432. Geometry problems with perpendicularities II.

by Virgil Nicula, Nov 16, 2015, 8:41 AM

PP1. Let a rectangle $ABCD$ . Let the midpoint $P$ of $[AB]$ P and $Q\in PD$ so that $CQ\perp PD $ . Prove that $\triangle BQC$ is isosceles. An easy extension. Let $ABCD$ be a

trapezoid so that $AD\parallel BC$ and $AB\perp AD$ . Denote the points $I\in AC\cap BD$ , $P\in AB$ so that $IP\parallel AD$ and $Q\in PD$ so that $CQ\perp PD$ . Prove that $BQ=BC$ .

Proof. Prove easily that $\frac {BP}{AP}=\frac {BC}{AD}$ . Observe that $BCQP$ is a cyclical quadrilateral and $\triangle BCP\sim\triangle ADP\implies$ $\widehat {BPC}\equiv\widehat{APD}\implies$ $\widehat {BQC}\equiv\widehat{BCQ}\implies$ $BQ=BC$ .

Generalization. Let $\{A,B,M,N\}$ be collinear points so that $M\in (AB)$ , $B\in (MN)$ and $\frac {MA}{MB}=\frac{NA}{NB}$ , i.e. the division $(A,B;M,N)$

is harmonically. For $P\not\in AB$ so that $PM\perp PN$ let the projections $R$ , $S$ of $B$ on the lines $PA$ , $PN$ respectively. Prove that $SR=SB$ .

Application. Let a circle $w=C(O,R)$ and $P$ so that $PO>R$ . Consider the tangent $PT$ to $w$ , where $T\in w$ and $\{A,B\}\subset w$ so that

$P\in AB$ . Let $\odot$ $\begin{array}{ccccc} \nearrow & M\in AB & ; & TM\perp OP & \searrow\\\\ \searrow & N\in PT & ; & MN\perp PT & \nearrow\end{array}$ $\odot$ and the projections $R$ , $S$ of $B$ on $NA$ , $PT$ respectively. Prove that $SR=SB$ .


PP2. Let $\triangle ABC$ with circumcircle $w=C(O)$ , $D$ so that $\triangle DAB\sim\triangle DCA$ , $S\in AD\cap BC$

and $\{A,E\}=\{A,D\}\cap w$ . Prove that the ray $[AS$ is the $A$-symmedian of $\triangle ABC$ and $DA=DE$ .

Proof 1. $\triangle DAB\sim\triangle DCA$ $\iff$ $\frac {DA}{DC}=$ $\frac {AB}{CA}=$ $\frac {BD}{AD}$ $\iff$ $\left\{\begin{array}{c} DB=\frac cb\cdot DA\\\\ DC=\frac bc\cdot DA\end{array}\right|$ $\implies$ $\boxed{\frac {DB}{DC}=\left(\frac cb\right)^2}\ (*)$ . Since $\widehat{SDB}\equiv$ $\widehat{SDC}\equiv$ $\widehat{BAC}$ obtain that

$\frac {SB}{SC}=$ $\frac {DB}{DC}$ $\stackrel{(*)}{\iff}$ $\frac {SB}{SC}=$ $\left(\frac cb\right)^2$ i.e. the ray $[AS$ is the $A$-symmedian of $\triangle ABC$ . Denote $T\in BB\cap CC$ , where denoted $XX$ - the tangent line to the circle

$w$ , where $X\in w$ . Thus, $m\left(\widehat{BDC}\right)=$ $m\left(\widehat{BOC}\right)=$ $2A\implies$ the point $D$ belongs to the circle with the diameter $[OT]$ , i.e. $DA\perp DO$ $\iff$ $DA=DE$ .

Proof 2 (Sunken rock). From $\angle BAD=\angle ACD$ and $\angle DAC=\angle BDA$ we get $\angle BDE=\angle CDE=\angle BAC\ (\ 1\ )$ and obviously, $\Delta BAD\sim\Delta ACD\ (\ 2\ )$ which implies

$AD^2=BD\cdot CD\ (\ 3\ )$ . If we prove $\frac{BD}{AD}=\frac{AB}{AC}\ (\ *\ )$ we are done. From $(2)$ we easily get $\frac{AB^2}{AC^2}=\frac{BD}{CD}$; with $(3)$ we get $\frac{AB^2}{AC^2}=\frac{BD^2}{AD^2}\iff\frac{AB}{AC}=\frac{BD}{AD}\ (\ 4\ )$ .

But with $(1)$ we have $\Delta BDE\sim\Delta BAC\implies\frac{BD}{AB}=\frac{DE}{AC}$ ; with $(4)$ we got the wanted $AD=DE$ , $BD$ symmedian, meaning $ABEC$ is an harmonic quadrilateral.

Remark. The division $(A,E;S,T)$ is a harmonical division and $DA^2=DS\cdot DT$ .


PP3. Let $XYZ$ be a right triangle for which $XY\perp XZ$. Denote the projections $U$, $V$ of the points $Y$, $Z$ respectively to a mobile

line $d$ for which $X\in d$. Define the point $W$ for which $WU\perp XY$ and $WV\perp XZ$. Ascertain the geometrical locus of the point $W$.

Proof. Denote the projections $A$ , $B$ of $U$ , $V$ on $XY$ , $XZ$ respectively and $M\in UA\cap YZ$ , $N\in VB\cap YZ$ . Thus,

$\triangle UXY\sim\triangle VZX\implies$ $\frac{MY}{YZ}=$ $\frac {AY}{XY}=$ $\frac {BZ}{XZ}=$ $\frac {NZ}{YZ}\implies$ $MY=NZ\implies$ $M\equiv N\equiv W\implies$ $W\in BC$ .


PP4. In $\triangle ABC$ altitude, angle bisector and median drawn from $A$ cut angle $A$ into four equal angles. Compute the angles of the triangle.

Proof 1. Let the midpoint $M$ of $[BC]$ , $\left\{\begin{array}{ccc} D\in BC & ; & AD\perp BC\\\\ L\in BC & ; & \widehat{LAB}\equiv\widehat{LAC}\end{array}\right\|$ , the circumcircle $w$ of $\triangle ABC$ and $\{A,S\}=\{A,S\}\cap w$ . Thus, $BD\parallel MS\implies$ $\triangle AMS$ is

$M$-isosceles $\implies$ $M$ belongs to the bisectors of $[AS]$ and $[AC]$ , i.e. $M\equiv O$ - the circumcenter of $\triangle ASC\implies$ $\triangle AOC$ is $O$-isosceles $\implies A=\frac {\pi}{2}$ , $C=\frac {\pi}{8}$ and $B=\frac {3\pi}{8}$ .

Proof 2. Denote $\left\{\begin{array}{ccc} A & = & 4x\\\\ B & = & 90^{\circ}-x\\\\ C & = & 90^{\circ}-3x\end{array}\right|$ . Thus, $AB=AL=c$ , $DB=DL=\frac 12\cdot LB=$ $\frac {ac}{2(b+c)}$ and $ML=\frac {a(b-c)}{2(b+c)}$ . Therefore, $\frac cb=$ $\frac {AL}{AC}=$ $\frac {ML}{MC}=$ $\frac {b-c}{b+c}\implies$

$b^2-2bc-c^2=0$ $\implies$ $\boxed{\frac bc=1+\sqrt 2}\ (*)$ . In conclusion, $\cos\frac A2=\cos \widehat{DAM}=$ $\frac {AD}{AM}=$ $\frac {LD}{LM}=$ $\frac {c}{b-c}\stackrel{(*)}{\implies}$ $\cos \frac A2=\frac {\sqrt 2}{2}\implies$ $A=\frac {\pi}{2}$ , $C=\frac {\pi}{8}$ and $B=\frac {3\pi}{8}$ .


PP5. Let $A$-right $\triangle ABC$ . Let the rectangle $MNPQ$ , where $\{M,N\}\subset (BC)$ , $P\in (AC)$ and $Q\in (AB)$ and $E\in BP\cap MQ$ and $F\in CQ\cap NP$ . Prove that $\widehat {BAE}\equiv\widehat {CAF}$ .

Proof. $MQAC$ and $NPAB$ are cyclically$\implies$ $\left\{\begin{array}{ccc} BM\cdot BC=BQ\cdot BA\\\ CN\cdot CB=CP\cdot CA\end{array}\right|\ (\div )\ \implies$ $\frac {BM}{CN}=$ $\frac {BQ}{CP}\cdot \frac {AB}{AC}$ $\implies$ $\boxed{\frac {BM}{CN}=\frac {AQ}{AP}\cdot \frac {AB}{AC}}\ (*)$ . Let $\left\{\begin{array}{c} m\left(\widehat{EAB}\right)=x\\\\ m\left(\widehat{FAC}\right)=y\end{array}\right|$ .

Thus, $\left\{\begin{array}{ccc} \frac {EB}{EP}=\frac {AB}{AP}\cdot\frac {\sin\widehat{EAB}}{\sin\widehat{EAP}} & \implies & \frac {MB}{MN}=\frac {AB}{AP}\cdot\tan x\\\\ \frac {FC}{FQ}=\frac {AC}{AQ}\cdot\frac {\sin\widehat{FAC}}{\sin\widehat{FAQ}} & \implies & \frac {NC}{NM}=\frac {AC}{AQ}\cdot\tan y\end{array}\right|\ (\div )\ \implies$ $\frac {MB}{NC}=$ $\frac {AQ}{AP}\cdot\frac {AB}{AC}\cdot \frac {\tan x}{\tan y}$ $\stackrel{(*)}{\implies}$ $\tan x=\tan y\implies$ $x=y\implies$ $\widehat {BAE}\equiv\widehat {CAF}$ .


PP6. Let $ABC$ be an acute triangle with the circumcenter $O$ . Denote $K\in AO\cap BC$ and $L\in (AB)$ , $M\in (AC)$ so that $KL=KB$ , $KM=KC$ . Prove that $LM\parallel BC$ .

Proof 1. $\frac {KB}{KC}=\frac {AB}{AC}\cdot\frac {\sin \widehat{KAB}}{\sin \widehat{KAC}}=$ $\frac cb\cdot \frac {\cos C}{\cos B}=\frac {2R\sin C\cos C}{2R\sin B\cos B}\implies$ $\boxed{\frac {KB}{KC}=\frac {\sin 2C}{\sin 2B}}\ (*)$ . Denote the distance $\delta_d(X)$

of the point $X$ to the line $d$ . Thus, $\left\{\begin{array}{c} \delta_{BC}(L)=KB\cdot \sin 2B\\\\ \delta_{BC}(M)=KC\sin 2C\end{array}\right|\stackrel{(*)}{\implies}$ $\delta_{BC}(L)=\delta_{BC}(M)\implies ML\parallel BC$ .

Proof 2. Denote $D\in BC$ so that $AD\perp BC$ and the midpoints $U$ , $V$ of $[BL]$ , $[CM]$ respectively. Therefore,

$\left\{\begin{array}{ccccc} \triangle AKV\sim\triangle ABD & \implies & \frac {AK}{AB}=\frac {AV}{AD} & \implies & \frac {AV}{AC}=\frac {AK\cdot AD}{AB\cdot AC}\\\\ \triangle AKU\sim\triangle ACD & \implies & \frac {AK}{AC}=\frac {AU}{AD} & \implies & \frac {AU}{AB}=\frac {AK\cdot AD}{AB\cdot AC}\end{array}\right|\implies$ $\frac {AV}{AC}=\frac {AU}{AB}\implies UV\parallel BC\implies LM\parallel BC$ .


PP7 (extension of China TST 1986). Let $ ABCD$ be a quadrilateral for which $ AB=AD\ ,\ CB=CD$ and $ B=D=90^{\circ}$ (a right deltoid).

Let $ P\in (AB)\ ,\ Q\in (AD)$ be two points so that $ PQ=PB+QD$ and $ m(\widehat{PAQ})=2\phi$ . Show that $ m(\widehat{PCQ})=90^{\circ}-\phi$ .

Proof. Observe that the semiperimeter of the triangle $ APQ$ is equal to $ AB\ ,\ CB\perp AP$ and $ C$ belongs to the bisector of the angle $ \widehat{PAQ}$ . Therefore,

the point $ C$ is the $ A$- exincenter of $\triangle APQ$ . Generally, $ m(\widehat{PCQ})=\frac{1}{2}\cdot \left(m(\widehat{APQ})+m(\widehat{AQP})\right)$ , i.e. $ \boxed{m(\widehat{PCQ})=90^{\circ}-\phi}$ .


PP8. Let $ABC$ be an acute triangle with the circumcircle $w=C(O)$ and $AH\perp BC$ , where $H\in BC$ . Denote the projections $K$ and

$L$ of $H$ on $AB$ and $AC$ respectively. Define $\{P,Q\}=KL\cap w$ and $\{A,T\}=AH\cap w$ . Prove that $H$ is the incenter of $\triangle PQT$ .

Proof. $APHL$ is cyclically $\implies$ $\widehat{AKL}\equiv\widehat{AHL}\equiv$ $\widehat{ACH}\equiv\widehat{ACB}$ $\implies$ $\widehat{AKL}\equiv\widehat{ACB}$ $\implies$ $BKLC$ is cyclically. Thus, the line $d\equiv \overline{PKLQ}$ (in this order) is an antiparallel

to $BC$ in $\triangle ABC$ $\implies $ $AO\perp d$ $\implies $ $AP=AQ$ $\implies$ $\widehat{ABP}\equiv$ $\widehat {AQP}\equiv$ $\widehat{APQ}\equiv$ $\widehat{ACQ}$ . Therefore, $\left\{\begin{array}{ccc} \widehat{ABP}\equiv\widehat{APQ} & \implies & AP^2=AK\cdot AB=AH^2\\\\ \widehat{ACQ}\equiv\widehat{AQP} & \implies & AQ^2=AL\cdot AC=AH^2\end{array}\right|$ $\implies$

$AP=AH=AQ=\rho$ $\iff H$ belongs to the intersection of the circle $C(A,\rho )$ with the bisector of $\widehat{PTQ}\iff$ $H$ is the incenter of $\triangle PQT$ .

Extension. Let $ABC$ be an acute triangle with the circumcircle $w=C(O)$ and $D\in (BC)\ ,\ L\in (CA)\ .\ K\in (AB)$

so that $\left\{\begin{array}{c} m\left(\widehat {ADK}\right)=B\\\\ m\left(\widehat {ADL}\right)=C\end{array}\right|$ . Denote $\{P,Q\}=KL\cap w$ and $\{A,T\}=AD\cap w$ . Prove that $D$ is the incenter of $\triangle PQT$ .

Proof. $\left\{\begin{array}{ccc} m\left(\widehat {ADK}\right)=B & \implies & AK\cdot AB=AD^2\\\\ m\left(\widehat {ADL}\right)=C & \implies & AL\cdot AC=AD^2\end{array}\right|$ $\implies$ $AK\cdot AB=AL\cdot AC\implies$ $BCLK$ is a cyclic $\implies$ the line $d\equiv \overline{PKLQ}$ (in this order) is an antiparallel

to $BC$ in $\triangle ABC$ $\implies $ $AO\perp d$ $\implies $ $AP=AQ$ $\implies$ $\widehat{ABP}\equiv$ $\widehat {AQP}\equiv$ $\widehat{APQ}\equiv$ $\widehat{ACQ}$ . Therefore, $\left\{\begin{array}{ccc} \widehat{ABP}\equiv\widehat{APQ} & \implies & AP^2=AK\cdot AB=AD^2\\\\ \widehat{ACQ}\equiv\widehat{AQP} & \implies & AQ^2=AL\cdot AC=AD^2\end{array}\right|$ $\implies$

$AP=AD=AQ=\rho$ $\iff D$ belongs to the intersection of the circle $C(A,\rho )$ with the bisector of $\widehat{PTQ}\iff$ $D$ is the incenter of $\triangle PQT$ .[/color]


PP9. Let $w=C(O)$ with diameter $[AB]$ , $Q\in w$ , $H\in AB$ for which $QH\perp AB\ ,\ \{C,D\}=w_1\cap w$ , where $w_1=C(Q,QH)$ and $M\in QH\cap CD$ . Prove that $MQ=MH$ .

Proof 1. Denote $N\in OQ\cap CD$ and the diameter $[QE]$ of $w$ . Then $OHMN$ is cyclically and $QH^2=QC^2=QN\cdot QE=$ $2\cdot QO\cdot QN=2\cdot QM\cdot QH\implies$ $QH=2\cdot QM$ .

Proof 2. Denote $\{Q,R\}=\{Q,H\}\cap w$ . Thus, $\widehat{QDC}\equiv\widehat{QCD}\equiv\widehat {QRC}\implies$ $QH^2 = QC^2 = QM\cdot QR =$ $ 2\cdot QM\cdot QH\implies$ $QH = 2\cdot QM$ .

Proof 3. Midpoint $S$ of $QH\ ,\ \{Q,R\}=QH\cap w$ and $\{H,P\}=QH\cap w_1\implies$ $SQ\cdot SR=SP\cdot SH\implies$

$p_w(S)=p_{w_1}(S)\implies$ $S\in CD\implies$ $S\equiv M\implies$ $M$ is the midpoint of $[QH]$ .


PP10. Let $AD$ be an altitude of the $\Delta ABC$ , where $D\in BC$ , $M\in\overline {AD}$ with $\frac{DM}{AM}=\frac{BD}{CD}$ and the projection $N$ of $D$ onto $BM$ . Prove that $AN\bot CN$ .

Proof. $\angle AMN=\angle CDN$ and $\frac{ND}{MN}=\frac{BD}{MD}=\frac{CD}{AM}$ implying $\Delta NDC\sim\Delta NMA$, i.e. does

exist a spiral similarity mapping the 2 triangles. Since $MN\bot ND, AM\bot DC$ obtain that $AN\bot CN$ .


PP11. Let a convex $ABCD$ with $\left\{\begin{array}{c} AB\perp AD\\\\ CB\perp CD\end{array}\right\|$ and for $\left\{\begin{array}{c} E\in (BA)\\\\ F\in (BC)\end{array}\right\|$ denote $P\in (EF)$ so that $DP\perp EF$ . Prove that $BA^2+PE^2+CF^2=BC^2+PF^2+AE^2$ .

Proof. $\left\{\begin{array}{ccccc} AB^2+\underline{AD}^2 & = & BD^2 & = & CB^2+\underline{\underline{CD}}^2\\\\ \underline{\underline{CD}}^2+CF^2 & = & DF^2 & = & \underline{\underline{\underline{PD}}}^2+PF^2\\\\ PE^2+\underline{\underline{\underline{PD}}}^2 & = & DE^2 & = & AE^2+\underline{AD}^2\end{array}\right\|\bigoplus\implies$ $\boxed{BA^2+PE^2+CF^2=\left(DB^2+DE^2+DF^2\right)-\left(\underline{DA}^2+\underline{\underline{DC}}^2+\underline{\underline{\underline{DP}}}^2\right)=BC^2+PF^2+AE^2}\ .$



PP12 (Swiss IMO Selection). Let $\triangle ABC$ be an acute-angled triangle with $AB \ne AC$. Let $H$ be the orthocenter of $\triangle ABC$, and let $M$ be the midpoint of $[BC]$ . Let $D\in (AB)$ and

$E\in (AC)$ such that $AE=AD$ and $H\in DE$ . Prove that the line $HM$ is perpendicular to the common chord of the circumcircle $w$ of $\triangle ABC$ and the circumcircle $\alpha$ of $\triangle ADE$ .

Proof. Denote $:$ the midpoint $F$ of $[AH]$ $;$ the circles $w=\mathbb C(O,R)$ and $\alpha =\mathbb C\left(O_a, R_a\right)$ $;$ the diameter $[AA']$ of $w$ and the midpoint $S$ of the arc $\overarc{BC}$ , where $BC$ separates $A$ , $S\ ;$

the intersections $N\in MH\cap AS$ , $K\in DE\cap AA'$ and $L\in DE\cap AS$ . Observe that $DE\perp AS$ , $L$ is the common midpoint of $[DE]$ and $[HK]$ . Thus, $AH\parallel MS\iff$

$\frac {NA}{NS}=\frac {AH}{MS}=\frac {2R\cos A}{R-R\cos A}=$ $\frac {2\cos A}{1-\cos A}\implies$ $\frac {NA}{AS}=\frac {2\cos A}{1+\cos A}\iff$ $\boxed{NA=\frac {2\cos A}{1+\cos A}\cdot {AS}}\ (1)$ . Therefore, from $\triangle ASA'$ obtain that $AS=AA'\cdot\cos\frac {B-C}2$ , i.e.

$\boxed{AS=2R\cos\frac {B-C}2}\ (*)$ . Hence $AH\cdot\cos\frac {B-C}2=AL=$ $AD\cos\frac A2\iff$ $AD\cdot\cos\frac A2=$ $2R\cos A\cos\frac {B-C}2\ \stackrel{(*)}{\implies}\ \boxed{AD=\frac {AS\cdot\cos A}{\cos \frac A2}}\ (2)\ .$ From the relations

$(1)$ and $(2)$ obtain that $\frac{AD}{NA}=\frac {1+\cos A}{2\cos\frac A2}=\cos\frac A2\implies$ $AD=AN\cdot\cos \frac A2\implies$ $ND\perp AB$ , i.e. $[AN]$ is the diameter of the circle $\alpha$ . Since $O_a\in FO$ and $O_a$ is the midpoint

of $[AN]$ , then get that $FO\parallel MH$ and $O_a\in FO$ $\implies$ $MH\parallel O_aO$ , i.e. the line $HM$ is perpendicular to the common chord of the circles $w$ and $\alpha$ .


PP13 (Bogomolny). Let the trapezoid $ABCD$ with $BC\parallel AD$ and $E\in AB\cap CD\ .$ Denote $F$ so that $CF\perp AC\ ,$ $DF\perp BD\ .$ Prove that $CF=DF\iff EF\perp BC\ .$

Proof 1. Let $\left\{\begin{array}{c} A(0,0)\ ;\ B(b,h)\\\\ C(c,h)\ ;\ D(d,0)\\\\ h>0\ ;\ 0<b<d<c\end{array}\right\|$ Thus, $E\in AB\cap CD\iff$ $\left\{\begin{array}{cccc} AB\ : & \frac yh & = & \frac xb\\\\ CD\ : & \frac yh & = & \frac {x-d}{c-d}\end{array}\right\|$ $\implies$ $\frac yh=\frac xb=\frac {x-d}{c-d}=\frac d{b+d-c}$ $\implies$ $\boxed{E\left(\frac {bd}{b+d-c},\frac {hd}{b+d-c}\right)}\ (1)\ .$

I"ll ascertain $F(x,y)\ :$ $\left\{\begin{array}{cc} CA\perp CF\iff\frac hc\cdot\frac {y-h}{x-c}+1=0\iff cx+hy=c^2+h^2 & (2)\\\\ DB\perp DF\iff\frac h{b-d}\cdot\frac y{x-d}+1=0\iff (d-b)x-hy=d(d-b) & (3)\end{array}\right\|$ $\implies$ $\boxed{x_F=\frac {c^2+d^2+h^2-bd}{c+d-b}}\ (4)\ .$ Therefore $:$

$\blacktriangleright\ \boxed{\ FC=FD\ }\ ,$ where $F(x,y)$ $\iff$ $(x-c)^2+(y-h)^2=(x-d)^2+y^2\iff$ $2(c-d)x+2hy=$ $c^2+h^2-d^2\ \stackrel{(2)}{\iff}$ $2\left(c^2+h^2\right)-2dx=$

$c^2+h^2-d^2\iff$ $2dx=c^2+d^2+h^2\ \stackrel{(4)}{\iff}\ 2d\left(c^2+d^2+h^2-bd\right)=(c+d-b)\left(c^2+d^2+h^2\right)\iff$ $\boxed{(d+b-c)\left(c^2+d^2+h^2\right)=2bd^2}\ (5)\ .$

$\blacktriangleright\ \boxed{\ EF\perp BC\ }\iff$ $x_E=x_F\ \stackrel{1\wedge 4}{\iff}$ $\frac {bd}{b+d-c}=\frac {c^2+d^2+h^2-bd}{c+d-b}\iff$ $bd(c+d-b)=(b+d-c)\left(c^2+d^2+h^2-bd\right)\iff$ $bd\left[(c+d-b)+(b+d-c)\right]=$

$(b+d-c)\left(c^2+d^2+h^2\right)\iff$ $\boxed{(d+b-c)\left(c^2+d^2+h^2\right)=2bd^2}\ (6)\ .$ In conclusion, the last relations $(5)$ and $(6)$ evinces that $\boxed{FC=FD\iff EF\perp BC}\ .$

Proof 2 (synthetically).


PP14. Let $\triangle ABC$ with $c<b\ ,$ its circumcircle $w$ and let $X\ ,$ $Y\ ,$ $Z$ be the points where the incircle is tangent to $BC$, $CA$, $AB$ respectively.

Let $U$ the midpoint of the arc $A\in \overarc{BC}\subset w\ .$ The line $UX$ meets $w$ again at $K\ .$ Let $T\in AK\cap YZ\ .$ Prove that $XT\perp YZ\ .$

Proof. Let $L\in YZ \cap BC\ .$ Then $ T \{BC,XL\} $ is harmonic division. So $K\{BC,XL\} $ is harmonic pencil too. By angle chasing, we have $\widehat{TKX}\equiv\widehat{AKU}\equiv\widehat{ACU}\ .$ Therefore,

$m\left(\angle UCB\right)-m\left(\angle ACB \right)= $ $\frac{B-C}{2} $ and also $ m\left(\angle TLX\right) = m\left(\angle ZLB\right) = 180^{\circ} -m\left( \angle LZB\right) -$ $ m\left(\angle LBZ\right) = 180^{\circ} - $ $\left(90^{\circ}-\frac A2\right) - $ $\left(180^{\circ}-B\right) = \frac{B-C}{2}\ .$

Therefore, $ ZLKX $ is cyclic. Since $ KU $ bisects $ \angle BKC $, it follows that $ KL$ is the external angle bisector of $ \angle BKC $ and hence $\widehat{XKL}\equiv\widehat{XTL}\equiv 90^{\circ}\ .$ That is $ XT \perp YZ\ .$


PP15. Let $\triangle ABC$ with the centroid $G\ ,$ the circumcircle $\Omega =\mathbb C(O,R)$ and the incircle $w=\mathbb C(I,r)\ .$ Prove that $:$

$1\ .\blacktriangleright\ \triangle AIO$ is right-angled $\iff$ $\frac {b+c}a\in\left\{2,\frac Rr\right\}\iff$ $h_a\in\left\{3r,R+r\right\}\ .$

$2\ .\blacktriangleright\ \triangle AGO$ is right-angled $\iff$ $\frac {b^2+c^2}{2a^2}\in\left\{1,\left(\frac {R\sqrt 3}{a}\right)^2\right\}\ .$

Proof.

$\boxed{\ \begin{array}{ccccccccccccc} \boxed{\ IA\perp IO\ } & \iff & IA^2+IO^2=OA^2 & \iff & \left(bc-4Rr\right)+\left(R^2-2Rr\right)=R^2 & \iff & 2Rh_a=6Rr & \iff & \boxed{h_a=3r} & \iff & 2sr=3ar & \iff & \boxed{\frac{b+c}a=2}\\\\ \boxed{\ OA\perp OI\ } & \iff & OA^2+OI^2=IA^2 & \iff & R^2+\left(R^2-2Rr\right)=bc-4Rr & \iff & 2R^2+2Rr=2Rh_a & \iff & \boxed{h_a=R+r} & \iff & r(2s-a)=aR & \iff & \boxed{\frac {b+c}a=\frac Rr}\end{array}\ }$

$\boxed{\ \begin{array}{ccccccccc} \boxed{\ GA\perp GO\ } & \iff & GA^2+GO^2=OA^2 & \iff & \frac {4m_a^2}9+R^2-\frac {a^2+b^2+c^2}9=R^2 & \iff & \cancel 2\left(b^2+c^2\right)-a^2-\left(a^2+\cancel{b^2+c^2}\right)=0 & \iff & \boxed{b^2+c^2=2a^2}\\\\ \boxed{\ OA\perp OG\ } & \iff & OA^2+OG^2=GA^2 & \iff & R^2+\left(R^2-\frac {a^2+b^2+c^2}9\right)=\frac {2\left(b^2+c^2\right)-a^2}9 & \iff & 18R^2=3\left(b^2+c^2\right) & \iff & \boxed{b^2+c^2=6R^2}\end{array}\ }$
This post has been edited 125 times. Last edited by Virgil Nicula, Nov 22, 2016, 8:40 AM

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430. Trigonometrical identities.
+ October 2015
429. Bretschneider's_formula
+ September 2015
428. Some metrical problems from the contests II.
+ July 2015
427. Selected nice or well-known geometry problems.
+ June 2015
426. Problems from and for baccalaureate.
+ May 2015
425. Geometrical extremity II.
424. Some nice problems of E. E. Q. Rodriguez and others.
+ March 2015
423. Geometry problems.
422. Geometrical problems.
420. Some geometry problems for high school II.
419. Some geometry problems for hight school I
418. Some nice problems with polynomials/equations III
417 <bis> 418 (educatie: "14.27.28.32.57"/\"127.248").
417. The incircle of ABC and other two incircles.
416. Cateva solutii ale unei probleme de tip slicing.
415. Geometry 8.
414. Geometry 7.
413. Geometry 6.
412. Geometry 5.
411. Geometry 4.
410. Geometry 3.
409. Geometry 2.
408. Geometry 1.
+ February 2015
407. Cyclical quadrilaterals.
+ November 2014
406. Trigonometry in geometry (middle or high school) II.
405. Art of problem solving (algebra).
403. Some nice geometry problems for the high school. II
402. Some interesting inequalities I.
401. Some nice problems with polynomials/equations II.
+ October 2014
400. Some nice geometry problems for the high school I.
+ September 2014
399. Algebra (I).
398. Geometry problems (II).
+ August 2014
397. Two similar trigonometrical identities.
+ July 2014
396. Own inegalities stronger than the Panaitopol's.
+ April 2014
395. Metrical relations in a quadrilateral.
394. A cevian in a triangle and two incircles.
393. Some problems with circles.
+ March 2014
392. Art of Problem Solving (geometry).
+ December 2013
391. CRUX Mathematicorum.
+ November 2013
390. Probleme de analiza matematica.
389. Some geometric properties in a triangle I.
+ October 2013
388. Some interesting "slicing" problems.
+ September 2013
387. Algebra (II).
+ August 2013
386. Barycentrical coordinates.
385. Some nice geometry problems.
384. Geometry problems (I).
383. Trigonometry in geometry (middle or high school) I.
+ July 2013
382. Peruan "jewels" (Israel Diaz, M. O. Sanchez a.s.o.)
381. Probleme date la diverse concursuri.
380. Functions. Range. Bijection and inverse.
+ June 2013
379. Geometrical extremity I.
378. Some metrical relations.
377. Some remarkable properties of the symmedian.
+ May 2013
376. Trigonometrical identities.
374. Some proofs of the Law of Cosines.
373. Extension of Chebyshev's inequality.
+ April 2013
372. The distancies : OI , OI_a , ON a.s.o.
371. Some problems from NMO (final stage) - 2013, Romania.
+ February 2013
370. Some metrical relations in a triangle.
+ January 2013
369. Mixtilinear circles of ABC.
364. Some interesting inequalities I
+ December 2012
363. Geometry problems with perpendicularities.
+ November 2012
362. Fermat's problem and other metrical problems.
+ October 2012
361. Arhimedes theorem of the broken chord in a circle.
360. Some problems for the middle school.
359. Ptolemy's theorem and generalized Ptolemy's relation.
358. Some properties of the remarkable lines in a triangle.
+ September 2012
357. Morley theorem.
356. Order relation between s & (2R+r) in an acute triangle.
+ August 2012
355. Some problems with polynomials or equations I
354. Gradul II - Univ. B.B. Cluj, aug. 1998 Prof I sau II.
353. Some problems of the analytical geometry.
+ July 2012
352. Some definite integrals.
351. Problems with areas.
350. Another (easy) integrals.
349. Another interesting limits.
348. Subiectele de la BAC 2012 /Mate-Info.
+ June 2012
347. Two isogonal lines in a triangle.
346. Some easy integrals.
345. Characteristic function of a set.
344. Some trigonometry problems.
343. Problems with collinear points and concurrent lines.
+ May 2012
342. Exponential or logarithmic equations/inequations.
+ April 2012
341. Some nice and interesting "slicing" problems.
340. Geometry problems for middle school.
339. Derivatives. Rolle's function.
338. ROU National Math Olympiad 2012.
+ February 2012
337. Nice metrical problems.
+ January 2012
336. Some problems of the analytical geometry.
335. Some difficult geometric/trigonometric inequalities.
334. 2011 Japan Mathematical Olympiad Finals Problem 1.
333. Problems with the common tangent for two circles.
332. Some metrical problems from the contests I.
+ December 2011
331. Some nice problems with polynomials/equations I.
+ November 2011
330. A triangle and an its transversal through a point.
329. Some problems with complex numbers.
328. A general identities with areas in a triangle ABC.
+ October 2011
327. Some geometrical/algebraic extremum problems.
326. Some problems from trigonometry.
325. Without the l'Hospital's rules.
324. Some determinants.
323. Easy, nice and interesting problems for high school.
322. Incenter, Gergonne, Nagel a.s.o. of ABC.
321. Harmonical quadrilateral.
320. Some problems with metrical relations.
+ September 2011
319. Some problems with induction.
318. IRAN - 2011 (geometrie).
317. Some interesting geometry problems for middle school.
316. Some properties of isogonal rays in an angle of ABC.
+ August 2011
315. Indian Team Selection Test 2010 ST1 P1 and extension.
314. USAMTS 2009 Round 2 #5.
313. For middle school - some nice problems.
312. Very nice (famous trigonometrical problems) !
311. Some problems from USAMO and other contests.
310. Some properties of the tangential quadrilateral.
309. Beautiful problem ! IRAN Selection Test for IMO, 2004.
308. Some nice and easy properties in a triangle.
+ July 2011
307. Very nice proofs !
306. Some easy and nice "slicing" problems.
305. Position of the roots w.r.t. a given real number.
304. Integrals ("brothers") III.
303. Integrals II.
302. Some problems with sequencies of the "roots" .
301. Some trigonometrico-geometrical inequalities.
300. Lagrange's multiplier. Examples and problems.
299. A class of the recurrent sequencies.
298. Some nice problems from Crux Mathematicorum.
297. Nice inequalities with concrete numbers.
296. Some nice Vittasko's problems.
295. M1 3th subject, 2c - School Graduate, ROMANIA.
294. The Octav Halunga's inequality in a triangle.
+ June 2011
293. A difficult trigonometric equation (third point !).
292. Length of the angled bisector and some means.
291. Some integrals.
290. Characterize "OP perpendicular on OA" in ABC a.s.o.
289. Nice identities in an algebra/ geometry/trigonometry.
288. An inequality with two A-cevians in a triangle ABC.
287. Some metrical problems with or without trigonometry.
286. ABC is right triangle iff s=2R+r and other similar pp.
285. Nice and easy inequalities. For example, IMAC - 2009.
284. Ellipse (definition).
283. Some nice and easy problems.
282. Factorize of polynomials.
+ May 2011
281. Identity with R in an acute triangle (ILL - 1974).
280. Common roots of two polynomial equations.
279. O.M. Holland , O.M. Ireland 1988 and TST USA 2000, etc.
278. In memoria profesorului Alexandru Lupas.
277. A nice and difficult relation with Lemoine's axis.
276. Chinese TST 2009 and Second Round 2006.
275. Some interesting algebraic equations and systems.
274. An inequality in a triangle for its interior point.
273. The area of the triangle IOH. When I belongs to OH ?
272. Outside of triangle. Extension of Fermat's point.
271. Relations concerning Fermat points.
+ April 2011
270. Morrocan M.O. 2010 and Croatia TST 2009.
269. Some nice and easy properties in a trapezoid.
268. Saraghin contest, 2011.
267. Germany 2002, Russia 2003, IMO 2003, Aus.-Polish 1992.
266. Rectangles erected on sides of a triangle. Concurrence.
265. Really nice problem !
264. Identities with complex numbers.
263. Newton's sums. Applications.
262. IMO Shortlist 2005, Geometry 6th.
261. IMO 2009, problem 2 and JBMO 2007, problems 1 & 2.
260. Some nice, simple identities and their applications.
259. ABC equilateral triangle ==> DEF equilateral triangle.
258. Some difficult and nice problems with complex numbers.
257. Seeking roots of 2th equation with complex coef.
256. An extremum problems with complex numbers.
+ March 2011
255. A line which is parallelly with OH.
254. An interesting C. Mateescu's geometrical inequality.
253. An old algebraic inequality.
252. OI=f(A,R) in certain conditions.
251. Problems of Analytical Geometry.
250. An application of the Euler's relation.
249. An usual and nice metrical relations in quadrilateral.
248. Pagina instructiva.
247. An remarkable inequality with median.
246. R1-R2=OI (nice).
245. An inequality in an acute triangle.
244. Some properties of cyclic orthodiagonal quadrilaterals.
243. Nice ! (a+b) is constant.
242. A general geometrical problems.
241. Areas in a triangle.
240. Some interesting geometrical inequalities.
239. Algebraic marathon.
238. Some metrical problems.
237. Some problems with ... problems.
236. A locus with the metrical relation.
235. Proofs of some geometry problems with complex numbers.
234. Some simple and nice problems from contests.
233. An interesting identity and an easy & nice inequality.
+ February 2011
232. Some geometrical inequalities (own).
Pagina libera
231. A conditioned minimum value.
230. Some usual synthetical properties in a triangle.
229. Concurs online MATEFBC, ed. 5 -subiect 2.
228. OJM - Romania, 2010 problem 3.
227. IMO ShortList 2003 (problem 5).
226. Some inequalities from Calculus (Real Analysis).
225. An interesting problems from Kunny.
224. Some nice and easy inequalities in a triangle.
223. An inequality with complex numbers-RMO shortlist 2010.
222. An extension of probl. 2 from IMO 2009.
221. Some nice and simple "slicing" problems.
220. Some nice geometry problems from contest.
+ January 2011
219. A very nice geometrical inequality.
218. Some problems from the Suiss IMO Selection Team 2006.
217. A nice and interesting Murray S. Klamkin's problem.
216. Nice ! Find the length of hypotenuse (Belarus - 2010)
215. Limit of the unique root for polynominal equations.
214. A range of some functions.
213. Another classical "slicing" problems.
212. A simple applications of the Casey's theorem.
211. Pretty hard problem !
210. Tangential circle of the triangle ABC.
209. Nice problem, nice proof !
208. An interesting geometrical inequality.
207. Some interesting problems of Toshio Seimiya, Japan.
206. Four problems with extremum in a quadrilateral.
205. A "slicing" problem with 11 methods.
204. Saraghin 2010 (three geometry problems).
203. An application of the Pascal's permutation theorem.
+ December 2010
202. Concurrent lines in a right triangle.
201. Applications of Desarques' and Pappus' theorems.
200. Some nice applications of the inversion.
199. Some properties of symmedian. Two extensions.
198. Generalized Carnot's lemma.
197. Triangles outside of a triangle and concurrence.
196. Similar rectangles outside of a triangle.
195. Semicircle. Nice application of harmonical division.
194. The Lalescu's sequence. Properties.
193. An inequality with n! .
192. An easy and nice problem of Valentin Vornicu.
191. Six nice problems with the incircle of a triangle.
190. Very nice ! Coculescu's Contest seniors 2010.
189. Pascal's theorem and some nice and easy applications.
188. Some problems from vectorial geometry.
187. Distancies.
186. Without the l'Hospital's rules.
185. The Durrande's relation.
184. An interesting geometry problem from RMO 2010.
183. A remarkable characterization of equilateral triangle.
182. Another two nice problems with complex numbers (own).
181. Two special inequalities from CRUX (own).
180. Own proposed problem from CRUX (with complex numbers).
+ November 2010
179. The range of |z| , |z^2+a|=|bz+c|. Particular cases.
178. Some nice and simple geometry problems.
177. A nice proof of one identity in a triangle.
176. Another nice problems for middle school.
175. A Geometric Proof (without trig.) of Heron's Formula.
174. Some interesting properties of modulus (middle school).
173. An interesting trigonometric equations.
172. The Zelinsky's problem.
171. Another "slicing" proposed problems (middle school).
+ October 2010
170. A nice "slicing" proposed problem for middle school.
169. Some metrical problems in a circle (III).
168. Some properties of the Lemoine's point.
167. Nice problems - OIM, 1995 and ... Toshio Seimyia.
166. Some constant geometrical expressions.
165. Some metrical problems (5) in a circle (II).
164. Some metrical problems (5) in a circle.
163. Applications of Ptolemy's inequality. Ex. - China,1998.
162. Generalization of proposed problem from Mongolia, 2000.
161. A difficult inequality (barycentrical coordinates)
160. A very nice "slicing" problem.
159. Multiple derivatives in a point.
158. Very nice and difficult limits.
157. Steinbart's theorem. Applications.
156. Some problems with perpendicularities.
155. A constant ratio in a triangle ABC.
154. Difficult inequality (without derivatives eventually).
153. Geometrical minimum/maximum.
152. Concurrence, cyclically, metrics, symmedian a.s.o.
151. Miscellaneous problems for the admission at math.
150. Some (5) problems with fixed points.
149. Some (8) problems with collinearity or concurrence.
148. A special class of systems.
147. A problem with two polynominals.
146. Synthetic, metric, trigonometric and analitycal proofs.
145. Spain MO P5 - Point on median.
144. A conditioned concurrency in a triangle.
143. Assess of a ratio (nice and difficult problem).
142. Inscribed/circumscribed quadrilateral w.r.t. ABC.
141. N is interior <==> 1/3<a/b<3 (CRUX Mathematicorum).
140. A synthetical property <==> linear/angled relation.
139. A caracterization of a right angle in a triangle.
138. Generalized "slicing" problems (own).
137. Stewart's relation.
136. A geometrical interpretation of a system (own).
135. Three strong geometrical inequalities.
+ September 2010
134. Some problems with projections/perpendiculars (2).
133. Some problems with projections/perpendiculars (1).
132. Asupra unei inegalitati intr-un triunghi.
131. A stronger Form of the Steiner-Lehmus Theorem (own).
130. A triangle and a fixed point.
129. BX=CY and three properties.
128. Value of IG and two inequalities in neighbourhood of 0.
127. Pagina educativa.
126. Routh's theorem.
125. I like very much this Darij Grinberg's message.
124. A nice and easy problem with a right-angled triangle.
123. My extension and its proof.
122. An identity for an equilateral triangle. Extension.
121. Viete's relations with cos A , cos B , cos C.
120. Two equal neighbouring angles/sides in a quadrilateral.
119. Inspired by a Miculita's problem.
118. An usual remarkable geometrical locus.
117. Some applications of harmonic division/quadrilateral.
116. The length of AE (nine methods !).
115. An difficult equation with one radical.
114. A property of tangential triangle for ABC.
113. Range for a function.
112. The equivalence relation ".s.s."
111. Simple, but interesting ...
110. German TST 2004 - Exam V , Problem 1
109. Two equal areas ==> find A.
108. Old, short and nice proof of Euler's relation.
107. XY parallel BC and X,Y are isogonal/izotomic points.
106. OLM - Bucuresti (geometrie).
105. A problem with three circles.
104. Concursul Revistei de Matematica din Galati (RMG).
103. Metrical usual relations in triangle. Applications.
102. Characterization of a metrical relation in a triangle.
101. O problema cu numere complexe.
100. Bobiller's theorem.
99. Some simple and nice geometry problems.
98. Symmedian & median (metrical relations).
97. Problem of butterfly.
+ August 2010
96. Integrals.
95. Three circles in an angle.
94. Equations of angle bisectors (interior and exterior).
93. The Appolonius' construct problems.
92. CA , CB tangent to parabola - OIMU 2008 Problem 4.
91. JBMO 2010, Problem 3.
90. Nice and difficult inequality in ABC (own).
89. Similarity (parallel/antiparallel).
88. A nice geometric locus conditioned metrically.
87. An interesting vectorial identity.
86. A very nice maxmin.
85. Problem "slicing" : 60-24-12.
84. IRAN national math olympiad - 2010
83. A metrical characterization of concurrence.
82. AA' , BN , CM concur in Gergonne point.
81. Sawayama and Thebault’s theorem.
80. Three concurrent perpendiculars (the Carnot's lemma).
79. A quadrilateral with many perpendicularities.
78. Jakobi's theorem.
77. Two equivalent perpendicularities.
76. A parallelogram and a circle (nice !).
75. Extended Steinbart Theorem.
74. Inegalitatea Erdos-Mordell.
73. A nice problems with a square and a perpendicularity.
+ July 2010
72. Colinearity in a triangle - H , P , M .
70. Colinearity with pole/polar - H\in PQ .
71. An angle questions (synthetic/trigonometric proofs).
69*1. Slicing problem 3.
69. Position of x1, x2 w.r.t. a given real number k.
68. Some nice applications of "pole/polar".
67. Remarkable algebraic/geometrical inequalities (1).
66. Remarkable perpendicularities in a triangle.
65. Parallel to BC through I .
64. Problems of the type "instant" (at most one minute).
63. A nice problem with radical center (livetolove212).
62. Nice and easy problem with a right triangle.
61. IMO Shortlist 2009 (very nice problem and its proof).
60. Mediteranean M.C. 2010 Problem 3.
59. IMO 2010, problem 4.
58. A cevian and two incircles.
57. Opinii si comentarii relativ la inv. rom. II
56. A extremum problem with areas.
55. Two important circles - mixtilinear incircle/excircle.
54. Cinci grade de libertate.
53. An remarkable concurrency.
52. Harmonical division.
51*1. JBTST-3/2010 Aplicatii la proprietatea fluturelui.
51. A nice problem with perpendicularity from Jaime.
50. IMAC 2010 seniors, the first day.
49. The midpoint of a cevian in an acute triangle.
+ June 2010
48. An inequality in a triangle : h_a+max{b,c} ...
47. Interesting and difficult identities in a triangle.
46. Outside of a quadrilateral.
45. \cos B+\cos C=1 <==> nice property.
44. Relatia IA=IO sau IA=IM intr-un triunghi ABC etc.
43. A "slicing" problem with a parameter.
42. IMAC 2010 Problema 2.
41. ONM 2010 Calarasi Problema 3.
+ May 2010
40. Lema lui Haruki.
39. A nice and remarkable problem with Brocard's point.
38. Minimum area of triangle touching with semicircle.
37. Own extension of the Steiner-Lehmus' problem.
36. ONM (juniori) - 2010 Iasi, problema 4.
35. Nice problem from Arqady (Eduard_Tur).
34. IMO Shortlist 2002 geometry problem 7.
+ April 2010
32. Linkuri diverse.
31.Some nice metrical problems.
29. Inegalitate integrala de la Kunny.
28. A fost odată ...
27.Problem of Darij Grinberg (I - centroid of triangle DEF).
26. Outside of a triangle.
25. Geometric equations.
24. A fine and cool inequalities.
23. A very nice exponential inequality (own - from CRUX).
22. Some interesting limits (sequence/function).
21 bis. Some problems with complex numbers.
21. Probleme de geometrie (Yetty & I).
20. O ineg.cu suma de AG/AI .
19. Own proposed problems in CRUX Mathematicorum - Canada.
18. O problema a lui Toshio Seimiya.
17. O inegalitate "tare" intr-un triunghi.
16. Length of the (interior) angle bisector (10 proofs).
15. Inequalities stronger than Panaitopol's.
14. Opinii si comentarii relativ la inv. rom.
13. Inegalitatea de la Sorin Borodi.
12. Inegalitatea I. Maftei & S. Radulescu (2).
11. Inegalitatea I. Maftei & S. Radulescu (1).
10. Walker's inequality and its extension.
9. Inegalitatea HO+HA+HB+HC >= 3R .
8. The first problem Shortlist ONM 2010 Romania.
7. Some interesting "slicing" problems.
6. Interesting problems from JBMO.
5. Theoretical preliminary for inequalities.
4. Remarkable geometrical inequalities (2).
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