postaffteff

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JetFire008

125 posts

JetFire008

#1 h Mar 15, 2025, 12:33 PM

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Let $P$ be the Fermat point of a $\triangle ABC$ . Prove that the Euler line of the triangles $PAB$ , $PBC$ , $PCA$ are concurrent and the point of concurrence is $G$ , the centroid of $\triangle ABC$ .

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JetFire008

125 posts

JetFire008

#2 h Mar 15, 2025, 12:39 PM

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The Fermat Point of a triangle is the interior point from which the sum of distance between vertices is minimum.

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JetFire008

125 posts

JetFire008

#3 h Mar 15, 2025, 12:42 PM

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Euler line is the straight line passing through the orthocenter, centroid, and circumcenter of a triangle.

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JetFire008

125 posts

JetFire008

#4 h Mar 15, 2025, 12:46 PM

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Orthocentre is the point of intersection of altitudes
Centroid is the point of intersection of medians.
Circumcentre is the point of intersection of perpendicular bisectors.

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JetFire008

125 posts

JetFire008

#5 h Mar 15, 2025, 12:51 PM

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If the Euler line of a $\triangle ABC$ is parallel to $BC$ , show that tan $B$ tan $C = 3$ .

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drmzjoseph

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drmzjoseph

#6 h Mar 16, 2025, 10:44 AM

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Old problem
Extend $PA$ until $X$ such that $BXC$ is equilateral now, taking as homothetic center the midpoint of $BC$ (3:1) sending X to circumcenter of $PBC$ , P to centroid of $PBC$ and A to centroid of $ABC$ that's enough

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JetFire008

125 posts

JetFire008

#7 h Mar 17, 2025, 3:52 PM

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Prove that the circumcircles of the four triangles formed by four lines have a common point.

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JetFire008

125 posts

JetFire008

#8 h Mar 17, 2025, 3:59 PM

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In $\triangle ABC$ , $BD$ and $CE$ are the bisectors of $\angle B$ , $\angle C$ cutting $CA$ , $AB$ at $D$ , $E$ respectively. If $\angle BDE = 24^{\circ}$ and $\angle CED = 18^{\circ}$ , find the angles of $\triangle ABC$

This post has been edited 1 time. Last edited by JetFire008, Mar 17, 2025, 4:02 PM

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drago.7437

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drago.7437

#9 h Mar 18, 2025, 2:23 AM

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JetFire008 wrote:

In $\triangle ABC$ , $BD$ and $CE$ are the bisectors of $\angle B$ , $\angle C$ cutting $CA$ , $AB$ at $D$ , $E$ respectively. If $\angle BDE = 24^{\circ}$ and $\angle CED = 18^{\circ}$ , find the angles of $\triangle ABC$

https://artofproblemsolving.com/community/c6h220396p1222521 here

This post has been edited 1 time. Last edited by drago.7437, Mar 18, 2025, 2:24 AM

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drago.7437

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drago.7437

#10 h Mar 18, 2025, 2:25 AM

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JetFire008 wrote:

Prove that the circumcircles of the four triangles formed by four lines have a common point.

Miquel

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JetFire008

125 posts

JetFire008

#11 h Mar 18, 2025, 12:57 PM

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Let $\triangle ABC$ , $D$ be the midpoint of $BC$ . Prove that
$AB^2+AC^2=2AD^2+2DC^2$ .
Or in other words, prove the Apollonius Theorem.

This post has been edited 1 time. Last edited by JetFire008, Mar 18, 2025, 12:59 PM

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JetFire008

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JetFire008

#12 h Mar 18, 2025, 1:02 PM

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In $\triangle ABC$ , $O$ is the circumcentre and $H$ is the orthocentre. Then, prove that $AH^2+BC^2=4AO^2$ .

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JetFire008

125 posts

JetFire008

#13 h Mar 18, 2025, 1:06 PM

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$P$ and $P'$ are points on the circumcircle of $\triangle ABC$ such that $PP'$ is parallel to $BC$ . Prove that $P'A$ is perpendicular to the Simson line of $P$ .

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JetFire008

125 posts

JetFire008

#14 h Mar 18, 2025, 1:18 PM

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The vertices of a triangle are on three straight lines which diverge from a point, and the sides are in fixed directions; find the locus of the center of the circumscribed circle.
Source:- Problems & Solutions In Euclidean Geometry written by M.N. Aref and William Wernick

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Korean_fish_Kaohsiung

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Korean_fish_Kaohsiung

#15 h Mar 19, 2025, 8:09 AM

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JetFire008 wrote:

Let $P$ be the Fermat point of a $\triangle ABC$ . Prove that the Euler line of the triangles $PAB$ , $PBC$ , $PCA$ are concurrent and the point of concurrence is $G$ , the centroid of $\triangle ABC$ .

If we don't have to show the point is $G$ then the problem is trivial by Liang-Zelich

Click to reveal hidden text

However we will show it is $G$ , consider the midpoint of $AC$ , $PC$ , as $M_B, M_P$ ,let the centroid of $PBC$ be $G_A$ then we know $\dfrac{BM_B}{BG}=\dfrac{BM_P}{BG_A}$ so $GG_A$ is parallel to $M_BM_P$ which is also parallel to $AP$ . now $AP$ meets the point outside of $BC$ , as $P_1$ such that $BP_1C$ is an equilateral triangle, and now by $\dfrac{P_1O_A}{O_AM_A}=\dfrac{M_AG}{GA}$ , where $M_A$ is the midpoint of $BC$ we have $GO_A$ is parallel to $GG_A$ so $G_A$ lies on $GO_A$ therefore it's done

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JetFire008

125 posts

JetFire008

#16 h Mar 20, 2025, 2:03 PM

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Let $ABC$ be an acute triangle whose incircle touches sides $AC$ and $AB$ at $E$ and $F$ , respectively. Let the angle bisectors of $\angle ABC$ and $\angle ACB$ meet $EF$ at $X$ and $Y$ , respectively, and let the midpoint of $BC$ be $Z$ . Show that $XYZ$ is equilateral if and only if $\angle A = 60^{\circ}$ .

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JetFire008

125 posts

JetFire008

#18 h Mar 21, 2025, 12:01 PM

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If from a point $O, OD, OE, OF$ are drawn perpendicular to the sides $BC, CA, AB$ respectively of $\triangle ABC$ then prove that
$BD^2-DC^2+CE^2-EA^2+AF^2-FB^2=0$ Not been able to solve this even though I know I can do this.

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Captainscrubz

65 posts

Captainscrubz

#19 h Apr 13, 2025, 12:54 PM

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JetFire008 wrote:

If the Euler line of a $\triangle ABC$ is parallel to $BC$ , show that tan $B$ tan $C = 3$ .

Bro wants to post every problem in one Topic :stretcher:

but nvm
Let $H$ be the orthocenter of $\triangle ABC$ and let $O$ be the circumcenter
Let the $\perp$ from $H$ and $O$ be $D$ and $M$
see that $HD=OM$
$\implies HD=2RcosCcosB=OM=RcosA$ then just use $cosA=-cos(B+C)$

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Captainscrubz

65 posts

Captainscrubz

#20 h Apr 13, 2025, 12:57 PM

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JetFire008 wrote:

In $\triangle ABC$ , $O$ is the circumcentre and $H$ is the orthocentre. Then, prove that $AH^2+BC^2=4AO^2$ .

Trigonometry bash or simply let $C'$ be the antipode and then $AC'BH$ is a rhombus

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JetFire008

125 posts

JetFire008

#21 h Apr 15, 2025, 4:55 PM

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Captainscrubz wrote:

JetFire008 wrote:

If the Euler line of a $\triangle ABC$ is parallel to $BC$ , show that tan $B$ tan $C = 3$ .

Bro wants to post every problem in one Topic :stretcher:

but nvm
Let $H$ be the orthocenter of $\triangle ABC$ and let $O$ be the circumcenter
Let the $\perp$ from $H$ and $O$ be $D$ and $M$
see that $HD=OM$
$\implies HD=2RcosCcosB=OM=RcosA$ then just use $cosA=-cos(B+C)$

Did it so more people bump to this post

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