237. Some problems with ... problems.

by Virgil Nicula, Mar 4, 2011, 9:37 AM

PP1. Let $ABC$ be a triangle and let $D\in (BC)$ be a point for which $\widehat{DAB}\equiv\widehat {DAC}$ . For a point $P\in (AD)$ define

the points $E\in BP\cap AC$ , $F\in CP\cap AB$ and $X\in BE\cap DF$ , $Y\in CF\cap DE$ . Prove that $\widehat {DAX}\equiv\widehat {DAY}$ .

Proof. Denote $\left\|\begin{array}{c} \frac {EA}{EC}=n\\\ \frac {FA}{FB}=m\end{array}\right\|$ and $\left\|\begin{array}{c} m(\widehat{XAB})=x \\\ m(\widehat{YAC})=y\end{array}\right\|$ . Prove easily that $\left\|\begin{array}{ccc} DB=\frac {ac}{b+c} & ; & DC=\frac {ab}{b+c}\\\ EC=\frac {b}{n+1} & ; & EA=\frac {nb}{n+1}\\\ FA=\frac {mc}{m+1} & ; & FB=\frac {c}{m+1}\end{array}\right\|$ . Applying the Ceva's theorem to the point $P$

in $\triangle ABC$ obtain that $mc=nb$ . Therefore, $\frac {AE}{AF}=$ $\frac {m+1}{n+1}$ and $\left\|\begin{array}{c} \frac {AD}{AF}\cdot \frac {\sin\left(\frac A2-x\right)}{\sin x}=\frac {XD}{XF}=\frac {EC}{EA}\cdot\frac {BD}{BF}\cdot \frac {BA}{BC}=\frac 1n\cdot\frac {a(m+1)}{b+c}\cdot\frac ca=\frac {(m+1)c}{n(b+c)}\\\\ \frac {AD}{AE}\cdot\frac {\sin\left(\frac A2-y\right)}{\sin y}=\frac {YD}{YE}=\frac {FB}{FA}\cdot\frac {CD}{CE}\cdot\frac {CA}{CB}=\frac 1m\cdot\frac {(n+1)a}{b+c}\cdot \frac ba=\frac {(n+1)b}{m(b+c)}\end{array}\right\|$ $\Longrightarrow$

$\left\|\begin{array}{c} \frac {AD}{AF}\cdot \frac {\sin\left(\frac A2-x\right)}{\sin x}=\frac {(m+1)c}{n(b+c)}\\\\ \frac {AD}{AE}\cdot\frac {\sin\left(\frac A2-y\right)}{\sin y}=\frac {(n+1)b}{m(b+c)}\end{array}\right\|$ $\Longrightarrow\ \frac {AE}{AF}\cdot \frac {\sin\left(\frac A2-x\right)}{\sin x}\cdot\frac {\sin y}{\sin\left(\frac A2-y\right)}=\frac {m(m+1)c}{n(n+1)b}=\frac {m+1}{n+1}$ $\Longrightarrow$

$\sin\left(\frac A2-x\right)\cdot \sin y=\sin\left(\frac A2-y\right)\cdot\sin x$ $\Longrightarrow$ $\cos\left(\frac A2+y-x\right)=$ $\cos\left(\frac A2+x-y\right)$ $\Longrightarrow x=y\ .$


PP2. Let $ABC$ be a triangle with the circumcircle $w$ . Let $\{P,Q\}\subset w$ be two points so that $PQ\parallel BC$ and the line $BC$

separates $A$ , $P$ . Prove that $\left\|\ \begin{array}{c} AP\cdot AQ=AB\cdot AC+PB\cdot PC\\\ AP^2=AB^\cdot AC+PB\cdot PC\ \Longleftrightarrow\ AB=AC\ \ \vee\ \ \widehat {PAB}\equiv\widehat {PAC}\end{array}\ \right\|\ .$

Proof. Suppose w.l.o.g. $PB\le PC\ .$ Denote $m(\angle PAB)=m(\angle PAC)=x\ .$ Observe that $\left\|\begin{array}{ccc} m(\angle APQ) & = & B+x\\\ m(\angle AQP) & = & C+x\end{array}\right\|$

and $AP\cdot AQ=AB\cdot AC+PB\cdot PC$ $\Longleftrightarrow$ $\sin (C+x)\cdot\sin (B+x)=$ $\sin C\cdot\sin B+\sin x\cdot\sin (A-x)$ $\Longleftrightarrow$

$\cos(C-B)-\cos (B+C+2x)=$ $\cos (C-B)-\cos (C+B)+\cos (2x-A)-\cos A$ $\Longleftrightarrow$ $\cos(C-B)+\cos (2x-A)=$

$\cos (C-B)+\cos A+\cos (2x-A)-\cos A$ , what is truly. Prove easily that the mentioned (in conclusion) equivalence is also truly.


PP3. Let $ ABCD$ be a convex quadrilateral , $ AC\not\perp BD$ and $ X\in (BD)$ so that $ XB^2\cdot (DC^2 - DA^2) - XD^2\cdot (BC^2 - BA^2)=$

$AB^2\cdot CD^2 - AD^2\cdot BC^2\ \ (*)$ . Denote reflection $R$ of $ A$ w.r.t. $ BD$ . Prove that $ ACRX$ is concyclically and $ XA = XR\implies$ $ \widehat {XCA}\equiv\widehat {XCR}$ .

Lemma (well-known or you can easily it). Are given two fixed points $ B$ , $ D$ and three real constants $ m$ , $ n$ , $ p$ so that $ m+n\ne 0$ .

Then the geometrical locus of the point $ X$ for which $ m\cdot XB^2+n\cdot XD^2=p$ is a circle $ C(O)$ , where $ O\in BD$ .

Proof. Denote $ \left\{\begin{array}{c} m=DC^2-DA^2\ ;\ n=BA^2-BC^2\\\\ p=B^2\cdot CD^2-AD^2\cdot BC^2\\\\ f(X)=m\cdot XB^2+n\cdot XD^2\end{array}\right\|$ . Observe that $ \left\{\begin{array}{c} AC\not\perp BD\Longleftrightarrow AB^2+CD^2\ne AD^2+BC^2\Longleftrightarrow m+n\ne 0\ .\\\\ f(X)=fC)=f(A)=f(R)=p\ .\end{array}\right\|$

Using the upper lemma results that $ ACRX$ is cyclically. Remark. $ XA=XR$ $ \implies$ the ray $ [CX$ is the bisector of the angle $ \widehat {ACR}$ .

Particular case. $ AB\cdot CD=AD\cdot BC$ , i.e. $ \frac {BA}{BC}=\frac {DA}{DC}$ . The relation $ (*)$ becomes $ \frac {XB^2}{XD^2}=$ $\frac {BC^2-BA^2}{DC^2-DA^2}=$

$\left(\frac {BC}{DC}\right)^2\cdot\frac {1-\left(\frac {BA}{BC}\right)^2}{1-\left(\frac {DA}{DC}\right)^2}$ $ \Longrightarrow$ $ \frac {XB}{XD}=\frac {BC}{DC}$ , i.e. the ray $ [CX$ is the bisector of the angle $ \widehat {ACR}$ .Thus, $ \widehat {RCB}\equiv\widehat {ACD}$ .

Remark. See the particular case $ AB\cdot CD = AD\cdot BC$ , i.e. $ \frac {XB}{CB} = \frac {XD}{CD}$ at http://www.artofproblemsolving.com/Forum/viewtopic.php?t=63420


PP4. The $ A$ - exincircle $ w_a = C(I_a,r_a)$ of $ \triangle ABC$ is tangent to $ AB$ , $ AC$ at the points $ B'$ , $ C'$ . Denote the incircle

$ w = C(I,r)$ of $ \triangle ABC$ and $ \begin{array}{ccc} P\in BI_{a}\cap B'C' & ; & Q\in CI_{a}\cap B'C' \\ \\ M\in PC\cap BQ & ; & L\in BC\ ,\ ML\perp BC\end{array}$ . Prove that $ ML = r$ .

Proof.

$ \blacktriangleright$ The point $ I$ is the orthocenter of the triangle $ I_aI_bI_c$ (evidently). Prove that the quadrilateral $ BPQC$ is cyclically.

Proof.

$ \blacktriangleright$ Denote $ T\in I_aM\cap BC$ . Show that $ \frac {TB}{TC} = \frac {p - c}{p - b}$ , i.e. $ \overline {I_aMT}\perp BC$ (the point $ T\equiv L$ and belongs to the circle $ w_a$ ).

Proof.

$ \blacktriangleright$ Apply lemma I to the triangle $ BCI_a$ , where $ I_aL\perp BC$ and for the point $ M\in I_aL$ the quadrilateral $ BPQC$

is cyclically. In conclusion, the point $ M$ is the orthocenter of $ \triangle BCI_a$ , i.e. $ CM\perp BI_a$ and $ BM\perp CI_a$ .

$ \blacktriangleright$ Apply lemma II to $ \triangle I_aI_bI_c$ with the orthocenter $ I$ , where $ M$ is the orthocenter of $ \triangle BCI_a$ .

In conclusion, $ ML$ is equally to the inradius of the orthic $ \triangle ABC$ w.r.t. the triangle $ I_aI_bI_c$ , i.e. $ ML = r$ .


PP5. Let $ ABC$ be a triangle with the circumcircle $ w = C(O,R)$ and let $ T$ , $ X$ be the intersections of $ BC$ , $ w$

respectively with the bisector of the angle $ \angle A$ . Through the vertex $ C$ we draw the perpendicular line to $ BC$

which intersects the external bisector of $ \angle A$ in the point $ P$ . Denote $ K\in OA\cap BP$ . Prove that $ KT\perp BC$ .

Proof I (metric). Suppose w.l.o.g. $ b > c$ . Observe that $ \left\|\begin{array}{c} m(\angle KAB) = 90^{\circ} - C \\ \\ m(\angle KAP) = 90^{\circ} - \frac {B - C}{2} \\ \\ m\left(\angle APC\right) = 90^{\circ} - \frac {B - C}{2}\end{array}\right\|$ . Denote $ \left\|\begin{array}{c} D\in BC \\ \ AD\perp BC\end{array}\right\|$ . Therefore,

$ \frac {KB}{KP} = \frac {AB}{AP}\cdot\frac {\sin\widehat {KAB}}{\sin\widehat {KAP}} =$ $ \left|\frac {c\cdot\sin \left(90^{\circ}-C\right)}{AP\cdot \cos\frac {B-C}2}\right|=$ $ \frac {c\cdot |\cos C|}{DC} = \frac cb$ . In conclusion, $ \frac {KB}{KP} = \frac {TB}{TC}$ $ \implies$ $ KT\parallel PC$ $ \implies$ $ KT\perp BC$ .

Proof II (synthetic). Denote $ L\in BC\cap AP$ , the diameter $ [XY]$ , the line $ d$ for which $ K\in d$ , $ d\parallel PC$, i.e. $ d\perp BC$ and the points $ U\in AP\cap d$ ,

$ V\in PT\cap LK$ , $ W\in LK\cap PC$ , $ T_1\in d\cap AX$ , $ T_2\in d\cap BC$ . Observe that $ KU=KT_1$ because $ K\in AO$ - the $ A$-median in $ \triangle AXY$

and $ \overline {UKT_1}\parallel \overline {XOY}$ . Thus, the division $ \{B,C;L,T\}$ is harmonically $ \Longleftrightarrow$ the pencil $ P\{B,C;L,T\}$ is harmonically $ \Longleftrightarrow$ a.s.o.


PP6. In an acute-angled triangle $ABC$ denote the midpoint $M$ of side $[BC]$ and the feet $D$ , $E$ , $F$ of the altitudes from $A$ , $B$ , $C$

respectively. Let $H$ be the orthocenter of $\Delta ABC$ , let $S$ be the midpoint of $[AH]$ , let $G\in EF\cap AH$ and let $N$ be the intersection

of the median $AM$ with the circumcircle of $\Delta BCH$ so that $BC$ doesn't separate $A$ , $N$ . Prove that $\widehat{HMA}\equiv\widehat{GNS}$ .

Proof 1. Denote $L\in EF\cap BC$ . From an well-known property obtain that $LH\perp AM$ . Denote $N'\in LH\cap AM$ . Therefore,

$AF\cdot AB=AH\cdot AD=AN'\cdot AM\implies $ $AF\cdot AB=AN'\cdot AM\implies BFN'M$ is cyclically $\implies m\left(\widehat{BN'M}\right)=$ $m\left(\widehat{BFM}\right)=B$ .

$AE\cdot AC=AH\cdot AD=AN'\cdot AM\implies $ $AE\cdot AC=AN'\cdot AM\implies CEN'M$ is cyclically $\implies m\left(\widehat{CN'M}\right)=$ $m\left(\widehat{CEM}\right)=C$ .

Therefore, $m\left(\widehat{BN'C}\right)=m\left(\widehat{BN'M}\right)+m\left(\widehat{CN'M}\right)=B+C=180^{\circ}-A\implies$ $BHN'C$ is cyclically $\implies N'\equiv N$ . Since the division

$(A,G,H,D)$ is harmonically and $NH\perp NA$ obtain that $\widehat{HNG}\equiv\widehat{HND}$ . Since the quadrilateral $DHNM$ is cyclically obtain that $\widehat{HND}\equiv \widehat{HMD}$ .

Thus, $\widehat{HNG}\equiv\widehat{HMD}$ . In conclusion, $m\left(\widehat{HMA}\right) =90^{\circ}-m\left(\widehat{DAM}\right)-m\left(\widehat{DMH}\right)=$ $90^{\circ}-m\left(\widehat{ANS}\right)-m\left(\widehat{HND}\right)=$

$90^{\circ}-m\left(\widehat{ANS}\right)-m\left(\widehat{HNG}\right)=m\left(\widehat{GNS}\right)$ $\implies$ $\widehat{HMA}\equiv \widehat{GNS}$ .

Proof 2. Observe taht $BCEF$ is cyclic and $M$ is the orthocenter of $\triangle AHK$ . Denote $N\in HK\cap AM$ . Since the division $(K,D;C,B)$ is harmonically and

$NMDH$ is cyclic obtain that $KH\cdot KN=KD\cdot KM=KC\cdot KB$ . Then $BCHN$ is cyclic. yje point $N$ is that of the problem. $m(\angle GNH)=$

$m(\angle HND)$ since the division $(D,G;H,A)$ is harmonically and $NH\perp NA$ . Since $SN=SH$ and $\widehat{SNH}\equiv\widehat{SHN}$ obtain that $m(\angle SNH)=$

$m(\angle SNG+$ $\angle GNH)$ and $m(\angle SHN)=$ $m(\angle HND)+$ $m(\angle HDN)=$ $m(\angle HND)+(\angle HMA)$ . Thus, $m(\angle SNG)=m(\angle HMA)$ .


PP7. Let $ABC$ be an acute triangle. Denote the midpoint $M$ of the side $[AB]$ . The bisectors of segments $[CB]$ ,

$[CA]$ intersect $CM$ at $Q$ , $P$ respectively. denote $I\in BQ\cap AP$ intersects $AP$ . Prove that $\widehat{ACI}\equiv\widehat{BCM}$ .

Proof. Let $D$ be the reflection of $C$ about $M$. Lines $AP,BQ$ cut $DB,DA$ at $U,V$ respectively, then $BCVD$ and $ACUD$ are isosceles trapezoids with legs

$CV=BD$ and $CU=AD,$ i.e. $\triangle ACV \sim \triangle BCU$ are isosceles with common apex $C$ $\Longrightarrow$ $\angle ACV=\angle BCU$. If the tangents to the circumcircle

of $\triangle ABC$ through $A,B$ intersect at $E,$ then we have $\angle CAV=\angle BAE=\angle C$ and $\angle CBU=\angle ABE=\angle C.$ Thus, by the Jacobi's theorem, it follows

that lines $CE,AU,BV$ concur at $I$ $\Longrightarrow$ $CI\equiv CE$ is the C-symmedian of $\triangle ABC$ $\Longrightarrow$ $\angle ACI=\angle BCM.$


PP8. In isosceles triangle $ ABC\ ,\ (CA = CB)$ consider the point $ M\in (AB)$ so that $ MA = 2\cdot MB$ , the midpoint $ F$

of the side $ [BC]$ and the orthogonal projection $ H$ of the point $ M$ on the line $ AF$ . Prove that $ \angle BHF = \angle ABC$ .

Proof. Denote the midpoint $ L$ of the side $ [AB]$ , the centroid $ G\in AF\cap CL$ of the triangle $ ABC$ , the midpoint $ N$ of the segment $ [AM]$ and the point $ D\in AF$

for which $ DB\perp AB$ . Thus, $ LN=LM$ , $ LA=LB$ , $ AN =$ $ NM =$ $ MB=$ $ 2\cdot NL$ , $ AG = GD$ , $ GC=$ $ DB=$ $ 2\cdot GL$ , $ AC\parallel GN\parallel DM$

and the quadrilateral $ MHDB$ is cyclically. Therefore, $ \widehat {BHF}\equiv$ $ \widehat {BHD}\equiv$ $ \widehat {BMD}\equiv$ $ \widehat {BAC}\equiv$ $ \widehat {ABC}$ , i.e. $ \widehat {BHF}\equiv\widehat {ABC}$ .

Remark. Observe that $ \frac {MB}{ML} = \frac {AB}{AL} = 2$ , i.e. the points $ A$ , $ M$ are harmonically conjugate w.r.t. the pair $ \{L,B\}$ .

Since $ HA\perp HM$ obtain $ \widehat {MHB}\equiv\widehat {MHL}$ . Thus, $ \widehat {LHA}\equiv\widehat {BHD}\equiv\widehat {FBL}$ , i.e. the quadrilateral $ LHFB$ is cyclically.


PP9 (Serbian JBMO TST, 2011). Let $ABC$ be a right triangle, $CA\perp CB$ and $a<b<2a$. Given are the points $M\in AC$ and

$N\in BC$ so that $AM=BC$ and $BN=MC$ . Denote the intersection $K\in AN\cap BM$. Find the value of the angle $\widehat{AKM}\ .$

Proof 1. Let $P$ , $Q$ such that $ACBQ$ is a rectangle and $PB=BC=AM=AQ\implies PQ=MC=BN$ . Then $\triangle CBM\cong$

$\triangle BPN\cong$ $\triangle QAP$ $\implies$ $PA=PN$ and $\angle APN=90^\circ$ . Thus, $AMBP$ is a parallelogram $\implies$ $\angle MKA=\angle KAP=45^\circ$ .

A preliminary metrical study. Prove easily that : $AM=a$ ; $NB=MC=b-a$ ; $CN=2a-b$ ;

$BM^{2}=$ $a^{2}+(b-a)^{2}\ ;\ AN=\sqrt 2\cdot BM\ .$ I'll establish the position of the point $K$ on the segments $[BM]$ and $[AN]$ .

Apply the Menelaus' theorem to the transversals $\overline{AKN}$, $\overline{BKM}$ for the triangles $BCM$ , $ACN$ respectively :

$\blacktriangleright\frac{AM}{AC}\cdot\frac{NC}{NB}\cdot\frac{KB}{KM}=1$ $\Longrightarrow$ $\frac{a}{b}\cdot\frac{2a-b}{b-a}\cdot\frac{KB}{KM}=1$ $\Longrightarrow$ $\Longrightarrow$ $\boxed{\ \frac{KB}{b(b-a)}=\frac{KM}{a(2a-b)}=\frac{1}{BM}\ }\ \ (1)\ .$

$\blacktriangleright\frac{BN}{BC}\cdot \frac{MC}{MA}\cdot\frac{KA}{KN}=1$ $\Longrightarrow$ $\frac{b-a}{a}\cdot\frac{b-a}{a}\cdot\frac{KA}{KN}=1$ $\Longrightarrow$ $\Longrightarrow$ $\boxed{\ \frac{KA}{a^{2}}=\frac{KN}{(a-b)^{2}}=\frac{\sqrt 2}{BM}\ }\ \ (2)\ .$

If denote $m(\widehat{AKM})=\phi$ , $m(\widehat{CBM})=u$ , $m(\widehat{ANC})=v$ , then $\phi =v-u\ .$

Proof 2.1 (metric). Apply the generalized Pythagoras' theorem to the side $AM$ in the triangle $AKM$ :

$2\cdot KA\cdot KM\cdot\cos \phi =KM^{2}+KA^{2}-AM^{2}=$ $2\cdot\frac{a(2a-b)}{BM}\cdot \frac{a^{2}\sqrt 2}{BM}\cdot\cos\phi =\frac{a^{2}(2a-b)^{2}}{BM^{2}}+\frac{2a^{4}}{BM^{2}}-a^{4}$ $\Longrightarrow$

$2\sqrt 2\cdot a(2a-b)\cos\phi=$ $(2a-b)^{2}+2a^{2}-BM^{2}=$ $2a(2a-b)$ $\Longrightarrow$ $\cos\phi =\frac{\sqrt 2}{2}$ $\Longrightarrow$ $\boxed{\ \phi=45^{\circ}\ }\ .$

Proof 2.2 (slight metric). Denote the intersection $P$ between the bisector of the angle $\widehat{BCM}$ and the line $BM$ .

Apply the theorem of the angle bisector to the ray $[CP$ in the triangle $BCM\ :\ \frac{PB}{CB}=\frac{PM}{CM}$ $\Longrightarrow$

$\frac{PB}{a}=\frac{PM}{b-a}=\frac{BM}{b}$ $\Longrightarrow$ $\boxed{\ BP=\frac{a}{b}\cdot BM\ }\ .$ Therefore, $\boxed{\ \phi=45^{\circ}\ }$ $\Longleftrightarrow$ the quadrilateral $CKPN$ is cyclically $\Longleftrightarrow$

$BP\cdot BK=BN\cdot BC$ $\Longleftrightarrow$ $\frac{a}{b}\cdot BM\cdot BK=a(b-a)$ $\Longleftrightarrow$ $BM\cdot BK=b(b-a)$ , what is truly - see the relation $(1)\ .$

Remark. Let $R$ be the middlepoint of the segment $[AN]$ . Then $\triangle ACN\sim\triangle BRM$ , where $\frac{AN}{BM}=\sqrt 2\ .$ Particularly,

$\boxed{\ RB\perp RM\ \ \wedge\ \ CP\parallel MR\ }$ . Let $S\in BC$ and $T\in BM$ so that $C\in (BS)$, $CS=BN$ and $ST\perp BM$. Then $\boxed{\ BK=ST\ }$ .

Proof 3.1 (slight trigonometric). Apply the Sinus' theorem in the triangle $AMK$ :

$\frac{a}{\sin \phi }=\frac{AM}{\sin\phi }=\frac{AK}{\sin (90^{\circ}+u)}=\frac{a^{2}\sqrt 2}{BM\cos u}=\frac{a^{2}\sqrt 2}{a}=a\sqrt 2$ $\Longrightarrow$ $\Longrightarrow$ $\boxed{\ \phi=45^{\circ}\ }\ .$

Proof 3.2 (trigonometric).[/size] $\tan u=\frac{CM}{CB}=\frac{b-a}{a}$ , $\tan v=\frac{CA}{CN}=\frac{b}{2a-b}$ and $\tan \phi=\tan (v-u)=$

$\frac{\tan v-\tan u}{1+\tan u\tan v}=$ $\frac{\frac{b}{2a-b}-\frac{b-a}{a}}{1+\frac{b}{2a-b}\cdot\frac{b-a}{a}}=$ $\frac{ab+2a^{2}-3ab+b^{2}}{2a^{2}-2ab+b^{2}}=1\ .$ Therefore, $\boxed{\ \phi =45^{\circ}\ }\ .$ In the my opinion, this proof is very quickly !

Proof 3.3 (trigonometric).[/size] $\sin u=\frac{b-a}{BM}$ , $\cos u=\frac{a}{BM}$ , $\sin v=\frac{b}{AN}$ , $\cos v=\frac{2a-b}{AN}\ .$ Thus, $\sin \phi =\sin (v-u)=$

$\sin v\cos u-\sin u\cos v=$ $\frac{b}{AN}\cdot\frac{a}{BM}-\frac{b-a}{BM}\cdot \frac{2a-b}{AN}=$ $\frac{2a^{2}-2ab+b^{2}}{AN\cdot BM}=\frac{BM^{2}}{AN\cdot BM}=\frac{BM}{AN}=\frac{\sqrt 2}{2}$ , i.e. $\boxed{\ \phi =45^{\circ}\ }\ .$

Proof 4 (analytic).[/size] In an analytical system denote $C(0,0)$ , $B(a,0)$ , $A(0,b)$ where $a<b<2a\ .$ Thus, the slope of the line $BM$ is

$m=\frac{a-b}{a}$ and the slope of the line $AN$ is $n=\frac{b}{b-2a}$ . Therefore, $\tan \phi =\left|\frac{m-n}{1+mn}\right|=$ $\left|\frac{\frac{a-b}{a}-\frac{b}{b-2a}}{1+\frac{a-b}{a}\cdot\frac{b}{b-2a}}\right|=1$ , i.e. $\boxed{\ \phi =45^{\circ}\ }\ .$


PP10. Fie $\triangle ABC$ neisoscel si $E\in(AB)$ , $D\in(AC)$ astfel incat $BCDE$ este inscriptibil. Notam $O\in EC\cap BD$ si mijloacele $N$ , $P$ pentru

$(DE)$ , $(BC)$ respectiv. Sa se arate ca $AO$ este tangenta la cercul circumscris pentru $\triangle PON$ (concurs Nicolae Coculescu, baraj seniori).

Demonstratie. Fie mijlocul $M$ al lui $[AO]$ . Atunci $P\in MN$ . Omotetizam: $OR=2ON$ , $OQ=2OP$ ; $A\in RQ$ . Deoarece $BQCO$ este paralelogram, atunci $CQ=BO$ . Cum $\triangle EOD\sim\triangle BOC$

si $\triangle ADE\sim\triangle ABC$ , deducem $\displaystyle\frac{CQ}{EO}=\frac{BO}{EO}=\frac{BC}{ED}=\frac{AC}{AE}$ . Din $\angle AEO\equiv\angle ACQ$ reiese $\triangle AEO\sim\triangle ACQ$ . Deci $\angle AOE\equiv\angle AQC$ . Deoarece $\angle CQP\equiv\angle BOP\equiv\angle EON$ ,

obtinem $\angle AON\equiv\angle OQA\equiv\angle OPN$ , de unde rezulta concluzia.


PP11. Let an $A$-right $\triangle ABC$ with the length $r$ of its inradius. Denote $\left\{\begin{array}{ccc} H\in BC & ; & AH\perp BC\\\\ D\in (HC) & ; & \widehat{DAH}\equiv\widehat{DAC}\\\\ P\in (AB) & ; & \mathrm m\left(\widehat {ADP}\right)=\phi\end{array}\right\|$ . Prove that $AP=2r\iff AP=AS\iff \phi =45^{\circ}$ , where $S\in DP\cap AH$ .

Proof. $\left\{\begin{array}{ccc} \mathrm m\left(\widehat {BDA}\right)=\mathrm m\left(\widehat {DAC}\right)+\mathrm m\left(\widehat {DCA}\right) & \implies & \mathrm m\left(\widehat {BDA}\right)=\frac A2+C\\\\ \mathrm m\left(\widehat {BAD}\right)=\mathrm m\left(\widehat {BAH}\right)+\mathrm m\left(\widehat {HAD}\right) & \implies & \mathrm m\left(\widehat {BAD}\right)=C+\frac A2\end{array}\right\|\implies$ $\widehat {BDA}\equiv \widehat {BAD}\implies$ $\triangle ABD$ is $B$-isosceles, i.e. $DB=c\ ,\ DC=a-c$ and $DH=HC-DC=$

$\frac {b^2}a-(a-c)=$ $\frac {b^2-a^2+ac}a=\frac {ac-c^2}a\implies$ $\boxed{DH=\frac {c(a-c)}a}\ (*)$ . Apply the Menelaus' theorem to the transversal $\overline{DSP}/\triangle ABH\ :\ \frac {DH}{DB}$ $\cdot\frac {PB}{PA}\cdot\frac {SA}{SH}=$ $1\ \stackrel{(*)}{\iff}\ \boxed{\frac {SH}{SA}=\frac {a-c}a\cdot\frac {PB}{PA}}\ (1)$ .

Therefore, $AP=2r\ \stackrel{(1)}{\iff}\ \frac {SH}{SA}=$ $\frac {a-c}a\cdot \frac {c-(b+c-a)}{b+c-a}$ $\iff$ $\frac {SH}{(a-c)(a-b)}=\frac {SA}{a(b+c-a)}=$ $\frac {AH}{(a-c)(a-b)+a(b+c-a)}=$ $\frac {\frac {bc}a}{a^2-a(b+c)+bc+a(b+c)-a^2}=$ $\frac 1a\iff$

$ SA=b+c-a\iff$ $ SA=2r$ . In conclusion, $\boxed{AP=2r\iff AP=AS}$ (the triangle $PAS$ is $A$-isosceles) $\iff$ $\widehat{APS}\equiv\widehat{ASP}\iff$ $2\left(\phi +\frac B2\right)+C=180^{\circ}\iff$ $2\phi =A\iff$ $\boxed{\phi =45^{\circ}}$ .
This post has been edited 58 times. Last edited by Virgil Nicula, Nov 22, 2015, 2:21 PM

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431. Trigonometry in geometry II.
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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