Nordic 2025 P3

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24 posts

#1 h Mar 25, 2025, 12:36 PM • 1 Y

Y by MS_asdfgzxcvb

Let $ABC$ be an acute triangle with orthocenter $H$ and circumcenter $O$ . Let $E$ and $F$ be points on the line segments $AC$ and $AB$ respectively such that $AEHF$ is a parallelogram. Prove that $\vert OE \vert = \vert OF \vert$ .

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#2 h Mar 25, 2025, 12:59 PM • 2 Y

Y by MS_asdfgzxcvb, Funcshun840

We prove a generalization. Let $P$ be a point and let $E$ and $F$ be on $AC$ and $AB$ , respectively, such that $AEPF$ is a parallelogram. The isogonal conjugate of $P$ in $AEF$ is $Q$ . We see that $\angle QEF=\angle PEA=\angle PFA=\angle QFE$ , so $QE=QF$ . Now, if $P$ is on $(BHC)$ , then $\angle BPC+\angle EPF=180^\circ-\angle A+\angle A=180^\circ$ , so $P$ has an isogonal conjugate in $BCFE$ . Hence $Q$ is the isogonal conjugate of $P$ in $ABC$ . The problem follows from $P=H$ .

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#4 h Mar 25, 2025, 1:23 PM • 3 Y

Y by amapstob, Funcshun840, Stuffybear

We have $\frac{BF}{FH}=\frac{CE}{EH}$ because of similar triangles. Use $FH=AE$ and $EH=AF$ and rewrite to $FB\cdot FA=EA\cdot EC$ and we are done by PoP.

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19 posts

#5 h Mar 25, 2025, 2:14 PM

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Consider the points in the complex plane with unit circle, lowercase letters etc. as usual.

Claim. $F$ is given by
$f = \frac{(a+b)(b+c)}{(b-c)}+a$ Proof. Denote the point corresponding to $f$ as defined above by $F'$ . We show $F'$ lies on $AB$ and $HF'$ is parallel to $AC$ , so we must have $F'=F$ . Firstly
$f = \frac{(a+b)(b+c)}{(a-b)(b-c)}(a-b) + a$ and $\overline{\frac{(a+b)(b+c)}{(a-b)(b-c)}}=\frac{\left(\frac{a+b}{ab}\right)\left(\frac{b+c}{bc}\right)}{\left(\frac{b-a}{ab}\right)\left(\frac{c-b}{bc}\right)}=\frac{(a+b)(b+c)}{(a-b)(b-c)}$ so $f = m\left(b-a\right)+a$ for some $m\in\mathbb R$ . So $F'$ lies on $AB$ . Now we show $HF'\parallel AC$ . This holds iff
$\frac{h-f}{a-c}=\overline{\frac{h-f}{a-c}}$ We compute these separately:
$\begin{align*} \frac{h-f}{a-c} &= \frac{a+b+c-\frac{(a+b)(b+c)}{(b-c)}-a}{a-c} \\ &= \frac{\left(b+c\right)\left(1-\frac{a+b}{b-c}\right)}{a-c} \\ &= \frac{(b+c)(c+a)}{(b-c)(c-a)} \end{align*}$ and
$\begin{align*} \overline{\frac{h-f}{a-c}} &= \frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{\left(\frac{1}{a}+\frac{1}{b}\right)\left(\frac{1}{b}+\frac{1}{c}\right)}{\frac{1}{b}-\frac{1}{c}}}{\frac{1}{a}-\frac{1}{c}} \\ &= \frac{\frac{b+c}{bc}-\frac{\frac{a+b}{ab}\cdot\frac{b+c}{bc}}{\frac{c-b}{bc}}}{\frac{c-a}{ac}} \\ &= \left(\frac{b+c}{bc}+\frac{a+b}{ab}\cdot\frac{b+c}{bc}\cdot\frac{bc}{b-c}\right)\cdot\frac{ac}{c-a} \\ &= \frac{1}{b(c-a)}\left(a\left(b+c\right)+\frac{c(a+b)(b+c)}{(b-c)}\right) \\ &= \frac{1}{b(c-a)}\left(\frac{ab^2 - ac^2 + c(ab+b^2+ac+bc)}{b-c}\right) \\ &= \frac{(c+a)(b+c)}{(c-a)(b-c)} \\ &= \frac{h-f}{a-c} \end{align*}$ As desired. So $F=F'$ , as desired, so the expression for $f$ as above is correct. $\square$

By the symmetric definitions of $F$ and $E$ w.r.t. $B$ and $C$ we have
$e=\frac{(a+c)(b+c)}{c-b}+a$ Since $O$ is the origin of our complex plane, we need to show $|f|=|e|$ , which is true since:
$\begin{align*} \left|f\right| &= \left|\frac{(a+b)(b+c)+ab-ac}{b-c}\right| \\ &= \frac{1}{\left|b-c\right|}\left|2ab+b^2+bc\right| \\ &= \frac{1}{\left|b-c\right|}\left|2a+b+c\right| \\ &= \frac{1}{|b-c|}\left|ac+ac+bc+c^2 + ab - ab\right| \\ &= \frac{1}{|b-c|}\left|(a+c)(b+c)+ac-ab\right| \\ &= \left|\frac{(a+c)(b+c)}{(b-c)}+a\right| \\ &= |e| \end{align*}$ as desired, so we are done. $\blacksquare$

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DottedCaculator

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DottedCaculator

#6 h Mar 25, 2025, 2:32 PM • 1 Y

Y by MS_asdfgzxcvb

Let $f_A=b^2+c^2-a^2$ . Then, $H=\left(\frac1{f_A}:\frac1{f_B}:\frac1{f_C}\right)$ , so $\frac{AE\cdot EC}{b^2}=\frac1{f_Af_C}+\frac1{f_Bf_C}=\frac{2c^2}{f_Af_Bf_C}$ . Therefore, $AE\cdot EC=\frac{2b^2c^2}{f_Af_Bf_C}=AF\cdot FB$ , which means $OE=OF$ .

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#7 h Mar 25, 2025, 2:57 PM • 1 Y

Y by MS_asdfgzxcvb

Let $E', F'$ be the feet from $B, C$ to $AC, AB$ , and let $D' = E'F' \cap BC$ . Since
$(B, C; D, D') = -1 = A(F, E; N, \infty_{EF})$ it follows that $AD' \parallel EF$ . Since we are done if $ON \perp EF$ , if $A'$ is the $A$ -antipode, scaling by two at $A$ means we want to show that $A'H \perp AD'$ . However, these two lines concur at the $A$ -queue point, which lies on $(ABC)$ so we are done.

This post has been edited 1 time. Last edited by YaoAOPS, Mar 25, 2025, 2:57 PM

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#8 h Mar 29, 2025, 6:30 PM

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Call $M$ the midpoint of $AH$ . Since $AFHE$ is a parallelogram, $M$ is also the midpoint of $EF$ . Define $X$ as the $A$ -antipode in $(ABC)$ . It is well known that $HBXC$ is a parallelogram, and since $\angle BHC = \angle BXC = \pi - \angle A$
$\Longrightarrow$ parallelogram $AFHE$ is similar to $HBXC$ .
In particular this implies $\Delta HXC \sim \Delta AEF$ , and since $\angle CHE = \pi - (\angle A) - (\frac{\pi}{2} - \angle A) = \frac{\pi}{2}$ , there is a rotation by $\frac{\pi}{2}$ centered at $H$ that sends $X\mapsto X’$ and $C\mapsto C’$ . Thus, $HX’ \parallel EF$ and rotating back gives $HX \perp EF$ . A homothety cantered at $A$ with factor $\frac{1}{2}$ sends $H\mapsto M$ and $X\mapsto O$ . Since $HX\perp EF\Longrightarrow OM\perp EF$ . Thus $O$ is on the perpendicular bisector of $EF$ which implies $\vert OE \vert = \vert OF \vert$ , as required. $\blacksquare$

This post has been edited 2 times. Last edited by anirbanbz, Mar 29, 2025, 6:32 PM

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Primeniyazidayi

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Primeniyazidayi

#9 h Mar 29, 2025, 7:49 PM

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For a further exercise,see https://artofproblemsolving.com/community/c6h288839p1561572

This post has been edited 1 time. Last edited by Primeniyazidayi, Mar 29, 2025, 7:50 PM

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#10 h Apr 1, 2025, 11:36 AM

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