta$ ).

$\blacktriangleright\ OMEP$ and $ONEQ$ are parallelograms $\implies$ the centroid $G$ of the quadrilateral $ABCD$ is midpoint of $[OE]$ and the center of $\beta$ .

With other words, the $E$ -median in $\triangle AEB$ is the $E$ -altitude in $\triangle CED$ and the $E$ -median in $\triangle CED$ is the $E$ -altitude in $\triangle AEB$ a.s.o.

Thus, the points $X$ , $Z$ belong to the circle with diameter $MP$ and the point $Y$ , $T$ belong analogously to the circle with diameter $NQ$ .

The quadrilateral $MNPQ$ is a rectangle with the center in $G\in MP\cap NQ$ . In conclusion, the points $\{X,Y,Z,T\}$ belong to the

circumcircle of the rectangle $MNPQ$ .

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333 posts

#10 h Jul 14, 2018, 4:26 PM • 1 Y

Y by Adventure10

By pythagorus theorem we get to prove- $AB^2+CD^2=4R^2$
now, in $\Delta ABC$ $\frac{AB}{sin BCA}$ = $2R$ = $\frac{CD}{sin CAD}$ in $\Delta ACD$ but cad=cbd=90°-bca so $sin CAD =cos BCA$
squaring and adding the above equations we get
$AB^2+CD^2=4R^2(sin^2CAD+cos^2CAD)$
which proves it.

This post has been edited 2 times. Last edited by mathetillica, Jul 14, 2018, 4:27 PM

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Math.Is.Beautiful

850 posts

Math.Is.Beautiful

#11 h Jul 14, 2018, 4:34 PM • 2 Y

Y by Adventure10, Mango247

https://youtu.be/mZaeingxjh4
I think this can now be easily formalized

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2125 posts

#12 h Nov 4, 2018, 9:25 AM • 2 Y

Y by Adventure10, Mango247

Let $O$ be the centre of the circle, Let $\angle AOB =x$ and Let $\angle COD =y$
Now, since, $\angle BCE=\frac{x}{2}$ and $\angle CBE = \frac{y}{2} \implies x+y=180^{\circ}$
Therefore, $AB^2=2R^2-2R^2\cos x \text{ and } CD^2=2R^2-2R^2\cos y= 2R^2+2R^2\cos x$ Hence, $EA^2+EB^2+EC^2+ED^2=AB^2+CD^2= 4R^2$ Proved!!

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328 posts

yshk

#13 h Jul 14, 2019, 8:34 PM • 2 Y

Y by Adventure10, Mango247

Math.Is.Beautiful wrote:

https://youtu.be/mZaeingxjh4
I think this can now be easily formalized

Elegant application of the constant inscribed angle given a fixed chord length.

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21 posts

EVS383

#14 h Apr 27, 2024, 11:22 PM

Y by

Solution using Parallel Axis Theorem
We want the quantity $EA^2 + EB^2 + EC^2+ED^2$ , so assign a mass of 1 to points $A, B, C, D$ . Our answer will then just be the moment of inertia at $E$ . Denote the center of mass of $A, B, C, D$ as $X$ and the center of the circle as $O$ . Then, by parallel axis theorem, we have:
$I_O = 4R^2$ $I_O = I_X + 4*OX^2$ $I_E = I_X + 4*EX^2$ Subtracting the first and second equations, we get $I_O - I_E = 4 * (OX^2-EX^2)$ . We then wish to prove that $OX = EX$ . This is true because if we let the altitudes from $O$ to $AC$ and $BD$ be $S$ and $T$ respectively, then $X$ is the midpoint of $ST$ .
Since $ESOT$ is a rectangle and the intersection between the diagonals of a rectangle bisects both diagonals, the intersection of the diagonals is just $X$ and $OX$ = $EX$ so $I_O = I_E$

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