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#1 h Sep 1, 2010, 11:33 AM • 2 Y

Y by Adventure10, Mango247

For an interior point $D$ of a triangle $ABC,$ let $\Gamma_D$ denote the circle passing through the points $A, \: E, \: D, \: F$ if these points are concyclic where $BD \cap AC=\{E\}$ and $CD \cap AB=\{F\}.$ Show that all circles $\Gamma_D$ pass through a second common point different from $A$ as $D$ varies.

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9775 posts

jayme

#2 h Sep 1, 2010, 2:14 PM • 1 Y

Y by Adventure10

Dear Mathlinkers,
the problem seems to me not clear. But, I can be wrong...
All the circles pass through A and D.???
Sincerely
Jean-Louis

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1345 posts

#3 h Sep 1, 2010, 2:52 PM • 2 Y

Y by Adventure10, Mango247

jayme wrote:

Dear Mathlinkers,
the problem seems to me not clear. But, I can be wrong...
All the circles pass through A and D.???
Sincerely
Jean-Louis

Dear Jean-Louis
The problem says that $D$ is not a constant point, it varies, we have to find a constant point different from $A.$

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shoki

#4 h Sep 1, 2010, 3:21 PM • 2 Y

Y by Adventure10, Mango247

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#5 h Sep 1, 2010, 5:56 PM • 2 Y

Y by Adventure10, Mango247

I have a solution for this problem.

First we will construct $S$ be constant point such that $S\in \Gamma_D$ .
Let $H\in AB$ such that $CA=CH$ and let $w$ be the circle pass through $A,H$ and tangent to $CA,CH$ . $w\cap (B,H,C)=S(S\neq H)$ .

Now we will prove that $ADSF$ is cyclic.
We have $\angle BHC=180^0-\angle AHC=180^0-\angle BAC=\angle BDC$ so $BHDC$ is cyclic but $BHSC$ is cyclic hence $BHDS$ is cyclic.
Thus $\angle SDF=180^0-\angle BDS=180^0-\angle BHS=\angle SHA=\angle SAF$ so $ADSF$ is cyclic and the problem is done

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#6 h Sep 1, 2010, 6:11 PM • 2 Y

Y by Adventure10, Mango247

$B \equiv AF \cap ED$ and $C \equiv AE \cap FD$ are conjugate points WRT $\Gamma_D.$ Then it follows that circle $(M)$ with diameter $BC$ is orthogonal to $\Gamma_D.$ Thus, inversion through pole $A$ with power equal to the power of $A$ WRT $(M)$ takes $(M)$ into itself and $\Gamma_D$ into a straight line $\gamma$ orthogonal to $(M),$ due to the conformity $\Longrightarrow$ $M \in \gamma.$ Hence, $\Gamma_D$ passes through the image of $M$ under the referred inversion, i.e. the orthogonal projection of the orthocenter of $\triangle ABC$ on $AM.$

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motal

#7 h Sep 2, 2010, 1:42 AM • 1 Y

Y by Adventure10

Luis González wrote:

$B \equiv AF \cap ED$ and $C \equiv AE \cap FD$ are conjugate points WRT $\Gamma_D$

What do you mean by 'conjugate points with respect to a circle' and why does it imply orthogonality?

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cnyd

#8 h Oct 21, 2010, 12:20 PM • 2 Y

Y by Adventure10, Mango247

shoki wrote:

Shoki ,could you explain better ?

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skytin

#9 h Oct 21, 2010, 4:40 PM • 2 Y

Y by Adventure10, Mango247

Let circle (AFE) intersect (BDC) at D and G and line AG intersect (BDC) at X easy to see that XB || AE and XC || AF , let AX intersect BC at point M easy to see that M is midpoint of BC and (BDC) is constant , so G is fixed

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shoki

#10 h Nov 1, 2010, 1:18 PM • 2 Y

Y by Adventure10, Mango247

Quote:

Let $H$ be the orthocenter.Let $H_b,H_c$ be the feet of perpendiculars from $B,C$ to $AC,AB$ . the problem is equivalent to prove that $\frac{H_bE}{H_cF}$ is constant.

@cnyd, see here

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#11 h May 5, 2013, 6:54 PM • 2 Y

Y by Adventure10, Mango247

Hi ;

This Point Is Also On $AM$ Where $M$ Is The Midpoint Of $BC$ Such That If We Connect It To $I$ Then We Have $IK$ Is Perpendicular To $AM$

Best Regard

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aZpElr68Cb51U51qy9OM

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aZpElr68Cb51U51qy9OM

#12 h May 5, 2013, 10:07 PM • 2 Y

Y by ahaanomegas, Adventure10

We use barycentric coordinates. Let $BC = a, CA = b, AB = c$ , and let $A = (1, 0, 0), B = (0, 1, 0), C = (0, 0, 1)$ .

Let $D = (p:q:r)$ . We claim that $\Gamma_D$ always passes through $P(a^2:b^2+c^2-a^2:b^2+c^2-a^2)$ , which is independent of $D$ . Since $E$ and $F$ are the traces of $D$ onto $CA$ and $AB$ , we have $E = (p:0:r)$ and $F = (p:q:0)$ . We are given that $A, E, D, F$ are concyclic; that is, the equations of the circumcircles of $AED$ and $AFD$ are equal.
$\begin{align*}\odot AED: a^2yz+b^2zx+c^2xy - (x+y+z)(\frac{a^2r(p+r) - b^2rp + c^2p(p+r)}{(p+q+r)(p+r)}y + \frac{b^2p}{p+r}z) = 0\end{align*}$ $\begin{align*}\odot AFD: a^2yz+b^2zx+c^2xy - (x+y+z)(\frac{c^2p}{p+q}y+\frac{a^2q(p+q) + b^2p(p+q) - c^2pq}{(p+q+r)(p+q)}z) = 0\end{align*}$
Therefore $(p+q)(a^2r(p+r)-b^2rp+c^2p(p+r)) = (p+q+r)(p+r)c^2p$ , or $c^2p(p+r) = (p+q)(a^2(p+r)-b^2p)$ . Similarly, $b^2p(p+q+r)(p+q) = (p+r)(a^2q(p+q)+b^2p(p+q)-c^2pq)$ , or $b^2p(p+q)= (p+r)(a^2(p+q) - c^2p)$ . Multiplying the two yields
$a^2(p+r)(p+q) = b^2p(p+q)+c^2p(p+r)$
Solving for $c^2p(p+r)$ and plugging this into the equation of the circumcircle of $AED$ yields
$a^2yz+b^2zx+c^2xy - (x+y+z)(a^2y + \frac{b^2p}{p+r}(z-y))) = 0$
Substituting $P$ into this equation yields $\begin{align*}a^2(b^2+c^2-a^2)(b^2+c^2-a^2)+b^2(b^2+c^2-a^2)(a^2)+c^2(a^2)(b^2+c^2-a^2) - (-a^2+2b^2+2c^2)(a^2(b^2+c^2-a^2)) = 0,\end{align*}$ which is true. Therefore the circumcircle of $AEDF$ always passes through $P$ , as claimed.

This was a lot more manageable than I'd thought. I didn't have to use Mathematica to expand the last expression, since the assumed condition $y = z$ allowed me to find the $x$ and $y$ -coordinates of $P$ in the first place.

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#13 h May 7, 2013, 3:39 AM • 1 Y

Y by Adventure10

Luis González wrote:

$B \equiv AF \cap ED$ and $C \equiv AE \cap FD$ are conjugate points WRT $\Gamma_D.$ Then it follows that circle $(M)$ with diameter $BC$ is orthogonal to $\Gamma_D.$

Alternatively, we can finish here by noting that $\Gamma_D$ passes through the second limit point of the coaxial system, different from $A.$ But of course a characterization of this point gives a nicer solution.

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leader

#14 h May 26, 2013, 1:04 PM • 2 Y

Y by Adventure10, Mango247

let $ABHC$ be a parallelogram and let $k=\odot HBC$ meet $AH$ again at a fixed point $L$ . $\angle BDC=\angle EDF=180-\angle BAC=180-\angle BHC$ so $D\in k$ . Now $\angle DLA=180-\angle DLH=\angle HBD=\angle AEB$ so $L\in \Gamma_{D}$ Q.E.D

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NewAlbionAcademy

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NewAlbionAcademy

#15 h Sep 7, 2013, 6:56 PM • 6 Y

Y by Binomial-theorem, sjaelee, dantx5, Adventure10, Mango247, and 1 other user

We do not use barycentric coordinates. Clearly the locus of $D$ is tha arc of a circle $\omega$ with chord $BC$ such that $\angle BDC=\pi -\angle A$ . Fix any point $D$ on this locus, and let $\Gamma_D$ intersect $\omega$ again at $X$ . I claim $X$ is the desired common point. Take another point $D'$ on the locus, and let the circumcircle of $\triangle AD'X$ intersect $AB$ and $AC$ at $F'$ and $E'$ respectively. If we show $C, D',$ and $F'$ are collinear then similarly $B, D',$ , and $E'$ are collinear, so $\Gamma_D'$ goes through $X$ .

We have $\angle F'D'X+\angle XD'C=\angle F'AX+\pi-\angle XDC=\pi$ , so $C, D',$ and $F'$ are collinear as desired, and we are done. QED.
(of course, leader's solution is much more elegant and gives a nice characterization of $X$ , but it is harder to find)

This was a lot more manageable than I'd thought. I didn't have to use Mathematica to expand the last expression, since I did not use barycentric coordinates in the first place.

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