In the triangle $A,B,C$, let $D$ be the middle of $BC$ and $E$ the projection of $C$ on $AD$. Suppose $\angle ACE = \angle ABC$. Show that the triangle $ABC$ is isosceles or rectangle.
Problem
Source: Swiss Imo Selection 2006
Tags: geometry, trigonometry, trig identities, Law of Sines, geometry proposed
stergiu
26.05.2006 09:49
Well , a) If point $E$ is the same with $D$ , then the triangle is isosceles, clearly! b) If $E$ does not coincide with $D$ , let $F$ the symmetric point of $C$ with respect to line $AD$.Then $DE$ is parallele $BF$ since in triangle $CBF$ the segment $DE$ joins two midpoints.This means that $\angle BFE = 90$. But $\angle ABC = \angle ACE = \angle AFC$ , so $ABFC$ is inscribed .Hench $\angle A = \angle BFC = 90$ and we are done ! Babis (Nice problem , but not of IMO level!)
BogG
26.05.2006 11:40
Waouw short solution ! Mine was really longer ! It was a problem to select the swiss team.. We had a harder geometry problem for the selection that I haven't solved if you want ... http://www.mathlinks.ro/Forum/viewtopic.php?p=519896#p519896
Virgil Nicula
27.05.2006 15:21
Bogdan Gheorghe wrote: A romanian in the swiss team ! Felicitari Bogdane si mult succes la O.I.M. ! This nice problem is very simply. $\mathrm {acute\ }\triangle ABC\ : \ D\in (BC)\ ,\ DB=DC\ ;\ E\in AD\ ,\ CE\perp AD\ .$ $\widehat {ABC}\equiv\widehat {ACE}\Longrightarrow m(\widehat {CAD})=90^{\circ}-B\ ,\ m(\widehat {BAD})=90^{\circ}-C\ .$ ===================================================== $1.\blacktriangleright$ The first method. $\frac{DB}{DC}=\frac{AB}{AC}\cdot \frac{\sin \widehat {BAD}}{\sin \widehat {CAD}}\Longleftrightarrow b\cdot\cos B=c\cdot\cos C\Longleftrightarrow$ $\sin 2B=\sin 2C\Longleftrightarrow B=C\ \ \vee\ \ B+C=90^{\circ}\Longleftrightarrow AB=AC\ \ \vee\ \ AB\perp AC\ .$ $2.\blacktriangleright$ The second method. Denote the circumcentre $O$ of the triangle $ABC\ .$ Then $m(\widehat {CAD})=90^{\circ}-B$ $\Longleftrightarrow$ $O\in AD\ .$ If $O\not\equiv D\ ,$ then $AB=AC ;$ if $O\equiv D\ ,$ then $AB\perp AC\ .$
BogG
27.05.2006 19:53
Waouw If I think that I have write 2 pages on the selection Take a look at this one ! http://www.mathlinks.ro/Forum/viewtopic.php?p=519896#p519896 Also geometry for the selection but really harder ^^
The QuattoMaster 6000
24.04.2008 07:47
BogG wrote: In the triangle $ A,B,C$, let $ D$ be the middle of $ BC$ and $ E$ the projection of $ C$ on $ AD$. Suppose $ \angle ACE = \angle ABC$. Show that the triangle $ ABC$ is isosceles or rectangle. All the problems from Swiss Imo Selection Team 2006 Here are a few other solutions to this problem:
Let $ F$ be the midpoint of $ AC$. It follows that $ DF$ is a midline of $ \triangle ABC$ and that $ \triangle EFC$ is isoceles since $ \angle AEC=90$, which means that $ \angle FDC=\angle ABC=\angle ACE=\angle FEC$, so $ DFCE$ is cyclic. This means that either $ D=E$, in which case, $ \angle ABC=\angle ACE=\angle ACB$, so $ \triangle ABC$ is isoceles or $ \angle BAC=\angle DFC=180-\angle DEC=90$, which means that $ \triangle ABC$ is right.
Let $ X$ be the foot of the altitude from $ C$ to $ AB$. Also, let $ Y$ be on $ BC$ such that $ YD\perp BC$. Let $ W$ be where $ YB$ meets $ AE$ and $ Z$ be where $ BY$ meets $ AC$. It follows that $ YD$ is a perpindicular bisector of $ BC$, so $ \angle YBC=\angle YCB=90-\angle ACB=90-\angle ACE=\angle DAZ$. It follows that $ ABDZ$ is cyclic. Furthermore, $ \angle YXB+\angle YDB=90+90=180$, so $ XYDB$ is cyclic. This means that $ AC\cdot ZC=BC\cdot CD=XC\cdot CY$, so $ AZYX$ is cyclic. This means that either $ X=A$, so $ \angle BAC=90$ or $ \angle AZB=180-\angle AXC=90$. This means that $ Y$ is the orthocenter of $ \triangle ABC$. Thus, $ YD$ passes through $ A$, so $ AY$ is the perpindicular bisector of $ BC$ and we conclude that $ \triangle ABC$ is isoceles.
fmasroor
04.02.2014 05:59
If you draw the altitude AX then by the condition AX is a symmedian, so bcosC/ccosB= b^2/c^2. expand with law of sines and factor, you get (a^2)(b-c)(b+c)=(b^2)(b-c)(b+c). Very easy problem