Given an acute triangle $ABC$ with $O$ as its circumcenter. Line $AO$ intersects $BC$ at $D$. Points $E$, $F$ are on $AB$, $AC$ respectively such that $A$, $E$, $D$, $F$ are concyclic. Prove that the length of the projection of line segment $EF$ on side $BC$ does not depend on the positions of $E$ and $F$.
Problem
Source: CGMO 2004 P6
Tags: trigonometry, geometry unsolved, geometry
feliz
28.12.2008 04:26
Let $ X$ and $ Y$ be the projections of $ D$ in $ AB$ and $ AC$. Since $ AEDF$ is cyclic, $ \angle DEA = \angle DFC$, i.e., $ \angle DEX = \angle DFY$, which means that $ DEX$ and $ DFY$ are similiar. If we proved that the projections of $ EX$ and $ FY$ have the same size, we would know that the projections of $ XY$ and $ EF$ have the same size, which suffices. Now let $ M = \frac {A + B}{2}$ and $ N = \frac {A + C}{2}$. Also, let $ R = BO = CO$. We have $ \frac {EX}{FY} = \frac {DX}{DY} = \frac {OM}{ON} = \frac {Rcos\angle BCA}{Rcos\angle ABC}$ Hence $ EXcos\angle BAC = FYcos\angle BCA$, and these are the measures of the desired projections. Obs.: $ D$ could be an arbitrary point in the line $ AO$.
panamath
01.10.2013 08:15
Let's take the attached picture as reference. First notice that $\angle{FED}=\angle{DAF}=\angle{BEX}$. Now we have \[XY=EF \sin{\angle{FEX}} = EF \sin(\angle{FED}+\angle{DEX})\] \[ = EF \left(\frac{BX \cdot EX + EX \cdot DX}{BE \cdot DE}\right)=DB\cos{\angle{BEX}} \cdot \frac{EF}{ED}\] \[= DB\cos{\angle{BEX}} \cdot \frac{\sin(180-\angle A)}{\sin \angle DAB}\]. Finally, all the terms of the $RHS$ are constant, so $LHS$ must as well be constant.
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AopsUser101
29.05.2020 00:58
After a hint to consider $T$: Define $\angle EAD = \angle EFD = a$ and $\angle FED = \angle DAF = b$. Let $T$ be the intersection of $(AEF)$ and $BC$. We know that $\angle ABO = a$ and $\angle BOC = 2a + 2b \Longleftrightarrow \angle OBD = 90 - a - b$. Therefore, $\angle ABD = 90 -b $ and $\angle ACD = 90 - a$. Noting that $\sin (90 + b - a) = \sin (90 + a -b)$, $\frac{BD}{BA}\cdot \sin (b) = \frac{DC}{CA} \cdot \sin (a)$ Note that $\frac{BD}{BA} = \frac{\sin (a)}{\sin (90 + b -a)}$ and $\frac{DC}{CA} = \frac{\sin (b)}{\sin (90 + a - b)}$. $$XB + YC = BE \sin (b) + CF \sin (a)$$Note that by Power of a Point, $BE \cdot BA = BT \cdot BD$, so $BE = \frac{BT \cdot BD}{BA}$. By symmetry, $\frac{CT \cdot CD}{AC} = AF$. Thus, $$XB + YC = \frac{BT \cdot BD}{BA} \sin (b) + \frac{CT \cdot CD}{AC} \sin (a) = \frac{1}{\sin (a) \sin (b)} \left(\frac{BT \cdot BD \sin (b)}{BA} + \frac{CT \cdot CD \sin (a)}{AC} \right)$$Using what we derived at the beginning, $$XB + YC = \frac{1}{\sin (a) \sin (b)} \cdot \frac{\sin (a) \sin (b) }{\sin (90 + b - a)} \left(BD + CD \right) = \frac{BC}{\sin (90 + b - a)}$$It follows that $XB + YC$ is independent of $E$ and $F$, so $BC - (XB + YC) = XY$ is also independent of $E,F$.
Pyramix
30.11.2022 21:15
Let the length of the projection for some $E$ and $F$ be $l_{E,F}$.
Let $X$ and $Y$ be the projections of $E$ and $F$ on $BC$ respectively.
Consider point $E'$ on $AC$ other than $E$ and $F'$ on $AB$ other than $F$ such that they satisfy the given properties as well.
Since $\square AEDF$ and $\square AE'DF'$ are cyclic quadrilaterals, we have $\angle DEF=\angle DAF=\angle DAF'=\angle DE'F'=\angle DAC=90^\circ-B$. Similarly, $\angle DFE=\angle DF'E'=90^\circ -C$. Let $K$ is the intersection of $EF$ and $E'F'$. Since $\angle DEF=\angle DEK=\angle DE'F'=\angle DE'K$, we get $\square EE'DK$ as cyclic. Similarly, $\square FF'DK$ is also cyclic.
Let $\angle EDE'=x\Rightarrow\angle EKE'=x$ (as $\square EE'DK$ is cyclic), $\Rightarrow\angle F'KF=x=F'DF$ (as $\square FF'DK$ is cyclic).
Hence, $\angle EDE'=\angle F'DF$.
Let $R_1$ be the circumradius $\square EE'DK$ and $R_2$ be the circumradius of $\square FF'DK$.
Applying Sine Rule,
(i) (in $\circ EE'DK$): $\frac{KD}{\sin (\angle KED)}=\frac{KD}{\sin(90^\circ-B)}=\frac{KD}{\cos B}=2R_1=\frac{EE'}{\sin x}$,
(ii) (in $\circ FF'DK$): $\frac{KD}{\cos C}=\frac{FF'}{\sin x}$.
Dividing both these equations gives $\frac{\cos C}{\cos B}=\frac{EE'}{FF'}\Rightarrow EE'\cos B=FF'\cos C$.
Now,
$EB\cos B=BX$ and $E'B\cos B=BX'$. Subtracting, we get $EE'\cos B=XX'$.
Similarly, we get $FF'\cos C=YY'$.
Hence, $XX'=YY'$. This means that $l_{E',F'}=l_{E,F}+(XX'-YY')=l_{E,F}$.
So, the projection length $l_{EF}$ is equal for two arbitrary pairs of points $E,F$, and hence is independent of them.
Hence proved.
$\textbf{Remark.}$ If the problem would have changed from using circumcenter $O$ to incentre $I$, then the result would be $EE'=FF'$.
Figure:
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math_comb01
03.06.2023 18:46
Though it is easy to find the trigonometric solution, does there exist a geometric solution to this?
djmathman
09.06.2023 09:57
Let $P$ be the foot of the altitude from $A$, and suppose $\overline{AP}$ intersects $\odot(AEF)$ for the second time again at $Q$. Further, let $X$ and $Y$ be the projections of $E$ and $F$ onto $\overline{BC}$. The given problem asks us to show $XY$ is constant. Because $AO$ and $AP$ are isogonal, $EF\parallel QD$. Thus \[ XY = \frac{XY}{PD}\cdot PD = \frac{EF}{QD}\cdot PD = \frac{\sin A}{\sin\angle PAD}\cdot PD, \]which is indeed constant. Yay.