279. O.M. Holland , O.M. Ireland 1988 and TST USA 2000, etc.

by Virgil Nicula, May 21, 2011, 2:19 AM

PP1 (Holland). Let $ABC$ be a triangle with $AB\perp AC$ . Denote $\left\{\begin{array}{c}
D\in BC\ ,\ AD\perp BC\\\
E\in (CD)\ ,\ EC=ED\end{array}\right\|$ and the symmetrical $F$ of $A$ w.r.t. $B$ . Prove that $FD\perp AE$ .

Proof 1. Denote the symmetrical point $K$ of the point $A$ w.r.t. $D$ and $L\in FD\cap AE$ . Since $\triangle AKF$ is $K$-right angled and

$\triangle AKF\sim \triangle CDA$ results that the median $FD$ from $\triangle AKF$ is homologously in similitude with the median $AE$ from

$\triangle CDA\Longrightarrow \widehat{DAE}\equiv\widehat{KFD}$ , i.e. $\widehat{KAL}\equiv\widehat{KFL}\Longleftrightarrow AFKL$ is cyclic $\iff\widehat{ALF}\equiv\widehat{AKF}\iff FD\perp AE$ .

Proof 2. Denote the midpoint $G$ of $[AD]$ . Observe that $GE\parallel AC\perp AB\implies$ $GE$ is $E$-altitude in

$\triangle ABE\implies$ the point $G$ is the orthocenter of $\triangle ABE\implies$ $FD\parallel BG\perp AE\implies$ $FD\perp AE$ .

Proof 3. Denote the points $K\in AD$ and $M\in AE$ so that $KF\parallel MF\parallel BC$ . Observe that $ACMD$ is a parallelogram $\implies$

$DM\parallel AC\implies DM\perp AF$ , i.e. $D$ is the orthocenter of $\triangle AFM$ $\implies$ $FD\perp AM\iff FD\perp AE$ .

Proof 4. Denote $\left\{\begin{array}{c}
\phi=m\left(\widehat{DAE}\right)\\\
\psi=m\left(\widehat{FDB}\right)\end{array}\right\|$ . Observe that $\tan\phi =\frac {DE}{DA}$ , i.e. $\boxed{\tan\phi =\frac {DC}{2\cdot DA}}\ (1)$ . From the well-known relation $\frac {FB}{FA}=\frac {DB}{DA}\cdot\frac {\sin\widehat{FDB}}{\sin\widehat{FDA}}\iff$

$\frac 12=\frac {DB}{DA}\cdot\tan\psi$ , i.e. $\boxed{\tan\psi =\frac {DA}{2\cdot DB}}\ (2)$ . Since $DA^2=DB\cdot DC$ from $(1)$ and $(2)$ obtain that $\tan\phi =\tan\psi\iff$ $\phi =\psi$ , i.e. $FD\perp AE$ .


An easy extension. Let $ABC$ be an $A$-right angled triangle. Denote the point $D\in BC$ for which $AD\perp BC$ .

Consider the points $E\in (DC)$ and $F\in AB$ so that $B\in (AF)$ and $\frac {FB}{FA}=\frac {DE}{DC}$ . Prove that $FD\perp AE$ .


Proof 1. Denote $\left\{\begin{array}{c}
\phi=m\left(\widehat{DAE}\right)\\\
\psi=m\left(\widehat{FDB}\right)\end{array}\right\|$ . Observe that $\boxed{\tan\phi =\frac {DE}{DA}}\ (1)$ . From the well-known relation $\frac {FB}{FA}=\frac {DB}{DA}\cdot\frac {\sin\widehat{FDB}}{\sin\widehat{FDA}}$

obtain that $\frac {FB}{FA}=\frac {DB}{DA}\cdot\tan\psi$ . Since $\frac {DA}{DB}=\frac {DC}{DA}$ obtain that $\boxed{\tan\psi =\frac {FB}{FA}\cdot\frac {DC}{DA}}\ (2)$ . From the relations $(1)$ and $(2)$

obtain that $FD\perp AE\iff$ $\psi =\phi\iff$ $\tan\psi =\tan\phi\iff$ $\frac {FB}{FA}\cdot\frac {DC}{DA}=\frac {DE}{DA}\iff$ $\frac {FB}{FA}=\frac {DE}{DC}$ , what is truly.

Proof 2. Let $H\in CF$ and $K\in AC$ for which $E\in HK$ and $HK\parallel AD$ . So $\frac {DE}{DC}=\frac {FB}{FA}=\frac {HE}{HK}\implies$ $\frac {DE}{DC}=\frac {HE}{HK}\implies$ $HD \parallel AC$ .

Denote $Q\in DH \cap AE$ . Then $\frac{QE}{QA}=\frac{DE}{DC}=\frac{FB}{FA}$ $\Longrightarrow QF \parallel BC$ . In $\triangle AFQ$ we have $QD \perp AF$ and $AD \perp QF$ $\Longrightarrow FD \perp AE$ .



PP2 (easy - Holland). Let $ABC$ be an $A$-isosceles triangle. Consider two points $P\in (AB)$ and $Q\in (AC)$

so that $AP=CQ$ . Prove that the circumcenter of $\triangle ABC$ belongs to the circumcircle of $\triangle PAQ$ .


Proof. Denote the circumcircle $w=C(O,R)$ of $\triangle ABC$ . Observe that $\triangle APO\equiv\triangle CQO\implies$

$\widehat{APO}\equiv\widehat{CQO}$ , i.e. the circumcenter $O$ of the triangle $ABC$ belongs to the circumcircle of $\triangle PAQ$ .



PP3 (Ireland - 1988). Let $P$ be a point of the small arc $CD$ in the circumircle of the square $ABCD$ . Prove that $PA^2+PA\cdot PC=PB^2+PB\cdot PD\ (*)$ .

Proof 1 (metric). Apply Ptolemy's theorem to $\left\{\begin{array}{cc}
CBAP\ : & AB\cdot (PA+PC)=PB\cdot AC\\\\\
DABP\ : & AB\cdot (PB+PD)=PA\cdot BD\end{array}\right\|\ \implies$

$\left\{\begin{array}{cc}
PA+PC=PB\cdot\sqrt 2 & \odot\ PA\\\\\
PB+PD=PA\cdot\sqrt 2 & \odot\ PB\end{array}\right\|\ \implies$ $PA\cdot (PA+PC)=PB\cdot (PB+PD)=PA\cdot PB\cdot\sqrt 2$ .

Proof 2 (trigonometric). Denote $\left\{\begin{array}{c}
m(\widehat{PDC})=x\\\\
m(\widehat{PCD})=y\end{array}\right\|$ . Observe that $x+y=45$ (degrees) and the required relation $(*)$ is equivalently with the trigonometrical relation

$\mathrm{LHS}\equiv \sin^2(45+y)+\sin(45+y)\sin x=$ $\sin^2(45+x)+\sin(45+x)\sin y\equiv\mathrm{RHS}$ . Since $(45+x)+(90+y)=180=(45+y)+(90+x)$

prove easily that $\mathrm{LHS}=\frac 12\cdot (1+\cos 2x+\sin 2x)$ and $\mathrm{RHS}=\frac 12\cdot (1+\cos 2y+\sin 2y)$ , i.e. $\mathrm{LHS}=\mathrm{RHS}=\sqrt 2\cdot \cos x\cos y=\sqrt 2\cdot PA\cdot PB$ .


An easy extension. Let $P$ be a point of the small arc $CD$ in the circumircle of the isosceles trapezoid

$ABCD$ with $AB\parallel CD$ and $AB=AD$ . Prove that $PA^2+PA\cdot PC=PB^2+PB\cdot PD\ (*)$ .


Proof (metric). Apply Ptolemy's theorem to $\left\{\begin{array}{cc}
CBAP\ : & AB\cdot (PA+PC)=PB\cdot AC\\\\\
DABP\ : & AB\cdot (PB+PD)=PA\cdot BD\end{array}\right\|\ \implies$

$\left\{\begin{array}{cc}
PA+PC=PB\cdot\frac {AC}{AB} & \odot\ PA\\\\\
PB+PD=PA\cdot\frac {BD}{AB} & \odot\ PB\end{array}\right\|\ \implies$ $PA\cdot (PA+PC)=PB\cdot (PB+PD)=PA\cdot PB\cdot\frac {AC}{AB}$ .



A similar problem. Let $ABC$ be an $A$-isosceles triangle with the circumcircle $w$ . Consider two mobile points $\{P,Q\}\subset w$

so that the sideline $BC$ separates $P$ , $Q$ and doesn't separate $P$ , $A$ . Prove that the ratio $\frac {[BPCQ]}{AQ^2-AP^2}$ is constant.


Proof. Denote $A'\in w$ so that $[AA']$ is a diameter of $w$ . Construct $\{R,S\}\subset w$ so that $AR\parallel PB$ , $A'S\parallel QB$ . Apply the Ptolemy's theorem

to the quadrilaterals $\left\{\begin{array}{cccc}
BRAP & \implies & AB^2=AP^2+PB\cdot PC & (1)\\\
BSA'Q & \implies & A'B^2=A'Q^2+QB\cdot QC & (2)\end{array}\right\|$ . Observe that the relation $(2)\iff$ $4R^2-AB^2=$

$4R^2-AQ^2+QB\cdot QC\iff$ $AB^2=AQ^2-QB\cdot QC\ (3)$ . Thus, $(1)\ \wedge\ (3)\implies \boxed{PB\cdot PC+QB\cdot QC=AQ^2-AP^2}$

and $\frac {[BPCQ]}{AQ^2-AP^2}=$ $\frac {[BPC]+[BQC]}{AQ^2-AP^2}=$ $\frac {(PB\cdot PC+QB\cdot QC)\cdot\sin A}{2\cdot \left(AQ^2-AP^2\right)}=$ $\frac 12\cdot \sin A$ (constant).


PP4 - Turkey TST 2011. Let $\triangle ABC$ with incenter $I$ and circumcircle $w$ . Denote diameter $[AD]$ of $w$ and $\left\{\begin{array}{cc}
E\in (AC\ ,\ AE=s-b\\\\
F\in (AC\ ,\ AF=s-c\end{array}\right\|$ . Show that $EF\perp DI$ .

Proof. Observe that $AD=2R\ ,\ b=$ $2R\sin B\ ,\ c=2R\sin C$ and $AI\cdot\cos\frac A2=s-a$. Apply the Pythagoras' relation in the mentioned triangles :

$\triangle ADE\ :\ DE^2=AD^2+AE^2-2\cdot AD\cdot AE\cdot\sin B $ $\iff DE^2=4R^2+(s-b)^2-2b(s-b)$

$\triangle ADF\ :\ DF^2=AD^2+AF^2-2\cdot AD\cdot AF\sin C $ $\iff DF^2=4R^2+(s-c)^2-2c(s-c)$

$\triangle AIE\ :\ IE^2=AI^2+AE^2-2\cdot AI\cdot AE\cdot\cos\frac A2 $ $\iff IE^2=AI^2+(s-b)^2-2(s-a)(s-b)$

$\triangle AIF\ :\ IF^2=AI^2+AF^2-2\cdot AI\cdot AF\cos\frac A2 $ $\iff IF^2=AI^2+(s-c)^2-2(s-a)(s-c)$

In conclusion, $DE^2-DF^2=IE^2-IF^2=(b-c)(b+c-2a)\implies DI\perp EF$ .



PP5 - USA TST 2000. Let $ ABCD$ be a cyclic quadrilateral and let $ E$ and $ F$ be the feet of perpendiculars from the intersection of diagonals

$ AC$ and $ BD$ to $ AB$ and $ CD$ respectively. Prove that $ EF$ is perpendicular to the line through the midpoints of $ AD$ and $ BC$ .

Lemma. Construct outside of $\triangle ABC$ the right triangles $\left\{\begin{array}{c}
ACE\ :\ EA\perp EC\\\
ABF\ :\ FA\perp FB\\\
\widehat {ACE}\equiv\widehat {ABF}\end{array}\right\|$ . Denote the midpoint $M$ of $[BC]$ , Prove that $ME=MF$ .


Proof 1 (synthetic)

Proof 2 (metric)

Proof. Denote $I\in AC\cap BD$ and the midpoints $M$ , $N$ of $[AD]$ , $[CD]$ respectively. Apply above lemma to $\triangle AID$ and its exterior right $\triangle IAE$ , $\triangle IDF$ .

Obtain $ME=MF$ . Apply again above lemma to $\triangle BIC$ and its exterior right $\triangle IEB$ , $\triangle IFC$ . Obtain $NE=NF$ .In conclusion, $MN\perp EF$ .

Another proof. Let $M$ , $N$ be the midpoints of $AD$ , $BC$ respectively. Let $P$ be the intersection of diagonals. Let $Q$ , $R$ be the reflections of $P$ over $AB$ , $CD$

respectively. We know that $AQBP$ and $DPCR$ are directly similar and are kites. Then the midpoints of the lines connecting corresponding points of the two

similar figures forms a figure similar to the original, i.e. $MENF$ is similar to both $AQBP$ and $DPCR$ . Therefore $MENF$ is a kite and $EF\perp MN$ .



PP6 - Polosh & Costa Rica 2006. Let $ABC$ be an acute triangle. Its incircle $w=C(I,r)$ touches it at the points $D\in BC$ , $E\in CA$ , $F\in AB$ . Denote:

$\blacktriangleright$ the centroid $G$ of the triangle $ABC$ ;

$\blacktriangleright$ the A-exincircle $w_a=C(I_a,r_a)$ , the B-exincircle $w_b=C(I_b,r_b)$ , the C-exincircle $w_c=C(I_c,r_c)$ ;

$\blacktriangleright$ the Nagel's point $N\in AD_1\cap BE_1\cap CF_1$ ; $n_a=AD_1$ , $n_b=BE_1$ , $n_c=CF_1$ ;

$\blacktriangleright$ the points $D_1\in BC\cap w_a$, $E_1\in CA\cap w_b$ , $F_1\in AB\cap w_c$ ;

$\blacktriangleright$ the reflections $D'$ , $E'$ , $F'$ of the points $D$ , $E$ , $F$ with respect to the point $I$ .

Lemma 1. $IG\perp BC\Longleftrightarrow b=c\ \ \vee\ \ b+c=3a$ .


Proof. $IG\perp BC$ $\Longleftrightarrow$ $IB^2-IC^2=GB^2-GC^2$ $\Longleftrightarrow$ $\frac{ac(p-b)}{p}-\frac{ab(p-c)}{p}=$

$\frac 19\left\{\left[2\left(a^2+c^2\right)-b^2\right]-\left[2\left(a^2+b^2\right)-c^2\right]\right\}$ $\Longleftrightarrow$ $3a(c-b)=c^2-b^2$ $\Longleftrightarrow$ $b=c\ \ \vee\ \ b+c=3a$ .


Remark. $N\in (IG)$ and $GN=2\cdot GI$ . Therefore, $IN\perp BC$ $\Longleftrightarrow$ $b+c=3a$ .

Lemma 2. $\boxed {a n^2_a+a(p-b)(p-c)=p(b^2+c^2)-bc(b+c)}\ \ (1)$ ;

$\frac{NA}{a}=\frac{ND_1}{p-a}=$ $\frac{n_a}{p}\ \ (2)$ ; $D'\in AD_1$ and $\frac{D'A}{p-a}=\frac{D'D_1}{a}=\frac{n_a}{p}\ \ (3)$ a.s.o.


Proof. Apply the Stewart's relation for the Nagel-ray $[AD_1$ in the triangle $ABC$ and obtain the relation $(1)$ .

Apply the van Aubel's relation $\frac{NA}{ND_1}=\frac{F_1A}{F_1B}+\frac{E_1A}{E_1C}$ and obtain the relation $(2)$ . $D'\in AD_1\cap BC$ $\Longleftrightarrow$

$\frac{D_1D'}{D_1A}=\frac{2r}{h_a}$ $\Longleftrightarrow$ $\frac{D_1D'}{D_1A}=\frac ap$ $\Longleftrightarrow$ $\frac{D_1D'}{a}=\frac{D_1A}{p}=$ $\frac{D'A}{p-a}=\frac{n_a}{p}$, i.e. the relations $(2)$ .

Remark. $\frac{AD'}{p-a}=\frac{ND_1}{p-a}=\boxed {\frac{ND'}{2a-p}=\frac{NA}{a}=\frac{n_a}{p}}\ \ (4)$.


An equivalent and complete enunciation of the proposed problem. Let $ABC$ be an acute triangle for which $b\ne c$ . Then

$B,C,E',F'$ are cyclically $\Longleftrightarrow$ $b+c=3a$ $\Longleftrightarrow$ $IN\perp BC$ $\Longleftrightarrow$ $N\equiv D'$ and in this case $I,B,C,E',F'$ belong to

the circle $C(M,MI)$ , where the point $M$ is the second intersection between the line $AI$ and the circumcircle of the triangle $ABC$ .


Proof (indication). The points $B,C,E',F'$ are cyclically $\Longleftrightarrow$ $NE'\cdot NB=NF'\cdot NC$ a.s.o.

Example. $a=3$, $b=4$, $c=5$.
Remark. $N\in w$ $\Longleftrightarrow$ $(b+c-3a)(c+a-3b)(a+b-3c)=0$ $\Longleftrightarrow$$N\in \{D',E',F'\}$ $\Longleftrightarrow$ $p^2+4r^2=16Rr$ .


PP7 - 32nd BMO 1996. Let $\{a, b, c, d\}\subset\mathbb R_+^*$ for which $\left\|\begin{array}{c}
s_1=a=b+c+d=12\\\
s_2=ab+ac+ad+bc+bd+cd\\\
s_4=abcd=27+s_2\end{array}\right\|\ \implies\ a=b=c=d=3$ .

Proof. $s_2\ge 6\sqrt {abcd}\implies$ $s_4=27+s_2\ge 27+6\sqrt {s_4}\implies$ $\sqrt{s_4}\ge 9\implies$ $4\sqrt[4]{s_4}\ge 12=s_1\ge 4\sqrt[4]{s_4}\implies$ $\frac {s_1}{4}=\sqrt[4]{s_4}\implies$ $a=b=c=d=3$ .


PP8 - 8th China Southeast MO. In $\triangle ABC$ , for an interior $P$ denote $\left\{\begin{array}{c}
D\in AP\cap BC\\\
E\in BP\cap CA\\\
F\in CP\cap AB\end{array}\right\|$ and $\left\{\begin{array}{cc}
 Y_a\in AC\ , & DY_a\parallel AB\\\
Z_a\in AB\ , & DZ_a\parallel AC\end{array}\right\|\ ,$

$X\in BC\cap Y_aZ_a$ . Construct similarly the points $Y\in CA\cap Z_bX_b$ and $Z\in AB\cap X_cY_c$ . Prove that $X\in YZ$ .


Proof. Denote $\frac {DB}{DC}=\frac zy$ , $\frac {EC}{EA}=\frac xz$ and $\frac{FA}{FB}=\frac yx$ . Suppose w.l.o.g. $x+y+z=1$ . Obtain that $\frac {Y_aA}{Y_aC}=\frac {z}{xy}$ and $\frac {Z_aA}{Z_aB}=\frac {y}{xz}$ . Apply Menelaus' teorem

to transversal $\overline{XZ_aY_a}\ :\ \frac {XB}{XC}\cdot\frac {Y_aC}{Y_aA}\cdot\frac {Z_aA}{Z_aB}=1\implies$ $\frac {XB}{XC}=\left(\frac zy\right)^2$ . Analogously $\frac {YC}{YA}=\left(\frac xz\right)^2$ and $\frac {ZA}{ZB}=\left(\frac yx\right)^2$ . Hence $X\in YZ$ .



PP9 (Iran TST 2007, Day 2). Let $\omega$ be the incircle of the triangle $ABC$. Let $P\in AB$ and $Q\in AC$ be two points such that $PQ\parallel BC$ and $PQ$ is tangent to the circle $\omega$. The

lines $AB$, $AC$ touch the circle $\omega$ at $F$, $E$ respectively. Denote the middlepoint $M$ of $[PQ]$ and $T\in EF\cap BC$. Prove that the line $TM$ is tangent to the circle $\omega$ (Ali Khezeli)


Proof. Denote the intersection $N\in IM\cap BC$ and the middlepoints $S$, $V$ of the segments $[AD]$, $[BC]$ respectively. Observe that $M\in AV$, $IM=IN$. From the well-known properties

$S\in IV$ and $TI\perp AD$ obtain immediatelly that $MN\parallel AD$, i.e. $TI\perp MN$ and the line $TM$ is tangent to the circle $w$ because the lines $TM$, $TN$ are symmetrically w.r.t the line $TI$.

Proof of the property - the point S belongs to the line IV.

Proof of the property - the lines TI and AD are perpendicularly.
This post has been edited 110 times. Last edited by Virgil Nicula, Nov 22, 2015, 6:55 AM

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