circumcenter is equidistant from B,C (HOMC 2018 S Team 9)

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circumcenter is equidistant from B,C (HOMC 2018 S Team 9)

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#1 h Feb 17, 2020, 10:19 PM • 1 Y

Y by Adventure10

Let $ABC$ be acute, non-isosceles triangle, inscribed in the circle $(O)$ . Let $D$ be perpendicular projection of $A$ onto $BC$ , and $E, F$ be perpendicular projections of $D$ onto $CA,AB$ respectively.
(a) Prove that $AO \perp EF$ .
(b) The line $AO$ intersects $DE,DF$ at $I,J$ respectively. Prove that $\vartriangle DIJ$ and $\vartriangle ABC$ are similar.
(c) Prove that circumcenter of $\vartriangle DIJ$ is equidistant from $B$ and $C$

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#2 h Feb 18, 2020, 3:49 AM • 1 Y

Y by Adventure10

a) Let AK be the diameter of (O).
$\widehat{OAC}+\widehat{CKA}=90^{\circ}$
$\widehat{BAD}+\widehat{ABC}=90^{\circ}$
$\widehat{AKC}=\widehat{ABC}$
$\Rightarrow \widehat{BAD}=\widehat{OAC}$
$\Rightarrow \widehat{BAO}=\widehat{DAC}$
$\Rightarrow \widehat{BAO}=\widehat{DFE}$
$\Rightarrow \widehat{AFE}+\widehat{BAO}=\widehat{AFE}+\widehat{DFE}=90^{\circ}$
$\Rightarrow AO\perp EF$
This LATEX thing costs me too much time

This post has been edited 1 time. Last edited by wassupbros, Feb 18, 2020, 3:53 AM

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#3 h Feb 18, 2020, 4:16 AM • 1 Y

Y by Adventure10

For part a, we can also draw a tangent of (O) which goes through A. I think it is easier to solve.

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#4 h Feb 18, 2020, 4:28 AM • 1 Y

Y by Adventure10

b) $\widehat{DIJ}=\widehat{AIE}=90^{\circ}-\widehat{DEF}=90^{\circ}-\widehat{FAD}=\widehat{ABC}$
$\widehat{IDJ}=180^{\circ}-\widehat{FDE}=\widehat{BAC}$
$\Rightarrow$ The two triangles are congruent

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#5 h Feb 18, 2020, 4:29 AM • 1 Y

Y by Adventure10

geo2006 wrote:

For part a, we can also draw a tangent of (O) which goes through A. I think it is easier to solve.

Good. Let $Ax$ be a tangent to $(O)$ so $\angle xAB=\angle ACB$ . Easy to prove $BCEF$ is cyclic $\Rightarrow$ $\angle ACB=\angle AFE\Rightarrow \angle xAB=\angle AFE\Rightarrow Ax\parallel EF\Rightarrow AO\perp EF$

This post has been edited 1 time. Last edited by BestChoice123, Feb 25, 2020, 8:04 AM

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#6 h Feb 18, 2020, 4:41 AM • 2 Y

Y by Adventure10, Mango247

parmenides51 wrote:

(c) Prove that circumcenter of $\vartriangle DIJ$ is equidistant from $B$ and $C$

omg the circumcenter of $\vartriangle DIJ$ is the midpoint of $BC$

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#7 h Feb 18, 2020, 4:43 AM • 2 Y

Y by Adventure10, Mango247

Can you post a screenshot (your diagram for part c) now? Did you draw a special case of the problem?

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#8 h Feb 18, 2020, 6:52 AM • 1 Y

Y by Adventure10

BestChoice123 wrote:

parmenides51 wrote:

(c) Prove that circumcenter of $\vartriangle DIJ$ is equidistant from $B$ and $C$

omg the circumcenter of $\vartriangle DIJ$ is the midpoint of $BC$

Great observation. It is actually

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#9 h Feb 18, 2020, 7:06 AM • 1 Y

Y by Adventure10

HintClick to reveal hidden text

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#10 h Feb 22, 2020, 8:58 PM • 1 Y

Y by Adventure10

BestChoice123 wrote:

geo2006 wrote:

For part a, we can also draw a tangent of (O) which goes through A. I think it is easier to solve.

Good. Let $Ax$ be a tangent of $(O)$ so $\angle xAB=\angle ACB$ . Easy to prove $BCEF$ is cyclic $\Rightarrow$ $\angle ACB=\angle AFE\Rightarrow \angle xAB=\angle AFE\Rightarrow Ax\parallel EF\Rightarrow AO\perp EF$

How did you deduce that BCEF is cyclic

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#11 h Feb 23, 2020, 3:57 AM

Y by

MikePerry wrote:

BestChoice123 wrote:

geo2006 wrote:

For part a, we can also draw a tangent of (O) which goes through A. I think it is easier to solve.

Good. Let $Ax$ be a tangent of $(O)$ so $\angle xAB=\angle ACB$ . Easy to prove $BCEF$ is cyclic $\Rightarrow$ $\angle ACB=\angle AFE\Rightarrow \angle xAB=\angle AFE\Rightarrow Ax\parallel EF\Rightarrow AO\perp EF$

How did you deduce that BCEF is cyclic

Note that AEF and ABC are two congruent trinagles

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#12 h Feb 23, 2020, 4:28 AM • 1 Y

Y by Mango247

Part c solution. Sorry for the bad image quality

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#13 h Feb 23, 2020, 7:36 AM • 1 Y

Y by Mango247

please resent after rotating the photo

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#14 h Feb 23, 2020, 8:17 AM

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Here's yours

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#15 h Feb 23, 2020, 3:01 PM

Y by

Even simpler:
As the triangles are similar, there are some cyclic quadrilaterals in the figure telling us that $I$ and $J$ are the projections of $C$ and $B$ onto $AO$ .
Now take the midpoint M of $BC$ and $N$ the midpoint of $IJ$ . By Thales, $MN//BJ//CI$ thus $M$ lies on the angle bissector of $IJ$ . Finally, notice $\angle BAO = \angle BDJ = \angle NDJ$ but as the triangles are similar, the circumcenter of $DIJ$ lies on $BC$ . So it must be $M$ .

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#16 h Feb 23, 2020, 3:17 PM

Y by

RodSalgDomPort wrote:

Even simpler:
As the triangles are similar, there are some cyclic quadrilaterals in the figure telling us that $I$ and $J$ are the projections of $C$ and $B$ onto $AO$ .
Now take the midpoint M of $BC$ and $N$ the midpoint of $IJ$ . By Thales, $MN//BJ//CI$ thus $M$ lies on the angle bissector of $IJ$ . Finally, notice $\angle BAO = \angle BDJ = \angle NDJ$ but as the triangles are similar, the circumcenter of $DIJ$ lies on $BC$ . So it must be $M$ .

I think you have mistaken B with C bro. And also I don't understand what are you trying to conclude at the end. Nothing wrong with your solution, I understand, it's just the symbols :-D

:-D

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#17 h Feb 24, 2020, 7:25 PM

Y by

How can we prove that AEF and ABC are congruent triangles

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#18 h Feb 25, 2020, 8:01 AM • 1 Y

Y by Mango247

MikePerry wrote:

How can we prove that AEF and ABC are congruent triangles

$DEAF$ is cyclic $\Rightarrow \angle AEF=\angle ADF$ . Since $\angle ADF=\angle ABD$ thus $\angle AEF=\angle ABD\Rightarrow \Delta AEF\cong \Delta ABC$

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