388. Some interesting "slicing" problems.

by Virgil Nicula, Oct 23, 2013, 12:16 PM

PP1. Let $\triangle ABC$ and an interior point $P$ so that $BP\perp AC\ ,\ PB=PC$ and $\left\{\begin{array}{c} m\left(\widehat{PAC}\right)=24\\\\ m\left(\widehat{PCA}\right)=6\end{array}\right\|$ . Denote $D\in BP\cap AC$ and suppose that $D\in (AC)$ . Find $x=m\left(\widehat{PAB}\right)$

Proof 1 (trigonometric). Apply the theorem of Sines in the triangles: $\left\{\begin{array}{cccc} ABC\ : & \frac {AB}{AC}=\frac {\sin C}{\sin B}=\frac {\sin 48^{\circ} }{\sin (108^{\circ}-x)} & \implies & \frac {AB}{AC}=\frac {\sin 48^{\circ}}{\sin (72^{\circ}+x)}\\\\ APB\ : & \frac {BA}{BP}=\frac {\sin \widehat{APB}}{\sin \widehat{PAB}}=\frac {\sin 114^{\circ} }{\sin x} & \implies & \frac {BA}{BP}=\frac {\cos 24^{\circ}}{\sin x}\\\\ APC\ : & \frac {CP}{CA}=\frac {\sin \widehat{PAC}}{\sin \widehat{APC}}=\frac {\sin 24^{\circ} }{\sin 150^{\circ}} & \implies & \frac {CP}{CA}=\frac {\sin 24^{\circ}}{\sin 30^{\circ}}\end{array}\right\|\ (*)$ . Observe that $PB=PC\iff$

$\frac {AB}{AC}=\frac {BA}{BP}\cdot\frac {CP}{CA}\stackrel{(*)}{\iff}$ $\frac {\sin 48^{\circ}}{\sin (72^{\circ}+x)}=\frac {\cos 24^{\circ}}{\sin x}\cdot \frac {\sin 24^{\circ}}{\sin 30^{\circ}}\iff$ $\sin \left(72^{\circ}+x\right)=\sin x\iff$ $\left(72^{\circ}+x\right)+x=180^{\circ}\iff$ $2x=108^{\circ}\iff$ $x=54^{\circ}$ .

PP2 (billiards). Let an $A$ -right-angled $\triangle ABC$ and two points $P\in (AC)\ ,\ Q\in (BC)$ so that $\left\{\begin{array}{c} \widehat {APB}\equiv\widehat{CPQ}\\\\ \widehat{BQA}\equiv\widehat{CQP}\end{array}\right\|$ . Prove that $\tan\widehat{BQA}=3\cot C$ .

Proof. Let $\left\{\begin{array}{c} m\left(\widehat {APB}\right)=m\left(\widehat{CPQ}\right)=y\\\\ m\left(\widehat{BQA}\right)=m\left(\widehat{CQP}\right)=x\\\\ (\ x+y+C=\pi\ )\end{array}\right\|$ and $\boxed{\frac {QP}{QC}=\frac {\sin C}{\sin y}=\frac {BP}{BC}}\ (*)$ . Apply an well-known relation:

$\left\{\begin{array}{ccc} \frac {AP}{AC}=\frac {BP}{BC}\cdot\frac {\sin\widehat{ABP}}{\sin\widehat{ABC}} & \implies & \frac {AP}{AC}=\frac {BP}{BC}\cdot\frac {\cos y}{\cos C}\\\\ \frac {AP}{AC}=\frac {QP}{QC}\cdot\frac {\sin\widehat{AQP}}{\sin\widehat{AQC}} & \implies & \frac {AP}{AC}=\frac {QP}{QC}\cdot\frac {\sin(C+y-x)}{\sin x}\end{array}\right\|\stackrel{(*)}{\implies}$ $\frac {\cos y}{\cos C}=\frac {\sin(C+y-x)}{\sin x}\iff$ $\cos y\sin x=\cos C\sin(C+y-x)\iff$

$\cos (x+C)\sin x+\cos C\sin 2x=0\iff$ $\cos (x+C)+2\cos x\cos C=0\iff$ $\boxed{\ \tan x\tan C=3\ }$ .

PP3. Let $ABC$ be an acute triangle with the circumcircle $w=C(O,R)$ and the orthocenter $H$ . Prove that $\boxed{A=60^{\circ}\ \implies\ OH=|b-c|\ }$ .

Proof 1. Suppose w.l.o.g. $c<b$ . Denote the second intersection $S$ of the $A$ -bisector with $w$ and $M\in OH\cap AB\ ,\ N\in OH\cap AC$ . . Observe that

$\left\{\begin{array}{ccccc} AH=2R\cos A & \implies & AH=R & \implies & AH=AO\\\\ \widehat{HAB}\equiv\widehat{OAC} & \implies & AS\perp OH & \implies & AM=AN\end{array}\right\|\implies$ $\triangle MAN$ is equilateral. Let $OH=x$ . Since $\left\{\begin{array}{ccccc} m\left(\widehat{BHC}\right) & = & 180^{\circ}-A & = & 120^{\circ}\\\\ m\left(\widehat{BOC}\right) & = & 2A & = & 120^{\circ}\end{array}\right\|$

obtain that the quadrilateral $BHOC$ is cyclically. Prove easily that $HM=MB=ON=m$ and $HN=NC=m+x$ . In conclusion,

$AM=MN=NA\iff$ $c-m=2m+x=b-(m+x)\iff$ $\left\{\begin{array}{c} x=b-c\\\\ 3m=2c-b\end{array}\right\|\implies$ $x=b-c\implies OH=|b-c|$ .

Proof 2. If $A=60^{\circ}$ , then $a^2=b^2+c^2-bc=3R^2$ and $OH^2=9OG^2 = 9R^2-\left(a^2+b^2+c^2\right) =$ $3a^2-\left(a^2+b^2+c^2\right) =$

$2a^2-\left(b^2+c^2\right) = 2\left(b^2+c^2-bc\right) - \left(b^2+c^2\right) =$ $b^2+c^2-2bc = (b-c)^2\implies$ $OH = |b-c|$ .

Proof 3. Suppose w.l.o.g. $c<b$ . Prove easily that the quadrilateral $BHOC$ is cyclically. Let $L\in w$ for which $OL\parallel HC$ . Observe that $LC=OH\ ,\ HB=OL$ and

$HL=OB=OC$ . Apply the Ptolemy's theorem to $HOLC\ :$ $HO\cdot LC+OL\cdot HC=OC\cdot HL\iff$ $\boxed{HO^2+HB\cdot HC=R^2}$ .Therefore, $a^2=b^2+c^2-bc=3R^2$

and $\left\{\begin{array}{ccccc} 2b\cos A=b & \implies & 2(c-a\cos B)=b & \implies & \cos B=\frac {2c-b}{2a}\\\\ 2c\cos A=c & \implies & 2(b-a\cos C)=c & \implies & \cos C=\frac {2b-c}{2a}\end{array}\right\|\ (*)$ . Thus, $HO^2+HB\cdot HC=R^2\iff$ $HO^2+4R^2\cos B\cos C=R^2\stackrel{(*)}{\implies}$

$3HO^2+4a^2\cdot\frac {2c-b}{2a}\cdot\frac {2b-c}{2a}=a^2\iff$ $3HO^2=\left(b^2+c^2-bc\right)-(2c-b)(2b-c)\iff$ $HO=|b-c|$ .

PP4. Three chords $AA_1$ , $BB_1$ , $CC_1$ of the circle $w=C(O)$ (with the order on the circle $A$ , $B_1$ , $C$ , $A_1$ , $B$ , $C_1$ ) pass

through a common point $K$ and form six angles of $60^{\circ}$ . Prove that $KA + KB + KC = KA_1 + KB_1 + KC_1$ .

Proof (Yetty). Common internal bisector $AKA_1$ of $\angle CKB, \angle C_1KB_1$ meets circumcircles of $BKC$ , $B_1KC_1$ again at $Z$ , $Z_1$ respectively. Observe that $ZB = ZC$

and $m(\angle BZC) =60^\circ$ $\Longrightarrow$ $\triangle BZC$ is equilateral and similarly $\triangle B_1Z_1C_1$ is equilateral. Apply the Ptolemy's theorem $\left\{\begin{array}{ccc} KBZC & : & KZ = KB + KC\\\\ KB_1Z_1C_1 & : & KZ_1 = KB_1+KC_1\end{array}\right\|$ .

Perpendicular bisectors of $BC$ , $B_1C_1$ through $Z$ , $Z_1$ respectively meet at $O$ . Since $KBCZ \sim KB_1Z_1C_1$ are similar $\widehat{OZK}\equiv \widehat{OZ_1K}$ $\Longrightarrow$ $\triangle OZZ_1$ is $O$ -isosceles,

just like $\triangle OAA_1$ $\Longrightarrow$ $\boxed{KA + KB + KC = ZK + KA = ZA = Z_1A_1 = Z_1K + KA_1 = KA_1 + KB_1 + KC_1}$ .

PP5. Let a convex $ABCD$ so that $AB=BD=DC$ and $\left\{\begin{array}{c} m\left(\widehat{ABD}\right)=36^{\circ}\\\\ m\left(\widehat{BDC}\right)=24^{\circ}\end{array}\right\|$ and $X\in AC\cap BD$ . Prove that $m\left(\widehat{AXD}\right)=54^{\circ}$ .

Proof. Suppose w.l.o.g. $AB=1$ and denote $\left\{\begin{array}{c} m\left(\widehat{AXD}\right)=x\\\\ m\left(\widehat{BAC}\right)=u\\\\ m\left(\widehat{ACD}\right)=v\end{array}\right\|$ . Observe that $x=u+36^{\circ}=v+24^{\circ}$ . Apply theorem of Sines in $\triangle AXB$ and $\triangle CXD\ :$

$\frac {XB}{\sin u}=$ $\frac 1{\sin x}=\frac {XD}{\sin v}=$ $\frac {XB+XD}{\sin u+\sin v}=\frac 1{\sin u+\sin v}$ . In conclusion, $\left\{\begin{array}{c} u+36^{\circ}=v+24^{\circ}=x\\\\ \sin u+\sin v=\sin x\end{array}\right\|\implies$ $\sin (v-12^{\circ})+\sin v=\sin(v+24^{\circ})\implies$

$\sin (24^{\circ}+v)-\sin (v-12^{\circ})=\sin v\implies$ $2\sin 18^{\circ}\cos(v+6^{\circ})=\sin v\implies$ $\cos (v+6^{\circ})=2\cos 36^{\circ}\sin v\implies$ $\cos (v+6^{\circ})=\sin (v+36^{\circ})-\sin (36^{\circ}-v)\implies$

$\sin (v+36^{\circ})-\sin (84^{\circ}-v)=\sin (36^{\circ}-v)\implies$ $2\sin (v-24^{\circ})\cos 60^{\circ}=\sin (36^{\circ}-v)\implies$ $\sin (v-24^{\circ})=\sin (36^{\circ}-v)\implies$ $v-24^{\circ}=36^{\circ}-v\implies$

$v=30^{\circ}\implies$ $x=v+24^{\circ}=54^{\circ}\implies$ $m\left(\widehat{AXD}\right)=54^{\circ}$ . I used the well-known relation $4\sin 18^{\circ}\cos 36^{\circ}=1\iff$ $\sin 72^{\circ}=\cos 18^{\circ}$ .

PP6. Let $\triangle ABC$ with $A=50^{\circ}$ and an interior $P$ , $E\in PB\cap AC$ and $F\in PC\cap AB$ . Suppose that $PB=PC$ , $m\left(\widehat{PEF}\right)=50^{\circ}$ and $m\left(\widehat{PFE}\right)=30^{\circ}$ . Find $m\left(\widehat {ACB}\right)$ .

Proof 1. Let $x=\angle{PBC}=\angle{PCB}$ . Then $\left\{\begin{array}{c} \angle{AFE}=50^\circ+B-x\\\\ \angle {AEF}=30^\circ+C-x\end{array}\right\|$ . It is given that $C>x$ and $B>x\ .$ Thus, from $\triangle{AFE}$ we get that $x=40^\circ$ and $B+C=130^\circ\ .$

Apply the law of Sines $:\ \left\{\begin{array}{cccc} \triangle{FEC}\ : & \frac{FE}{CE} & = & 2\sin (C-40^{\circ})\\\\ \triangle{BEC}\ : & \frac {CE}{BE} & = & \frac {\sin{40^\circ}}{\sin C}\\\\ \triangle{BFC}\ : & \frac {BE}{FE} & = & \frac {\sin(C+40^\circ)}{\cos C}\end{array}\right\|\ \bigodot$ $\implies$ $\sin C\cos C=2\sin{40^\circ}\sin (C-40^\circ)\sin (C+40^\circ)$ $\implies\ C=60^\circ\ .$

PP7 (Julio Orihuela,Peru). Let a convex quadrilateral $ABCD$ with $\left\{\begin{array}{cc} m\left(\widehat{CAB}\right)=10^{\circ}\ ; & m\left(\widehat{CAD}\right)=20^{\circ}\\\\ m\left(\widehat{ADB}\right)=100^{\circ}\ ; & m\left(\widehat{ACD}\right)=40^{\circ}\end{array}\right\|$ . Find $x=m\left(\widehat{ACB}\right)$ .

Proof 1. Law of Sines $:\ \left\{\begin{array}{cc} \triangle ABD\ : & \frac {BA}{BD}=\frac {\sin\widehat{BDA}}{\sin\widehat{BAD}}=\frac {\sin 100^{\circ}}{\sin 30^{\circ}}\\\\ \triangle DBC\ : & \frac {BD}{BC}=\frac {\sin\widehat{BCD}}{\sin\widehat{BDC}}=\frac {\sin (x+40^{\circ})}{\sin 20^{\circ}}\\\\ \triangle CBA\ : & \frac {BC}{BA}=\frac {\sin\widehat{BAC}}{\sin\widehat{BCA}}=\frac {\sin 10^{\circ}}{\sin x}\end{array}\right\|\bigodot\iff$ $2\cos 10^{\circ}\sin (x+40^{\circ})\sin 10^{\circ}=\sin 20^{\circ}\sin x\iff$ $\sin (x+40^{\circ})=\sin x\iff$ $\boxed{x=70^{\circ}}$ .

Proof 2. Law of Sines $:\ \left\{\begin{array}{cc} \triangle ABD\ : & \frac {AD}{BD}=\frac {\sin 50^{\circ}}{\sin 30^{\circ}}\\\\ \triangle BCD\ : & \frac {BD}{CD}=\frac {\sin (x+40^{\circ})}{\sin (x+60^{\circ})}\\\\ \triangle ACD\ : & \frac {CD}{AD}=\frac {\sin 20^{\circ}}{\sin 40^{\circ}}\end{array}\right\|\bigodot\iff$ $2\sin 50^{\circ}\sin (x+40^{\circ})\sin 20^{\circ}=\sin (x+60^{\circ})\sin 40^{\circ}\iff$ $\sin 50^{\circ}\sin (x+40^{\circ})=$

$\sin (x+60^{\circ})\cos 20^{\circ}\iff$ $\cos (x-10)-\cos(x+90)=\sin (x+80)+\sin (x+40)\iff$ $\sin x=\sin (x+40)\iff$ $x+(x+40)=180\iff$ $\boxed{x=70}$ .

Proof 3. $\left\{\begin{array}{cc} m\left(\widehat{BDC}\right)=20^{\circ}\ ; & m\left(\widehat{ABD}\right)=50^{\circ}\\\\ m\left(\widehat{ACB}\right)=x\ ; & m\left(\widehat{CBD}\right)=120^{\circ}-x\end{array}\right\|$ . Apply the trigonometric form of the Ceva's theorem :

$\sin\widehat{DBA}\sin\widehat{CAD}\sin\widehat{BDC}\sin\widehat{ACB}=\sin\widehat{BAC}\sin\widehat{ADB}\sin\widehat{DCA} \sin\widehat{CBD}\iff$ $\sin 50\sin 20\sin 20\sin x=\sin 10 \sin 100\sin 40\sin (120-x)\iff$

$\cos 40\sin^220\sin x=\sin 10\cos 10\sin 40\cos (30-x)\iff$ $2\cos 40\sin 20\sin x=\sin 40\cos (x-30)\iff$ $\cos 40\sin x=\cos 20\cos(x-30)\iff$

$\sin (x+40)+\sin (x-40)=\cos (50-x)+\cos (x-10)\iff$ $\sin (x-40)=\cos (x-10)\iff$ $\cos (130-x)=\cos (x-10)\iff$ $130-x=x-10\iff$ $\boxed{x=70}$ .

Proof 4 (synthetic).

PP8. Let a $B$ -right $\triangle ABC$ let $\{D,E\}\subset (BC)$ so that $D\in (BE)$ and $\widehat{BAD}\equiv\widehat{DAE}\equiv\widehat{EAC}$ . Let $H\in AC$ so that $EH\perp AC$ . Prove that $AD=2\cdot HC\iff$ $C=36^{\circ}$ .

Proof 1. Suppose w.l.o.g. $HC=1$ and let $m\left(\widehat{BAD}\right)=x$ . Thus, $\left\{\begin{array}{cccc} \triangle ABD\ : & AB=AD\cdot \cos\widehat{BAD} & \implies & AB=2\cos x\\\\\ \triangle ABE\ : & AB=AE\cdot \cos\widehat{BAE} & \implies & AE=\frac {2\cos x}{\cos 2x}\\\\ \triangle ABC\ : & AB=AC\cdot \cos\widehat{BAC} & \implies & AC=\frac {2\cos x}{\cos 3x}\\\\ \triangle AEH\ : & AH=AE\cdot \cos\widehat{EAH} & \implies & AH=\frac {2\cos^2x}{\cos 2x}\end{array}\right\|$ . So $AC=HC+HA\iff$

$\frac {2\cos x}{\cos 3x}=1+\frac {2\cos^2x}{\cos 2x}\iff$ $2\cos x\cos 2x=2\cos 2x\cos 3x+\cos 3x\iff$ $\cos x=2\cos 2x\cos 3x\iff$ $\cos x=\cos 5x+\cos x\iff$ $5x=90^{\circ}\iff$ $\boxed{x=18^{\circ}}$ .

Proof 2. Suppose w.l.o.g. $HC=1$ and let $m\left(\widehat{BAD}\right)=x$ $\implies$ $\left\{\begin{array}{c} BE=c\cdot\tan 2x\\\\ BC=c\cdot\tan 3x\end{array}\right\|$ and $CE=BC-BE\implies$ $CE=c(\tan 3x-\tan 2x)$ . Thus, $\boxed{c=2\cos x}\ (*)$ and

$HC=CE\cdot \sin 3x\implies$ $1=c(\tan 3x-\tan 2x)\sin 3x\stackrel{(*)}{\implies}$ $2\cos x\cdot \sin 3x\cdot\frac {\sin x}{\cos 2x\cos 3x}=1\implies$ $\sin 2x\sin 3x=\cos 2x\cos 3x\implies$ $\cos 5x=0\implies$ $\boxed{x=18^{\circ}}$ .

Proof 3 (synthetic).

PP9 (Ruben Huillca). Let $A$ -right $\triangle ABC$ . Suppose exist $\{a,b\}\subset\mathbb R^*_+\ ,\ \left\{\begin{array}{cccc} E\in (AC)\ : & EA=b & , & EC=a+b\\\\ F\in (AB)\ : & FA=a & , & FB=b\end{array}\right\|$ and $P\in BE\cap CF$ . Find $\alpha =m\left(\widehat {CPE}\right)$ .

Proof 1. Let $\left\{\begin{array}{c} m\left(\widehat{ACF}\right)=x\\\\ m\left(\widehat{AEB}\right)=y\end{array}\right\|$ . So $\left\{\begin{array}{ccc} \alpha & = & y-x\\\\ \tan x & = & \frac a{a+2b}\\\\ \tan y & = & \frac {a+b}b\end{array}\right\|$ $\implies\tan \alpha=\tan (y-x)=$ $\frac {\tan y-\tan x}{1+\tan x\tan y}=$ $\frac {\frac {a+b}b-\frac a{a+2b}}{1+\frac a{a+2b}\cdot\frac {a+b}b}=$ $\frac {a^2+3ab+2b^2-ab}{ab+2b^2+a^2+ab}=1\implies$ $\alpha=45^{\circ}$ .

Proof 2. Construct $D$ so that $CD\perp CA\ ,\ CD=b$ and $CF$ separates $D$ , $A$ . So $CDBF$ is a parallelogram $,\ ABE\equiv CED\implies EB=DE$ and $EB\perp ED\implies\alpha =45^{\circ}$ .

PP10 (Ruben Auqui). Show that $\tan\frac {\pi}{24}=\left(\sqrt 2-1\right)\left(\sqrt 3-\sqrt 2\right)$ .

Proof 1. Let $\triangle ABC$ with $A=45^{\circ}$ and $AB=1+\sqrt 3\ ,\ AC=2+\sqrt 2$ . Let $\left\{\begin{array}{ccccc} D\in (AC) & : & DA=\sqrt 2 & ; & DC=2\\\\ E\in (AB) & : & EA=1 & ; & EB=\sqrt 3\end{array}\right\|$ .

Prove easily that $\left\{\begin{array}{c} DE=AE=1\\\\ DB=DC=2\end{array}\right\|$ and $\left\{\begin{array}{c} m\left(\widehat{ABD}\right)=\frac {\pi}{6}\\\\ m\left(\widehat{CBD}\right)=C=\frac {7\pi}{24}\end{array}\right\|\implies$ $B=\frac {11\pi}{24}$ . Let the projection $P$ of $C$ on $AC$ . Thus, $PC=PA=1+\sqrt 2$ ,

$PB=\sqrt 3-\sqrt 2$ and $m\left(\widehat{BCP}\right)=\frac {\pi}{24}$ . Hence $\tan\frac {\pi}{24}=\frac {PB}{PC}=\frac {\sqrt 3-\sqrt 2}{1+\sqrt 2}\implies$ $\tan\frac {\pi}{24}=\left(\sqrt 2-1\right)\left(\sqrt 3-\sqrt 2\right)$ .

Proof 2. $\tan\frac {\pi}{24}=\frac {\sin\frac {\pi}{12}}{1+\cos\frac {\pi}{12}}=$ $\frac {\sqrt 6-\sqrt 2}{4+\sqrt 6+\sqrt 2}=$ $\frac {\sqrt 3-1}{2\sqrt 2+\sqrt 3+1}=$ $\frac {\left(\sqrt 3-1\right)\left[2\sqrt 2-\left(\sqrt 3+1\right)\right]}{8-\left(4+2\sqrt 3\right)}=$

$\frac {\left(\sqrt 3-1\right)\left[2\sqrt 2-\left(\sqrt 3+1\right)\right]}{2\left(2-\sqrt 3\right)}=$ $\frac {\left(2+\sqrt 3\right)\left(\sqrt 3-1\right)\left[2\sqrt 2-\left(\sqrt 3+1\right)\right]}{2}=$ $\sqrt2\left(\sqrt 3+2\right)\left(\sqrt 3-1\right)-\left(2+\sqrt 3\right)=$

$\sqrt 2\left(1+\sqrt 3\right)-\left(2+\sqrt 3\right)=$ $\sqrt 3\left(\sqrt 2-1\right)-\sqrt 2\left(\sqrt 2-1\right)=$ $\left(\sqrt 2-1\right)\left(\sqrt 3-\sqrt 2\right)$ .

PP11. Let $\triangle ABC$ with $C=30^{\circ}$ and there is $D\in (BC)$ so that $AD=BC$ and $\left(\widehat{ADB}\right)=40^{\circ}$ . Prove that $B=100^{\circ}$ .

Proof 1. Apply law of Sines $\odot$ $\begin{array}{ccccc} \nearrow & \triangle ABD\ :\ \frac {AD}{\sin B}=\frac {BD}{\sin\widehat{DAB}} & \implies & DB=\frac {AD\cdot\sin \left(40^{\circ}+B\right)}{\sin B} & \searrow\\\\ \searrow & \triangle ACD\ :\ \frac {AD}{\sin C}=\frac {CD}{\sin\widehat{CAD}} & \implies & DC=AD\cdot 2\sin 10^{\circ} & \nearrow\end{array}\ \bigoplus\implies 1=$ $\frac {\sin \left(40^{\circ}+B\right)}{\sin B}+2\sin 10^{\circ}$ $\iff$

$2\sin 10^{\circ}\sin B+\sin \left(40^{\circ}+B\right)-\sin B=0\iff$ $2\sin 10^{\circ}\sin B+2\sin 20^{\circ}\cos\left(20^{\circ}+B\right)=0\iff$ $\sin B+2\cos 10^{\circ}\cos\left(20^{\circ}+B\right)=0\iff$

$\cos\left(90^{\circ}-B\right)+\cos\left(30^{\circ}+B\right)+\cos\left(10^{\circ}+B\right)=0\iff$ $\cos\left(30^{\circ}-B\right)+\cos\left(10^{\circ}+B\right)=0\iff$ $2\cos 20^{\circ}\cos\left(10^{\circ}-B\right)=0\iff$ $B=100^{\circ}$ .

Proof 2 (slicing - Carlos Abanto). Construct equilateral $\triangle ADE$ , where $AD$ separates $B$ and $E$ . Thus $m\left(\widehat{AED}\right)=2C$ , i.e. $E$ is the circumcenter of $\triangle ADC$ .

Thus $\left\{\begin{array}{c} m\left(\widehat{EAC}\right)=50^{\circ}\\\\ m\left(\widehat{EDC}\right)=80^{\circ}\end{array}\right\|$ and $BC=AD=EA=ED=EC\implies$ $\left\{\begin{array}{ccc} EA=EC & \implies & m\left(\widehat{ECA}\right)=50^{\circ}\\\\ ED=EC & \implies & m\left(\widehat{DEC}\right)=20^{\circ}\\\\ CB=CE & \implies & m\left(\widehat{EBC}\right)=50^{\circ}\end{array}\right\|\implies$ $m\left(\widehat{AEC}\right)=80^{\circ}$

and $\ m\left(\widehat{EBC}\right)=m\left(\widehat{EAC}\right)=50^{\circ}\implies ABCE$ is a cyclical quadrilateral $\implies$ $B+m\left(\widehat {AEC}\right)=180^{\circ}\implies$ $B=100^{\circ}$ .

PP12 (Miguel Ochoa Sanchez). Let $\triangle ABC$ with $M\in (AC)$ so that $AM=BC$ and $\widehat{MBC}\equiv\widehat{BAC}$ . The $C$ -bisector of $\triangle ABC$ cut $AB$ and $BM$ in $N$ , $P$ respectively. Find $\frac {PN}{PC}$ .

Proof. $\left\{\begin{array}{ccc} MC=b-a\ ,\ AM=a\\\\ \frac {NA}b=\frac {NB}a=\frac {c}{a+b}\end{array}\right\|\implies$ $\triangle CBM\sim\triangle CAB\implies$ $\frac {CB}{CA}=\frac {CM}{CB}\implies$ $\frac ab=\frac {b-a}a\implies$ $\boxed{b^2-a^2=ab}\ (*)\ \ \wedge\ \ \boxed{\frac ab=\frac {-1+\sqrt 5}{2}}\ (1)$ . Apply the

Menelaus' theorem to $\overline{BPM}/\triangle ANC\ : \frac {BN}{BA}\cdot\frac {MA}{MC}\cdot\frac {PC}{PN}=1\iff$ $\frac {PN}{PC}=\frac {BN}{BA}\cdot\frac {MA}{MC}=$ $\frac a{a+b}\cdot \frac a{b-a}=\frac {a^2}{b^2-a^2}\stackrel{(*)}{=}\frac {a^2}{ab}=\frac ab\implies$ $\frac {PN}{PC}=\frac ab\ \stackrel{(1)}{=}\ \frac {-1+\sqrt 5}{2}$ .

PP13 Let $\triangle ABC$ and $M$ the midpoint $M$ of $[BC]$ so that $\left\{\begin{array}{c} m\left(\widehat {MAB}\right)=105^{\circ}\\\\ m\left(\widehat {MAC}\right)=30^{\circ}\end{array}\right\|$ . Find the value of the angle $\widehat{ACB}$ .

Proof 1. Let $N$ for which $ACNB$ is a parallelogram, i.e. $M$ is the common midpoint for $[BC]$ , $[AN]$ and the projection $P$ of $N$ on $AC$ . Thus $MA=MN=MP=PN$ ,

i.e. $MNP$ is an equilateral triangle. Since $m\left(\widehat{CNP}\right)=45^{\circ}$ obtain that $PN=PC$ , i.e. $PC=PM$ $\iff$ $m\left(\widehat{PMC}\right)=m\left(\widehat{PCM}\right)=15^{\circ}$ . In conclusion, $m\left(\widehat{ACB}\right)=15^{\circ}$ .

Proof 2. Let $m\left(\widehat{ACB}\right)=x$ . Hence $m\left(\widehat{ABC}\right)=45^{\circ}-x$ . Apply the law of Sines $:\ MB=MC\iff$ $\frac {AM}{MB}=\frac{AM}{MC}\iff$ $\frac {\sin\widehat{ABM}}{\sin\widehat{BAM}}=\frac {\sin\widehat{ACM}}{\sin\widehat{CAM}}\iff$ $\frac {\sin \left(45^{\circ}-x\right)}{\sin 105^{\circ}}=$

$\frac {\sin x}{\sin 30^{\circ}}\iff$ $\sin \left(45^{\circ}-x\right)=2\sin x\cos 15^{\circ}\iff$ $\sin \left(45^{\circ}-x\right)=2\sin x\cos\left(45^{\circ}-30^{\circ}\right)\iff$ $\frac {\sqrt 2}{2}\cdot (\cos x-\sin x)=2\cdot \frac {\sqrt 2}{2}\cdot\left(\frac {\sqrt 3}{2}+\frac 12\right)\cdot \sin x\iff$

$\left(1+\sqrt 3\right)\sin x=\cos x -\sin x\iff$ $\tan x=\frac 1{2+\sqrt 3}=2-\sqrt 3\iff$ $\tan x=2-\sqrt 3\iff$ $x=15^{\circ}$ . Indeed, $\tan 15^{\circ}=\frac {\sin 30^{\circ}}{1+\cos 30^{\circ}}=$ $\frac {\frac 12}{1+\frac {\sqrt 3}2}=$ $\frac 1{2+\sqrt 3}=2-\sqrt 3$

PP14. Let a square $ABCD$ and $M\in (AD)$ , $N\in (BC)$ so that $AM=BN$ . Let $P\in (AC)\ ,\ NP\perp AC$ so that $[APM]=[CPN]$ . Find $m\left(\widehat{APM}\right)$ .

Proof. Suppose that $AB=1$ . Let $:\ \left\{\begin{array}{ccc} R\in (AC) & ; & MR\perp AC\\\\\ AM=BN=x & ; & m\left(\widehat{APM}\right)=\phi\end{array}\right\|$ . Thus $ABNPM$ is cyclic $\implies$ $m\left(\widehat{BAN}\right)=\phi$ and $\tan\phi =\frac {BN}{BA}\implies \boxed{\ \tan\phi =x\ }\ (*)$ .

Prove easily that $[AMP]=\frac {x(x+1)}4$ and $[CNP]=\frac {(x-1)^2}4$ . In conclusion, $[AMP]=[CNP]\iff x(x+1)=$ $(x-1)^2\iff x=$ $\frac 13\ \stackrel{(*)}{\iff}\ \tan\phi =$ $\frac 13$ .

PP15. Let $ABCD$ be a convex quadrilateral with $\angle{DAC}= \angle{BDC}= 36^\circ$ , $\angle{CBD}= 18^\circ$ and $\angle{BAC}= 72^\circ$ . Determine the measure of $\angle{APD}$ , where $P\in AC\cap BD$ .

Lemma. Let $ABCD$ be a convex quadrilateral. Prove that $\left\{\begin{array}{ccc} m(\widehat{CBD})=x & ; & m(\widehat{BDC})=2x\\\\ m(\widehat{CAD})=y & ; & m(\widehat{CAB})=2y\\\\ P\in AC\cap BD & ; & m(\widehat{APD})=z\end{array}\right\|$ $\implies$ $\boxed{\cos (z+x-y)+\cos (z+y-x)+\cos (x+y-z)=0}\ \ (*)$ .

Proof of the lemma. $\left\{\begin{array}{ccc} m(\widehat{ABD})=z-2y & ; & m(\widehat{ACD})=z-2x\\\\ m(\widehat{ACB})=180^{\circ}-x-z & ; & m(\widehat{ADB})=180^{\circ}-y-z\\\\ \sin\widehat{ABD}\sin\widehat{DAC}\sin\widehat{CDB}\sin\widehat{BCA} & = & \sin\widehat{CBD}\sin\widehat{BAC}\sin\widehat{ADB}\sin\widehat{DCA}\end{array}\right\|$ $\Longleftrightarrow$ $\sin (z-2y)\cdot\sin y\cdot\sin 2x\cdot\sin (z+x)=$

$\sin x\cdot\sin 2y\cdot\sin (z+y)\cdot\sin (z-2x)$ $\Longleftrightarrow$ $\cos x\cdot [\cos (x+2y)-\cos (2z+x-2y)]=\cos y\cdot [\cos (2x+y)-\cos (y+2z-2x)]$ $\Longleftrightarrow$

$\cos (2x+2y)+\cos 2y-\cos (2x+2z-2y)-\cos (2z-2y)=$ $\cos (2x+2y)+\cos 2x-\cos (2y+2z-2x)-\cos (2z-2x)$ $\Longleftrightarrow$

$\sin (x+y)\cdot\sin (x-y)+\sin 2z\cdot\sin (2x-2y)+\sin (2z-x-y)\cdot\sin (x-y)=0$ $\Longleftrightarrow$ $\sin (x+y)+2\sin 2z\cdot\cos (x-y)+\sin (x+y-2z)=0$ $\Longleftrightarrow$

$\sin z\cdot\cos (x+y-z)+\sin 2z\cdot\cos (x-y)=0$ $\Longleftrightarrow$ $\cos (x+y-z)+2\cos z\cdot\cos (x-y)=0$ $\Longleftrightarrow$ $\cos (x+y-z)+\cos (z+x-y)+\cos (z+y-x)=0$ .

Remark. The relation $(*)$ is symmetrically in the variables $x$ , $y$ . Thus, I can propose the following problem :

PP. Let $ABCD$ be a convex quadrilateral with $\angle{DAC}=18^{\circ}$ , $\angle{BDC}= 72^\circ$ and $\angle{CBD}=\angle{BAC}= 36^\circ$ . Determine the measure of $\angle{APD}$ , where $P\in AC\cap BD$ .

Proof of the proposed problem. For $\left\{\begin{array}{c} x=18^{\circ}\\\\ y=36^{\circ}\end{array}\right\|$ the relation $(*)$ becomes $\cos (54-z)+\cos (z-18)+\cos (z+18)=0$ , i.e. $f(z)\equiv\sin (36+z)+2\cos z\cos 18=0$ .

Observe that $z>90^{\circ}$ and on the interval $\left(\frac{\pi}{2},\pi \right)$ the function $f$ is strict decreasing and $f\left(108^{\circ}\right)=0$ . In conclusion, $m(\widehat{APD})=108^{\circ}$ .

PP16. Let an $A$ -isosceles $\triangle ABC$ and an interior point $P$ for which denote $R\in BP\cap AC$ . Suppose that $m\left(\widehat{PBC}\right)=m\left(\widehat{APR}\right)=30^{\circ}$ . Prove that $PC=AB$ .

Proof 1. If denote $\left\{\begin{array}{ccc} m\left(\widehat{PBA}\right) & = & \alpha\\\\ m\left(\widehat{ACP}\right) & = & m\end{array}\right\|$ , then $\left\{\begin{array}{ccc} m\left(\widehat{PAB}\right) & = & 30^{\circ}-\alpha\\\\ m\left(\widehat{PAC}\right) & = & 90^{\circ}-\alpha\\\\ m\left(\widehat{PCB}\right) & = & 30^{\circ}+\alpha -m\end{array}\right\|$ . Trigonometrical form of the Ceva' theorem $:\ \sin\widehat{PAB}\sin\widehat{PBC}\sin\widehat{PCA}=$

$\sin\widehat{PAC}\sin\widehat{PBA}\sin\widehat{PCB}\iff$ $\sin (30^{\circ}-\alpha )\sin 30^{\circ}\sin m=$ $\sin (90^{\circ}-\alpha )\sin \alpha\sin (30^{\circ}+\alpha -m)\iff$ $\sin (30^{\circ}-\alpha )\sin m=$

$2\cos\alpha\sin \alpha\sin (30^{\circ}+\alpha -m)\iff$ $\sin (30^{\circ}-\alpha )\sin m=$ $\sin 2\alpha\sin (30^{\circ}+\alpha -m)\iff$ $\cos (\alpha +m-30^{\circ})-\cos (m+30^{\circ}-\alpha)=$

$\cos (m+\alpha -30^{\circ})-\cos (30^{\circ}+3\alpha -m)\iff$ $\cos (m+30^{\circ}-\alpha)=\cos (30^{\circ}+3\alpha -m)\iff$ $m+30^{\circ}-\alpha =30^{\circ}+3\alpha -m\iff$ $\boxed{\ m=2\alpha\ }$ .

Proof 2. I"ll use same notations from the above proof. Denote $S\in AP\cap BC$ and prove easily that $SB=SP$ . Apply the theorem of Sines $:$

$\left\{\begin{array}{cccc} \triangle ABC\ : & \frac {SB}{SC}=\frac {\sin\widehat{SAB}}{\sin\widehat{SAC}} & \implies & \frac {SB}{SC}=\frac {\sin (30-\alpha )}{\sin (90-\alpha )}\\\\ \triangle PSC\ : & \frac {SP}{SC}=\frac {\sin\widehat{SCP}}{\sin\widehat{SPC}} & \implies & \frac {SB}{SC}=\frac {\sin (30+\alpha -m)}{\sin (90+m-\alpha )}\end{array}\right\|$ . Therefore, $\frac {SB}{SC}=\frac {\sin (30-\alpha )}{\sin (90-\alpha )}=$ $\frac {\sin [30-(m-\alpha )]}{\sin [90+(m-\alpha )]}\iff$ $\frac {\sin (30-\alpha )}{\cos \alpha}=$

$\frac {\sin [30-(m-\alpha )]}{\cos (m-\alpha )}\iff$ $\frac 12-\frac {\sqrt 3}2\cdot \tan\alpha =\frac 12-\frac {\sqrt 3}2\cdot \tan (m-\alpha )\iff$ $\tan\alpha =\tan (m-\alpha )\iff$ $\boxed{\ m=2\alpha\ }$ .

Proof 3 (synthetic). Very nice problem! I could not restrain myself of posting something. Let the equilateral $\triangle ABD$ outside the given triangle. Easy angle chasing shows

that $D$ is the circumcenter of $\triangle ABP$ and $AP$ is the bisector of $\angle DAC$ . So $DAPC$ is a kite. $DP = AD$ makes it a rhombus, done. Best regards, sunken rock.

PP17. Let $\triangle ABC$ with $D\in (AC)$ so that $AD=BC$ and $\left\{\begin{array}{ccc} A & = & 40^{\circ}\\\\ C & = & 20^{\circ}\end{array}\right\|$ . Prove that $x\equiv m\left(\widehat{DBC}\right)=10^{\circ}$ .

Proof 1. Observe that $\frac {DA}{DB}=\frac {BC}{BD}\iff$ $\frac {\sin\widehat{DBA}}{\sin\widehat{DAB}}=\frac {\sin\widehat{BDC}}{\sin\widehat{BCD}}\iff$ $\frac {\sin (120^{\circ}-x)}{\sin 40^{\circ}}=\frac {\sin (20^{\circ}+x)}{\sin 20^{\circ}}\iff$ $\frac {\sin (60^{\circ}+x)}{2\sin 20^{\circ}\cos 20^{\circ}}=\frac {\sin (20^{\circ}+x)}{\sin 20^{\circ}}\iff$

$\cos (30^{\circ} -x)=2\sin (20^{\circ}+x)\cos 20^{\circ}\iff$ $\cos (30^{\circ}-x)=\sin (40^{\circ}+x)+\sin x\iff$ $\cos (30^{\circ}-x)-\cos (90^{\circ}-x)=\sin (40^{\circ}+x)\iff$ $2\sin (60^{\circ}-x)\sin 30^{\circ}=$

$\sin (40^{\circ}+x)\iff$ $\sin (60^{\circ}-x)=\sin (40^{\circ}+x)\iff$ $60^{\circ}-x=40^{\circ}+x\iff$ $2x=20^{\circ}\iff x=10^{\circ}$ . Otherwise. $\frac {BC}{BA}=\frac {AD}{AB}\iff$ $\frac {\sin\widehat{BAC}}{\sin\widehat{BCA}}=\frac {\sin\widehat{ABD}}{\sin\widehat{ADB}}\iff$

$\frac {\sin 40^{\circ}}{\sin 20^{\circ}}=\frac {\sin (120^{\circ}-x)}{\sin (20^{\circ} +x)}\iff$ $2\cos 20^{\circ}\sin (20^{\circ}+x)=\sin (60^{\circ}+x)\iff$ $\sin (40^{\circ}+x)+\sin x=\sin (60^{\circ}+x)\iff$ $\sin (40^{\circ}+x)=\sin (60^{\circ}+x)-\sin x\iff$

$\sin (40^{\circ}+x)=2\sin 30^{\circ}\cos (30^{\circ}+x)\iff$ $\cos(50^{\circ}-x)=\cos (30^{\circ}+x)\iff$ $50^{\circ}-x=30^{\circ}+x\iff$ $2x=20^{\circ}\iff x=10^{\circ}$ .

Proof 2. Let $E\in (AD)$ so that $EA=EB$ . Thus, $\triangle CBE$ is $C$ -isosceles with $C=20^{\circ}\implies$ $AD=BC=CE\implies$ $AD=CE\implies$

$BE=AE=CD\implies$ $CD=BE$ . In conclusion, $\triangle CBE$ satisfies the hypothesis of the following $(\Downarrow )$ proposed problem PP18 from where obtain that $x=10^{\circ}$

PP18. Let a $B$ -isosceles $\triangle ABC$ with $B=20^{\circ}$ and $D\in (BC)$ so that $BD=AC$ . Prove that $x\equiv m\left(\widehat{BAD}\right)=10^{\circ}$ .

Proof. $\frac {AC}{AB}=\frac {BD}{BA}\iff$ $\frac {\sin 20^{\circ}}{\cos 10^{\circ}}=\frac {\sin x}{\sin (20^{\circ}+x)}\iff$ $\sin x=2\sin (20^{\circ}+x)\sin 10^{\circ}\iff$ $\sin x=\cos (10^{\circ}+x)-\cos (30^{\circ}+x)\iff$

$\cos (90^{\circ}-x)+\cos (30^{\circ}+x)=\cos (10^{\circ}+x)\iff$ $2\cos 60^{\circ}\cos (30^{\circ}-x)=\cos (10^{\circ}+x)\iff$ $30^{\circ}-x=10^{\circ}+x\iff$ $x=10^{\circ}$ . See PP1 from here.

PP19. Let a convex $ABCD$ with $I\in AC\cap BD$ so that $AB=BC=CD$ and $IA=ID$ . Prove that $IB\ne IC\implies$ $m\left(\widehat{AIB}\right)=60^{\circ}$ .

Proof. Denote $\left\{\begin{array}{c} AB=BC=CD=x\\\\ IA=ID=y\\\\ IB=u\ ;\ IC=v\end{array}\right\|$ . Apply the Stewart's theorem to the isosceles $:\ \left\{\begin{array}{cccc} \triangle ABC\ : & BI^2+IA\cdot IC=BA^2 & \implies & u^2+yv=x^2\\\\ \triangle DCB\ : & CI^2+IB\cdot ID=CD^2 & \implies & v^2+yu=x^2\end{array}\right\|$ $\implies$

$u^2+yv=x^2=v^2+yu\iff$ $\left(u^2-v^2\right)=y(u-v)\ \stackrel{u\ne v}{\iff}\ y=u+v\iff$ $x^2=u^2+v(u+v)\iff$ $x^2=u^2+uv+v^2\iff$ $m\left(\widehat{AIB}\right)=60^{\circ}$ .

PP20. Let $\triangle ABC$ with $A = 60^{\circ}$ . Let $AP$ bisect $\widehat { BAC}$ and let $BQ$ bisect $\widehat {ABC}$ , where $P\in BC$ and $Q\in AC$ . If $AB + BP = AQ + QB$ . Find the angles of $\triangle ABC$ .

Proof I (metric).. Let $l_b = BQ$ . Thus, $l_b = \frac {2ac}{a + c}\cdot\cos\frac B2$ . Therefore, $AB + BP = AQ + QB$ $\Longleftrightarrow$ $c + \frac {ac}{b + c} = \frac {bc}{a + c} + l_b$ $\Longleftrightarrow$ $c(c + a)(c + b) + ac(a + c) =$

$bc(b + c) + 2ac(b + c)\cdot\cos\frac B2$ $\Longleftrightarrow$ $ab + (a + c)^2 - b^2 = 2a(b + c)\cdot\cos\frac B2$ $\Longleftrightarrow$ $ab + 4p(p - b) = 2a(b + c)\cdot\cos\frac B2$ $\Longleftrightarrow$ $ab + 4ac\cdot\cos^2\frac B2 = 2a(b + c)\cdot \cos\frac B2$ $\Longleftrightarrow$

$b\left(1 - 2\cdot\cos\frac B2\right) = 2c\cdot\cos\frac B2\cdot\left(1 - 2\cdot\cos\frac B2\right)$ $\Longleftrightarrow$ $b = 2c\cdot\cos\frac B2$ $\Longleftrightarrow$ $\sin B = 2\cdot\sin C\cdot\cos\frac B2$ $\Longleftrightarrow$ $2\sin\frac B2\cos\frac B2 = 2\sin C\cos\frac B2$ $\Longleftrightarrow$ $\sin\frac B2 = \sin C$ $\Longleftrightarrow$

$B = 2C$ $\Longleftrightarrow$ $C = 40^{\circ}$ , $B = 80^{\circ}$ because $A = 60^{\circ}$ .

Proof II (trigonometric). Denote ${ x=m(\widehat ABQ})$ . I"ll apply the Sinus' theorem in the triangles $ABP$ , $ABQ$ : $AB+BP=AQ+QB$ $\Longleftrightarrow$ $1+\frac {BP}{BA}=\frac {AQ+QB}{AB}$ $\Longleftrightarrow$

$1+\frac {\sin 30}{\sin (30+2x)}=\frac {\sin x+\sin 60}{\sin (60+x)}$ $\Longleftrightarrow$ $\sin (60+x)\left[\sin (30+2x)+\frac 12\right]=\sin (30+2x)(\sin x+\sin 60)$ $\Longleftrightarrow$ $\cos (x-30)-\cos (90+3x)+\cos (30-x)=$

$\cos (x+30)-\cos (30+3x)+\cos (2x-30)-\cos (90+2x)$ $\Longleftrightarrow$ $\underline {\cos (x-30)}+\underline {\underline {\sin 3x}}+\underline {\underline {\underline {\cos (x-30)}}}=$ $\underline {\underline {\underline {\cos (x+30)}}}-\underline {\underline {\cos (30+3x)}}+\cos (2x-30)+\underline {\sin 2x}$ $\Longleftrightarrow$

$[\cos (x-30)-\cos (90-2x)]+[\sin 3x+\sin (60-3x)]=$ $[\cos (x+30)-\cos (x-30)]+\cos (2x-30)$ $\Longleftrightarrow$ $2\sin \left(30-\frac x2\right)\sin\left(60-\frac {3x}{2}\right)+2\sin 30\cos (3x-30)=$

$\cos (2x-30)-2\sin 30\sin x$ $\Longleftrightarrow$ $\cos (3x-30)+2\sin\left(30-\frac x2\right)\sin \left(60-\frac {3x}{2}\right)=$ $\cos (2x-30)-\cos (90-x)$ $\Longleftrightarrow$ $\cos (3x-30)+2\sin\left(30-\frac x2\right)\sin \left(60-\frac {3x}{2}\right)$

$=2\sin \left(30+\frac x2\right)\sin\left(60-\frac {3x}{2}\right)$ $\Longleftrightarrow$ $\sin (120-3x)=2\sin\left(60-\frac {3x}{2}\right)\left[\sin\left(30+\frac x2\right)-\sin\left(30-\frac x2\right)\right]$ $\Longleftrightarrow$ $2\sin\left(60-\frac {3x}{2}\right)\cos\left(60-\frac {3x}{2}\right)=$

$2\sin\left(60-\frac {3x}{2}\right)\left[\sin\left(30+\frac x2\right)-\sin\left(30-\frac x2\right)\right]$ $\Longleftrightarrow$ $\sin\left(60-\frac {3x}{2}\right)=0\ \ \vee\ \ \cos\left(60-\frac {3x}{2}\right)=\sin\left(30+\frac x2\right)-\sin\left(30-\frac x2\right)$ .

Thus, $\boxed {\ x=40\ }$ or $\sin\left(30+\frac {3x}{2}\right)+\sin\left(30-\frac x2\right)=\sin\left(30+\frac x2\right)$ $\Longleftrightarrow$ $2\sin\left(30+\frac x2\right)\cos x=\sin\left(30+\frac x2\right)$ $\Longleftrightarrow$ $x\in\emptyset$ .

PP21 (Edson Curahua Ortega). Let an $A$ -isosceles $\triangle ABC$ and its interior point $P$ so that $AP=BC$ and $\left\{\begin{array}{ccc} m\left(\widehat{PAB}\right) & = & 24\\\\ m\left(\widehat{PAC}\right) & = & 12\end{array}\right\|$ . Prove that $m\left(\widehat{ABP}\right)=30^{\circ}$ .

Proof 1 (İxtiyar İsmayilov). Let $\left\{\begin{array}{c} AB=AC=b\\\\ BC=AP=a\\\\ m\left(\widehat{ABP}\right)=x\end{array}\right\|$ . Apply th. of Sines in $\left\{\begin{array}{cccc} \triangle ABC\ : & \frac {BC}{AB}=\frac {\sin\widehat{BAC}}{\sin\widehat{ACB}} & \implies & \frac ab=\frac {\sin 36^{\circ}}{\sin 72^{\circ}}\\\\ \triangle ABP\ : & \frac {AP}{AB}=\frac {\sin\widehat{ABP}}{\sin\widehat{APB}} & \implies & \frac ab=\frac {\sin x}{\sin (24^{\circ}+x)}\end{array}\right\|$ $\implies$ $\frac {\sin 36^{\circ}}{\sin 72^{\circ}}=$ $\frac {\sin x}{\sin (24^{\circ}+x)}$

$\iff$ $2\sin x\cos 36^{\circ}=\sin (24^{\circ}+x)$ $\iff$ $\sin (36^{\circ}+x)-\sin (36^{\circ}-x)=$ $\sin (24^{\circ}+x)$ $\iff$ $\sin (36^{\circ}+x)=$ $\sin (36^{\circ}-x)+$ $\sin (24^{\circ}+x)$ $\iff$

$\sin (36^{\circ}+x)=$ $2\sin 30^{\circ}\cos (6^{\circ}-x)$ $\iff$ $\cos (54^{\circ}-x)=$ $\cos (6^{\circ}-x)$ $\iff$ $(54^{\circ}-x)+(6^{\circ}-x)=0$ $\iff 60^{\circ}-2x=0\iff$ $x=30^{\circ}\ .$

Proof 2.

This post has been edited 217 times. Last edited by Virgil Nicula, Mar 19, 2016, 8:57 PM

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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.

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).

Submit

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391345 views moment
by ryanbear, May 9, 2023, 6:10 AM
We need virgil nicula to return to aops, this blog is top 10 all time.
by OlympusHero, Sep 14, 2022, 4:44 AM
blog
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by OlympusHero, May 13, 2021, 10:23 PM
the visits tho
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Holy- Darn this is good. shame it's inactive now
by the_mathmagician, Jan 17, 2021, 7:43 PM
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by OlympusHero, Dec 30, 2020, 6:08 PM
long blog
by MrMustache, Nov 11, 2020, 4:52 PM
372554 views!
by mrmath0720, Sep 28, 2020, 1:11 AM
wow... i am lost.

369302 views!

-piphi
by piphi, Jun 10, 2020, 11:44 PM
That was a lot! But, really good solutions and format! Nice blog!!!!
by CSPAL, May 27, 2020, 4:17 PM
impressive
awesome. 358,000 visits?????
by OlympusHero, May 14, 2020, 8:43 PM
Nice blog.
The element of Mathematics is dense here.
I love the style of your posts.
Thanks for writing the blog! I can learn Geometry from here too.
by epsilonist, Jun 27, 2011, 6:00 AM
Thank you very much for this blog.
by Affine, May 7, 2011, 6:20 PM
You seem to have solved every existing problem.
by Goutham, May 2, 2011, 10:59 AM
I love your posts, man (those which I understand)! It's interesting how you punish integrals with recurrence relations and sometimes make seemingly unrelated integrals evaluate each other!
by Caspar, Apr 11, 2011, 2:36 PM
Thank you a lot dear Virgil for an interesting and instructive blog.
by vittasko, Mar 12, 2011, 9:42 AM
Interesting Blog
by ShahinBJK, Mar 12, 2011, 5:07 AM
Interesant, d-le Nicula.
by silent_control, Mar 3, 2011, 6:00 PM
Yours is the largest blog full of math equations as far as I have known! Congrats!
by Goutham, Jan 30, 2011, 2:06 PM
15001th view.
by fries4guys, Jan 20, 2011, 4:57 PM
Thanks to all !
by Virgil Nicula, Dec 11, 2010, 8:32 PM
El_Ectric:
You can either copy some parts from the blog post, paste it at "Search Blogs" (which is lower on the page, after "Blog Stats") and find the whole post.

You're welcome.
by Thomas_, Dec 8, 2010, 6:42 PM
Thanks Thomas_.
by El_Ectric, Dec 8, 2010, 4:21 PM
El_Ectric: Try to open this blog with Internet Explorer (browser). If it still doesn't work, just copy the post you're interested in, and than paste it into Word Document, or almost any kind of reading documents and you'll have all the details.

Nice blog!
by Thomas_, Nov 24, 2010, 4:59 PM
Dear Virgil !

Congratulations for this blog !
Babis stergiou - Greece
by stergiu, Nov 12, 2010, 6:15 AM

72 shouts

My (math)blog

388. Some interesting "slicing" problems.

by Virgil Nicula, Oct 23, 2013, 12:16 PM

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