341. Some nice and interesting "slicing" problems.

by Virgil Nicula, Apr 29, 2012, 3:02 AM

PP1. Let $ABC$ be an $A$ -isosceles triangle. The bisector of $\widehat{ACB}$ meet $AB$ in $M$ . Prove that $AM+MC=BC\iff A=100^{\circ}$ .

Proof 1 (trigonometric). Denote $m\left(\widehat{ACM}\right)=x$ . Sinus' theorem in the triangles $AMC$ and $BMC\ :$ $AM+MC=BC\iff$ $\frac {AM}{MC}+1=\frac {BC}{MC}\iff$

$\frac {\sin x}{\sin 4x}+1=\frac {\sin 3x}{\sin 2x}\iff$ $\sin 2x(\sin x+\sin 4x)=\sin 3x\sin 4x\iff$ $\sin\frac {5x}{2}\cos\frac {3x}{2}=\sin 3x\cos 2x\iff$ $\sin\frac {5x}{2}=2\sin\frac {3x}{2}\cos 2x\iff$

$\sin\frac {5x}{2}=\sin\frac {7x}{2}-\sin\frac x2\iff$ $\sin\frac {7x}{2}-\sin\frac {5x}{2}=\sin \frac x2\iff$ $2\sin\frac x2\cos3x=\sin\frac x2\iff$ $\cos 3x=\frac 12\iff$ $3x=60^{\circ}\iff$ $x=20^{\circ}\iff$ $A=100^{\circ}$ .

Proof 2 (trigonometric). Observe that $\frac {BM}{AM}=\frac {BC}{AC}$ . Denote $N\in CM$ so that $M\in (CN)$ and $MN=MA$ . Thus, $BC=AM+MC\iff$ $CB=CN\iff$ $m\left(\widehat {MNB}\right)=$

$m\left(\widehat {CNB}\right)=90^{\circ}-\frac B4$ and $m\left(\widehat {MBN}\right)=90^{\circ}-\frac {5B}{4}\iff$ $\frac {MN}{MB}=\frac {MA}{MB}=$ $\frac {CA}{CB}\iff$ $\frac {\cos\frac {5B}{4}}{\cos\frac B4}=\frac {\sin B}{\sin 2B}\iff$ $2\cos B\cos \frac {5B}{4}=\cos \frac B4\iff$

$\cos\frac {9B}{4}+\cos\frac B4=\cos\frac B4\iff$ $\cos\frac {9B}{4}=0^{\circ}\iff$ $\frac {9B}{4}=90^{\circ}\iff$ $B=40^{\circ}\iff$ $A=100^{\circ}$ .

Proof 3. Thus, $\frac {BM}{AM}=\frac {BC}{AC}$ . Let $E\in (BC)$ so that $CE=CM$ . So $BC=AM+MC\iff$ $BE=AM\iff$ $\frac {BM}{BE}=\frac {BM}{AM}=\frac {BC}{AC}=\frac {BC}{AB}\iff$ $\frac {BM}{BE}=\frac {BC}{AB}\iff$

$\triangle ABC\sim\triangle EMB\iff$ $\left(\widehat{BME}\right)=B\iff$ $m\left(\widehat{CEM}\right)=m\left(\widehat{CME}\right)=2B\iff$ $2B+2B+\frac B2=180^{\circ}\iff$ $B=40^{\circ}\iff$ $A=100^{\circ}$ .

PP2. Let $ABC$ be the $C$ -right-angled isosceles triangle whose equal sides have length $1$ . For $P\in [AB]$ let the feet of the perpendiculars from $P$ to the other sides are $Q\in CA$ and

$R\in CB$ . Consider the areas of the triangles $APQ$ , $PBR$ and the area of the rectangle $QCRP$ . Prove that regardless of how $P$ is chosen, the largest of these three areas is at least $\frac 29$ .

Proof. Denote $\left\{\begin{array}{c} QA=QP=CR=x\\\\ RB=RP=CQ=1-x\end{array}\right\|$ and $\left\{\begin{array}{ccc} m=[AQP] & = & \frac {x^2}{2}\\\\ n=[BRP] & = & \frac {(1-x)^2}{2}\\\\ p=[CQPR] & = & x(1-x)\end{array}\right\|$ . Prove easily that $\max\{m,n,p\}=$

$\left\{\begin{array}{ccc} n & \mathrm{if} & 0\le x\le \frac 13\\\\ p & \mathrm{if} & \frac 13\le x\le \frac 23\\\\ m & \mathrm{if} & \frac 23\le x \le 1\end{array}\right\|$ and $\left\{\begin{array}{ccccc} n\ge \frac 29 & \iff & 9x^2-18x+5\ge 0 & \iff & x\le\frac 13\\\\ p\ge\frac 29 & \iff & 9x^2-9x+2\le 0 & \iff & \frac 13\le x\le \frac 23\\\\ m\ge \frac 29 & \iff & 9x^2-4\ge 0 & \iff & \frac 23\le x\end{array}\right\|$ . Thus, $\max\{m,n,p\}\ge \frac 29$ .

PP3. $\triangle{ABC}$ is an equilateral triangle. Extend $BC$ to $D$ and extend $BA$ to $E$ such that $AE=BD$ . Prove that the triangle $CED$ is $E$ -isosceles.

Proof 1 (own). Construct the isosceles trapezoid $BCEH$ , where $HE\parallel BC\ \wedge\ \boxed{EC=HB}\ (*)$ . Observe that $HE=AE=BD$

i.e. $HE\parallel BD$ and $HE=BD$ , what means that the quadrilateral $BDEH$ is a parallelogram, i.e. $HB=ED\stackrel{(*)}{\implies} EC=ED$ .

Proof 2. Denote $\left\{\begin{array}{c} AB=BC=CA=a\\\ CD=AE=x\end{array}\right\|$ . Thus, $BE=BA+AE=a+x$ . Appear two cases : $\left\{\begin{array}{ccccc} 0<x< a & \iff & D\in (BC) & \iff & DC=a-x\\\\ a<x & \iff & C\in (BD) & \iff & CD=x-a\end{array}\right\|$ .

If denote the midpoint $F$ of $[CD]$ , then $BF=\frac {a+x}{2}$ , i.e. $BE=2\cdot BF$ . Since $m\left(\widehat{EBF}\right)=60^{\circ}\implies$ $EF\perp BF\iff$ $\triangle CED$ is $E$ -isosceles.

Proof 3. Let $F\in [BC$ (ray) so that $\overline{DF}=\overline{BC}$ . Since $\triangle{ABC}$ is equilateral, so $DF=BC=BA$ . Hence $BF=BD+DBA+AE=BE$ . So $\triangle{BEF}$ is isosceles with

$m(\angle{EBF})=60^{o}$ , which means it is also an equilateral triangle. Now, we have $\widehat{EBF}\equiv\angle{EFB}\ ,\ EB=EF\ ,\ BC=DF$ . So $\triangle{EBC} \cong \triangle{EFD}$ . So finally $EC=ED$ .

Proof 4. $\implies\left\{\begin{array}{cccc} EC/\triangle EBC\ : & EC^2=BE^2+BC^2-BE\cdot BC=(a+x)^2+a^2-a(a+x) & \implies & EC^2=a^2+ax+x^2\\\\ ED/\triangle EBD\ : & ED^2=BE^2+BD^2-BE\cdot BD=(a+x)^2+x^2-x(a+x) & \implies & ED^2=a^2+ax+x^2\end{array}\right\|$ $\implies$ $EC=ED$ .

Remark. Let $G\in BC$ so that $B\in (GC)$ and $BG=x$ . Prove easily that $\triangle ABG\equiv\triangle CAE$ , i.e. $CE^2=AG^2=a^2+x^2+ax$ .

PP4. In $A$ -isosceles $\triangle ABC$ a trisector of angle $\widehat{BAC}$ intersects $(BC)$ at $D$ . Suppose that $\{a,b\}\subset \mathbb Z$ and $AD=b-7$ . Ascertain $BC$ .

Proof 1. Suppose w.l.o.g. $BD<DC$ . Thus, $\left(\widehat{DAC}\right)=2\cdot \left(\widehat{DAB}\right)\implies$ exists $r>0$ so that $BD=rb$ and $DC=r(2b-7)$ . Apply the Stewart's relation to the cevian $AD$ in

$\triangle ABC\ :\ b^2\cdot r(2b-7)+b^2\cdot rb=(b-7)^2\cdot r(3b-7)+rb\cdot r(2b-7)\cdot r(3b-7)\iff$ $b^2(3b-7)=(b-7)^2(3b-7)+r^2b(2b-7)(3b-7)\iff$

$b^2=(b-7)^2+r^2b(2b-7)\iff$ $7(2b-7)=r^2b(2b-7)\iff$ $b=\frac {7}{r^2}$ . Since $rb\in\mathbb Z$ obtain that $r=\frac 17\implies$ $b=343\ ,\ AD=336\ ,\ a=\boxed{BC=146}$ .

PP5. Let $D$ be an inside point of $\triangle ABC$ and $m(\widehat{BAD})=m(\widehat{ACD})=m(\widehat{DCB})=20^{\circ}$ and $m(\widehat{DAC})=80^{\circ}$ . Prove that $m(\widehat{DBC})=30^{\circ}$ .

Proof 1 (trigonometric). Denote $m\left(\widehat{CBD}\right)=x$ . Apply the trigonometrical form of the Ceva's theorem:

$\boxed{\begin{array}{c} \sin\widehat{DAB}\cdot\sin\widehat{DBC}\cdot\sin\widehat{DCA}=\sin\widehat{DBA}\cdot\sin\widehat{DCB}\cdot\sin\widehat{DAC}\\\\ \sin 20^{\circ}\sin x\sin 20^{\circ}=\sin (40^{\circ}-x)\sin 20^{\circ}\sin 80^{\circ}\\\\ 2\sin x\sin 10^{\circ}\cos 10^{\circ}=\sin (40^{\circ}-x)\cos 10^{\circ}\\\\ \cos (x-10^{\circ})-\cos (x+10^{\circ})=\cos (50^{\circ}+x)\\\\ \cos (x-10^{\circ})-\cos (50^{\circ}+x)==\cos (x+10^{\circ})\\\\ 2\sin 30^{\circ}\sin (x+20^{\circ})=\cos (x+10^{\circ})\\\\ \cos (70^{\circ}-x)=\cos (x+10^{\circ})\\\\ 70^{\circ}-x=x+10^{\circ}\\\\ x=30^{\circ}\end{array}}$

Proof 2 (synthetic). Denote $E\in AD\cap BC$ and $F\in (BC)$ so that $m\left(\widehat{EDF}\right)=60^{\circ}$ . Observe that $\triangle AEB\stackrel{(a.s.a.)}{\equiv}\triangle CFD$ and

$\triangle EDF$ is equilateral. Thus, $BE=DF=DE$ , i.e. the triangle $BED$ is $E$ -isosceles and $m\left(\widehat{CBD}\right)=30^{\circ}$ , i.e. $m\left(\widehat{ABD}\right)=10^{\circ}$ .

Extension. For $\phi\in \left(0,30^{\circ}\right)$ let the $A$ -isosceles triangle $ABC$ with $A=60^{\circ}+2\phi$ and its interior $D$ so that $m\left(\widehat{DAB}\right) =m\left(\widehat{DCB}\right)=\phi$ . Prove that $m\left(\widehat{DBC}\right)=30^{\circ}$ .

Proof 1. Denote $m\left(\widehat{DBC}\right)=x$ . Apply the trigonometrical form of the Ceva's theorem:

$\boxed{\begin{array}{c} \sin\phi\sin x\sin (60-2\phi )=\sin (60+\phi )\sin (60-\phi -x)\sin\phi\\\\ \sin x\sin (120+2\phi )=\sin (60+\phi )\sin (60-\phi -x)\\\\ 2\sin x\cos (60+\phi )=\sin (60-\phi -x)\\\\ \sin (60+\phi +x)-\sin (60+\phi -x)=\sin (60-\phi -x)\\\\ \sin (60+\phi +x)-\sin (60-\phi -x)=\sin (60+\phi -x)\\\\ 2\sin (\phi +x)\cos 60=\sin (60+\phi -x)\\\\ \sin (\phi +x)=\sin (60+\phi -x)\\\\ \phi +x=60+\phi -x\\\\ x=30\end{array}}$ .

Proof 2 (synthetic - own). Denote $E\in AD\cap BC$ and $F\in (BC)$ so that $m\left(\widehat{EDF}\right)=60^{\circ}$ . Observe that $\triangle AEB\stackrel{(a.s.a.)}{\equiv}\triangle CFD$

and $\triangle EDF$ is equilateral. Thus, $BE=DF=DE$ , i.e. the triangle $BED$ is $E$ -isosceles and $m\left(\widehat{CBD}\right)=30^{\circ}$ . Nice extension !

PP6. Let $ABC$ be an $A$ -isosceles triangle with the $A$ -altitude $AD$ where $D\in BC$ and with the $B$ -angle bisector $BE$ where $E\in AC$ so that $BE=2\cdot AD$ . Find $m\left(\widehat{BAC}\right)$ .

Proof 1 (trigonometric). Denote $B=2x$ . Thus, $BE=2\cdot AD\iff$ $\frac {BE}{BC}=\frac {2\cdot AD}{2\cdot DC}\iff$ $\frac {\sin 2x}{\sin 3x}=\tan 2x\iff$ $\sin 3x=\cos 2x\iff$ $x=18^{\circ}\iff$ $B=36^{\circ}$ .

Proof 2 (metric). $BE=2\cdot AD\iff$ $BE^2=4\cdot AD^2\iff$ $\frac {a^2b(a+2b)}{(a+b)^2}=4b^2-a^2\iff$ $a^2b=(2b-a)(a+b)^2\iff$ $a^3+a^2b-3ab^2-2b^3=0\iff$

$(a+2b)(a^2-ab-b^2)=0\iff$ $a^2-ab-b^2=0\iff$ $\frac ab=\frac {1+\sqrt 5}{2}\iff$ $2\cdot\frac {DB}{AB}=\frac {1+\sqrt 5}{2}\iff$ $\cos B=\frac {1+\sqrt 5}{4}\iff$ $B=36^{\circ}$ .

Proof 3. Let $B=2x$ and $F\in AC$ so that $FB\perp BC$ . Thus, $BE=2\cdot AD\iff$ $BE=BF$ $\iff$ $m\left(\widehat{BFE}\right)=m\left(\widehat{BEF}\right)=3x\iff$ $3x+2x=90^{\circ}\iff$ $B=36^{\circ}$ .

Proof 4. Denote $B=2x$ and $F\in BC$ so that $EF\perp BC$ . Thus, $\frac {EF}{AD}=\frac {CE}{CA}=\frac {ab}{a+b}\cdot\frac 1b=\frac {a}{a+b}$ . Thus, $BE=2\cdot AD$ $\iff$ $BE=2\cdot \frac {a+b}{a}\cdot EF\iff$

$\sin x=\frac {EF}{BE}=\frac {a}{2(a+b)}\iff$ $2\sin x=\frac {a}{a+b}\iff$ $2\sin x=\frac {\sin 4x}{\sin 4x+\sin 2x}\iff$ $2\sin x=\frac {4\cos 2x\sin x\cos x}{2\sin 3x\cos x}\iff$ $\sin 3x=\cos 2x\iff B=36^{\circ}$ .

Proof 5 (synthetic). Denote $I\in BE\cap AD$ and $F\in AD$ so that $ABFC$ is a rhombus. Thus, $BE=2\cdot AD\iff$ $BE=AF\iff$ the trapezoid

$ABFE$ is isosceles $\iff$ $IA=IE\iff$ $m\left(\widehat{DAC}\right)=m\left(\widehat{AEB}\right)=3x\iff$ $2x+3x=90^{\circ}\iff$ $x=18^{\circ}\iff$ $B=36^{\circ}$ .

Proof 6 (metric). Denote $B=2x$ , i.e. $m\left(\widehat{AEB}\right)=3x$ . Suppose $BE=2\cdot AD$ . Therefore, $BE\cdot AC\cdot \sin 3x=AD\cdot BC\iff$

$BE\cdot AC\cdot\sin 3x=2\cdot AD\cdot DC\iff$ $AC\cdot \sin 3x= DC\iff$ $\sin 3x=\cos 2x\iff$ $x=18^{\circ}\iff$ $B=36^{\circ}$ .

Proof 7 (metric). I"ll use the van Aubel's relation $\frac {IA}{b+c}=\frac {ID}{a}=\frac {AD}{2s}$ in $\triangle ABC$ with the incenter $I$ and $D\in AI\cap BC$ . Therefore, $\left\{\begin{array}{c} IA=\frac {2b}{2b+a}\cdot AD\\\\ IE=\frac {b}{2b+a}\cdot BE\end{array}\right\|$ .

In conclusion, $BE=2\cdot AD\iff$ $IA=IE\iff$ $\widehat{IAE}\equiv\widehat{IEA}\iff$ $3x+2x=90^{\circ}\iff$ $x=18^{\circ}\iff$ $B=36^{\circ}$ .

An easy extension. Let $ABC$ be a triangle and $\left\{\begin{array}{ccc} D\in (BC) & ; & \widehat{DAB}\equiv\widehat{DAC}\\\\ E\in (BC) & ; & \widehat{EBA}\equiv\widehat{EBC}\end{array}\right\|$ . Prove that $\boxed{\ \frac {BE}{AD}=1+\frac {AB}{AC}\iff a^2=c^2+ab\ }$ .

Proof. $a^2=c^2+ab\iff$ $\sin^2A-\sin^2C=\sin A\sin B\iff$ $\sin (A-C)\sin (A+C)=\sin A\sin B\iff$ $\sin (A-C)=\sin A\iff$ $2A-C=180^{\circ}\iff$

$2A-C=A+B+C\iff$ $A=B+2C\iff$ $\frac A2=\frac B2+C\iff$ $\widehat{IAE}\equiv\widehat{IEA}\iff$ $IA=IE\stackrel{(\mathrm{van'Aubel})}{\iff}$ $\frac {b+c}{2s}\cdot AD=\frac {b}{2s}\cdot BE\iff$ $\frac {BE}{AD}=1+\frac {AB}{AC}$ .

PP7. Let $\triangle ABC$ and the midpoint $D$ of $[BC]$ . Suppose that $\widehat{ BAD}\equiv\widehat{ACB}$ and $m(\angle DAC)=15^o$ . Findthe measure of $\angle ACB$ .

Proof VN1. $DB=DC\iff$ $AB\cdot\sin\widehat{DAB}=AC\cdot\sin\widehat{DAC}\iff$ $c\cdot\sin C=b\cdot \sin 15^{\circ}\iff$ $\sin^2C=\sin B\sin 15^{\circ}\iff$ $\sin^2C=\sin (2C+15^{\circ})\sin 15^{\circ}\iff$

$1-\cos 2C=$ $\cos 2C-\cos \left(2C+30^{\circ}\right)\iff$ $1+\cos \left(2C+30^{\circ}\right)=$ $2\cos 2C\iff$ $1+\frac {\sqrt 3}{2}\cdot\cos 2C-\frac 12\cdot\sin 2C=$ $2\cdot\cos 2C\iff$

$\sin 2C+\left(4-\sqrt 3\right)\cdot\cos 2C=2$ $\iff$ $2\sin C\cos C+\left(4-\sqrt 3\right)\cdot\left(\cos^2C-\sin^2C\right)=$ $2\left(\sin^2C+\cos^2C\right)\stackrel{(t=\tan C)}{\iff}$

$2t+\left(4-\sqrt 3\right)\left(1-t^2\right)=$ $2\left(t^2+1\right)\iff$ $\left(6-\sqrt 3\right)t^2-2t+\left(\sqrt 3-2\right)=0\iff$ $t=\frac {1}{\sqrt 3}$ because $C<90^{\circ}\iff$ $\tan C=\frac {1}{\sqrt 3}$ $\iff C=30^{\circ}$ .

Proof VN2. $BAD\sim BCA\iff$ $\frac ca=\frac {m_a}{b}=\frac {a}{2c}\iff$ $\left\{\begin{array}{c} a=c\sqrt 2\\\\ b=m_a\sqrt 2\end{array}\right\|$ . Thus $\frac {\sqrt 3+1}{2\sqrt 2}=\cos 15^{\circ}=$ $\frac {b^2+m_a^2-\frac {a^2}{4}}{2bm_a}=$ $\frac {b^2+\frac {b^2}{2}-\frac {c^2}{2}}{2b\cdot \frac{b}{\sqrt 2}}=$ $\frac {3b^2-c^2}{2b^2\sqrt 2}$ $\implies$ $\sqrt 3+1=\frac {3b^2-c^2}{b^2}$

$\iff$ $c^2=b^2\left(2-\sqrt 3\right)\iff$ $b=c\cdot \frac {\sqrt 6+\sqrt 2}{2}$ . In conclusion, $\boxed{\frac a2=\frac {b}{\sqrt 3+1}=\frac {c}{\sqrt 2}}\implies$ $\cos C=\frac {a^2+b^2-c^2}{2ab}=$ $\frac {4+\left(\sqrt 3+1\right)^2-2}{4\left(\sqrt 3+1\right)}\implies$ $\cos C=\frac {\sqrt 3}{2}\implies C=30^{\circ}$ .

Proof (synthetical - Sunken Rock). From $\widehat{BAD}\equiv\widehat{BCA}$ it yields that $BAD\sim BCA\iff$ $a=c\sqrt 2$ and $b=m_a\sqrt 2$ . Construct the $O$ -right and isosceles $\triangle AOC$ so that $AC$ separates $O$ and $D$ . Prove easily that $OA=OD=OC$ , i.e. the point $O$ is the circumcenter of $\triangle ADC$ . In conclusion, $C=\frac 12\cdot m(\widehat{AOD})=30^\circ$

PP8. Let $ABC$ be a triangle with $\left\{\begin{array}{ccc} m\left(\widehat{PBC}\right)=30^{\circ}\ ;\ m\left(\widehat{PBA}\right)=8^{\circ}\\\\ m\left(\widehat{PAB}\right)=m\left(\widehat{PAC}\right)=22^{\circ}\end{array}\right\|$ . Prove that $m\left(\widehat{APC}\right)=142^{\circ}$ .

Proof 1 (trigonometric). Denote $m\left(\widehat{PCB}\right)=x$ , i.e. $m\left(\widehat{PCA}\right)=98^{\circ}-x$ and apply the trigonometric form of the Ceva's theorem : $\sin\widehat{PAB}\cdot\sin\widehat{PBC}\cdot\sin\widehat{PCA}=$

$\sin\widehat{PAC}\cdot\sin\widehat{PBA}\cdot\sin\widehat{PCB}$ $\iff$ $\sin 22^{\circ}\sin 30^{\circ}\sin (98^{\circ}-x)=$ $\sin 22^{\circ}\sin 8^{\circ}\sin x$ $\iff$ $\sin (82^{\circ}+x)=$ $2\sin 8^{\circ}\sin x\iff$

$\cos (8^{\circ}-x)=$ $\cos (8^{\circ}-x)-\cos (8^{\circ}+x)$ $\iff$ $\cos (8^{\circ}+x)=0\iff$ $x=82^{\circ}\iff$ $m\left(\widehat{APC}\right)=142^{\circ}$ .

Proof 2 (synthetic). Denote $D\in AP$ , $BD\perp AP$ and the symetrical point $E$ of $B$ w.r.t. $D$ . Observe that $m\left(\widehat{BPD}\right)=30^{\circ}$ , $\triangle BPE$ is equilateral , $BC\perp PE$ and

$\triangle PCE$ is $C$ -isosceles. Therefore, $m\left(\widehat{PEC}\right)=$ $m\left(\widehat{PEA}\right)=$ $m\left(\widehat{PBA}\right)=8^{\circ}$ and $m\left(\widehat{PCA}\right)=2\cdot m\left(\widehat{PEC}\right)= 16^{\circ}$ . In conclusion, $m\left(\widehat{APC}\right)=142^{\circ}$ .

An easy extension. Let $ABC$ be a triangle with $\left\{\begin{array}{ccc} m\left(\widehat{PBC}\right)=30^{\circ}\ ;\ m\left(\widehat{PBA}\right)=x\\\\ m\left(\widehat{PAB}\right)=m\left(\widehat{PAC}\right)=30^{\circ}-x\end{array}\right\|$ , where $x\in\left(0,30^{\circ}\right)$ . Prove that $m\left(\widehat{APC}\right)=150^{\circ}-x$ .

PP9. Let $\triangle ABC$ with $C=45^{\circ}$ . Denote $D\in (AC)$ so that $AD=2\cdot DC$ . Suppose that $m\left(\widehat{ADB}\right)=60^{\circ}$ . Find $A$ .

Proof 1 ("slicing"). Denote $E\in (AD)$ and $F\in (BD)$ so that $\triangle DEF$ is equilateral. Prove easly that $AE=ED=DC=$

$EF=FD$ and $FA=FB=FC$ . Therefore, $F$ is the circumcenter of $\triangle ABC$ and $AF\perp BD$ . In conclusion, $\boxed{A=75^{\circ}}$ .

Proof 2 (trigonometric). Denote $x=m\left(\widehat{ABD}\right)$ , i.e. $\boxed{A=120^{\circ}-x}\ (*)$ . Apply an well-known relation $\frac {DA}{DC}=\frac {BA}{BC}\cdot\frac {\sin\widehat{DBA}}{\sin\widehat{DBC}}\iff$ $2\sin A\sin 15^{\circ}=\sin C\sin x\iff$

$\boxed{2\sin 15\sin (60^{\circ}+x)=\sin 45^{\circ}\sin x}\iff$ $2\cdot\frac {\sqrt 2(\sqrt 3-1)}{4}\cdot \left(\frac {\sqrt 3}{2}\cdot\cos x+\frac 12\cdot \sin x\right)=\frac {\sqrt 2}{2}\cdot \sin x\iff$ $\left(\sqrt 3-1\right)\cdot\left(\sqrt 3\cos x+\sin x\right)=2\sin x\iff$

$\sqrt 3\left(\sqrt 3-1\right)\cos x=\left(3-\sqrt 3\right)\sin x\iff$ $\tan x=1\iff x=45^{\circ}\stackrel{(*)}{\iff}\boxed{A=75^{\circ}}$ .

PP10. Let $ABC$ be an $A$ -right triangle and $D$ be a point $D\in (AC)$ is a point such that $DC = 2\cdot AD$ . Let $E\in (BD)$ be a point such that

$\widehat{CED}\equiv\widehat{ABC}$ and let $F\in AC$ be a point such that $A$ is the midpoint of $[FC]$ .opposite $C$ pass $A$ . Prove that $m\left(\widehat{DEF}\right)=2\cdot\left(\widehat{ABC}\right)$ .

Proof. Problem asks to prove $\angle FBC=\angle FED\ (\ 1\ )$ . Also, as given, we easily can see that $FD=2CD\ (\ 2\ )$ . Take $M$ on $BE$ such that $CM\bot CE\implies$

$\triangle CEM\sim\triangle ABC\implies$ $\angle EMC=\angle BFC$ , so $BFMC$ is cyclic and $MD$ bisector of $\angle CMF\implies\frac{FM}{MC}=\frac{FD}{CD}=2$ . Taking $N$ midpoint of $|MF|$ we infer that

$\triangle ENM\cong\triangle ECM\ (s.a.s.)$ , so $EN\bot FM$ and $\angle MEN=\angle MEC=\angle ABC$ , but $EN\bot FM$ and $N$ midpoint of $FM$ implies $\triangle FEM$ isosceles with

$\angle FEM=2\angle MEN=2\angle ABC=\angle FBC$ . Prove easily that $EF=2\cdot BE$ .

Equivalent enunciation. Let an $A$ -isosceles $\triangle ABC$ with $A=2\alpha$ and let $\left\{\begin{array}{ccc} D\in (BC) & ; & BD=2\cdot DC\\\\ E\in (AD) & ; & m\left(\widehat{CED}\right)=\alpha\end{array}\right\|$ . Prove that $m\left(\widehat{BED}\right)=2\alpha$ .

Proof. Denote $u=m\left(\widehat{BAD}\right)$ and $x=m\left(\widehat{BED}\right)$ . Thus, $m\left(\widehat{CAD}\right)=2\alpha -u$ and $\left\{\begin{array}{ccc} m\left(\widehat{ABE}\right)=x-u & ; & m\left(\widehat{CBE}\right)=90^{\circ}-\alpha +u-x\\\\ m\left(\widehat{ACE}\right)=u-\alpha & ; & m\left(\widehat{BCE}\right)=90^{\circ}-u\end{array}\right\|$ . Apply in $\triangle ABC$ the

well-known relation $\frac {DB}{DC}=$ $\frac {AB}{AC}\cdot\frac {\sin\widehat{DAB}}{\sin\widehat{DAC}}\iff$ $2\sin(2\alpha -u)=\sin u\iff$ $\tan u=\frac {2\sin 2\alpha}{1+2\cos 2\alpha}\iff$ $\tan u=\frac {2\sin \alpha\sin 2\alpha}{\sin 3\alpha}\iff$ $\boxed{\tan u=\frac {\cos\alpha -\cos 3\alpha}{\sin 3\alpha}}\ (*)$ .

Apply analogously same relation in $\triangle BEC\ :\ \frac {DB}{DC}=$ $\frac {EB}{EC}\cdot\frac {\sin\widehat{DEB}}{\sin\widehat{DEC}}$ $\iff$ $\frac {DB}{DC}=\frac {\sin\widehat{BCE}}{\sin\widehat{CBE}}\cdot\frac {\sin\widehat{DEB}}{\sin\widehat{DEC}}\iff$ $2\sin\alpha\cos (\alpha +x-u)=\cos u\sin x\iff$

$\tan x=\frac {2\sin\alpha \cos (\alpha -u)}{\cos u+2\sin\alpha \sin (\alpha -u)}\implies$ $\boxed{\tan x=\frac {\sin (2\alpha -u)+\sin u}{2\cos u-\cos (2\alpha -u)}}=$ $\frac {\sin 2\alpha -\cos 2\alpha \tan u+\tan u}{2-\cos 2\alpha -\sin 2\alpha\tan u}=$ $\frac {\sin 2\alpha +(1-\cos 2\alpha )\tan u}{(2-\cos 2\alpha )-\sin 2\alpha \tan u}\ \stackrel{(*)}{=}$

$\frac {\sin 2\alpha \sin 3\alpha +2(1-\cos 2\alpha )\sin \alpha\sin 2\alpha}{(2-\cos 2\alpha )\sin 3\alpha -2\sin \alpha\sin^22\alpha}=$ $\frac {\sin 2\alpha (\sin 3\alpha +2\sin\alpha -2\sin\alpha\cos 2\alpha)}{-\cos 2\alpha \sin 3\alpha +2\sin 3\alpha -2\sin \alpha \left(1-\cos^22\alpha\right)}=$ $\frac {\sin 2\alpha (\sin 3\alpha +2\sin \alpha -\sin 3\alpha +\sin \alpha)}{\cos 2\alpha (2\sin\alpha \cos 2\alpha -\sin 3\alpha)+2(\sin 3\alpha -\sin\alpha)}=$

$\frac {3\sin\alpha\sin 2\alpha}{\cos 2\alpha (\sin 3\alpha -\sin\alpha -\sin 3\alpha )+4\sin\alpha\cos 2\alpha}=\tan 2\alpha$ . In conclusion, $\tan x=\tan 2\alpha\iff$ $x=2\alpha$ , i.e. $m\left(\widehat{BED}\right)=2\alpha$ .

PP11. Let $ABCD$ be a parallelogram. For an interior point $P$ of $\triangle ACD$ denote $L\in BP\cap CD$ and $K\in DP\cap BC$ . Prove that $BK\cdot BC=DL\cdot DC\iff \widehat{DAP}\equiv\widehat{CAB}$ .

Proof. Denote $\left\{\begin{array}{c} O\in AC\cap BD\\\ S\in KO\cap AD\\\ T\in LS\cap AB\end{array}\right\|$ and $\left\{\begin{array}{ccccc} AD=BC=a & : & BK=u & ; & KC=a-u\\\\ AB=CD=b & ; & DL=v & ; & LC=b-v\end{array}\right\|$ . Thus, $\left\{\begin{array}{c} AS=a-u\\\ SD=u\end{array}\right\|$ and $AT\parallel DL\iff\frac {AT}{DL}=\frac {SA}{SD}\iff$

$AT=\frac {v(a-u)}{u}\implies$ $\boxed{\frac {AT}{AB}=\frac {v(a-u)}{bu}}\ (1)$ . Apply Menelaus' theorem to $\overline{DPK}/\triangle BCL\ :\ \frac {DL}{DC}\cdot$ $\frac {KC}{KB}\cdot\frac {PB}{PL}=1\iff$ $\frac vb\cdot\frac {a-u}{u}\cdot\frac {PB}{PL}=1\iff$ $\boxed{\frac {PL}{PB}=\frac {v(a-u)}{bu}}\ (2)$ .

From $(1)\ \wedge\ (2)$ obtain that $\frac {PL}{PB}=\frac {AT}{AB}=\frac {v(a-u)}{bu}$ , i.e. $\boxed{LS\parallel AP}\ (3)$ . Therefore, $BK\cdot BC=DL\cdot DC\iff$ $DS\cdot DA=DL\cdot DC\iff$

$ACLS$ is cyclically $\iff$ $\widehat{ACL}\equiv\widehat{LSD}\stackrel{(3)}{\iff}$ $\widehat{CAB}\equiv \widehat{DAP}$ .

PP12. Let $A$ -isosceles $\triangle ABC\ ,\ A<90^{\circ}$ and bisector $[BD$ of $\widehat{ABC}\ ,\ D\in AC$ . Let $E\in (BC)$ so that $CE=CD$ and circumcenter $O$ of $\triangle BDC$ . Prove that $\left(\widehat{BOE}\right) =A$ .

Proof 1. Let the midpoint $M$ of $[BC]$ . Thus, $O\in AM\cap DE$ because $m\left(\widehat{EDC}\right)=90^{\circ}-\frac B2=90^{\circ}-m\left(\widehat{DBC}\right)$ $\implies$

$m\left(\widehat{ODB}\right)=90^{\circ}-C=m\left(\widehat{OAD}\right)=m\left(\widehat{OAB}\right)\implies$ $\widehat{ODB}\equiv\widehat{OAB}\implies$ $ABOD$ is cyclic $\implies$ $m\left(\widehat{BOE}\right) =A$ .

Proof 2. Denote $BC=a$ , $AB=AC=b$ and $A=2\phi$ , $B=C=90^{\circ}-\phi$ . Let the midpoint $M$ of $[BC]$ and $m\left(\widehat{BOE}\right)=x$ . Thus, $O\in AM$ and $m\left(\widehat{BOC}\right)=$

$2\cdot \left(\widehat{BDC}\right)=$ $2\left(A+\frac B2\right)=$ $2A+B=$ $90^{\circ}+3\phi$ . Therefore, $\frac {DA}{b}=\frac {DC}{a}=\frac {b}{a+b}\implies$ $EC=DC=\frac {ab}{a+b}$ and $EB=a-EC=$ $a-\frac {ab}{a+b}=$ $\frac {a^2}{a+b}\implies$ $\frac {EB}{EC}=\frac ab$ .

Apply an well-known relation $\frac {EB}{EC}=\frac {OB}{OC}\cdot\frac {\sin\widehat{EOB}}{\sin\widehat{EOC}}\iff$ $\frac ab=\frac {\sin x}{\sin (90^{\circ}+3\phi -x)}\iff$ $\frac {\sin 2\phi}{\cos \phi}=\frac {\sin x}{\cos (x-3\phi )}\iff$ $\sin 2\phi\cos (x-3\phi )=\cos\phi\sin x\iff$

$\sin (x-\phi )+\sin (5\phi -x)=\sin (x+\phi )+\sin (x-\phi )\iff$ $\sin (5\phi -x)=\sin (x+\phi )\iff$ $5\phi -x=x+\phi\iff$ $x=2\phi\iff$ $m\left(\widehat{BOE}\right) =A$ .

PP13. Let equilateral $\triangle ABC$ and $E\in (AC)$ . The perpendicular line from $E$ to $AB$ cut the perpendicular line from $C$ to $CB$ at $D$ . Let the mid-point $K$ of $AE$ . Find $\left(\widehat{KBD}\right)$ .

Proof. Denote the midpoints $M$ and $S$ of $[BC]$ and $[BD]$ respectively. Thus, $S\in AM$ , $KF\parallel BC$ , $SB=SD=$

$SC=SF=SK$ $\implies$ $\{C,K,F\}\subset w$ - the circle with diameter $[BD]\implies$ $m\left(\widehat{KBD}\right)=m\left(\widehat{KCD}\right)=$ $30^{\circ}$ .

PP14. Let $ABC$ be an $A$ -isosceles triangle with $A=20^{\circ}$ . Construct the $D$ -isosceles $\triangle ACD$ so that $D=100^{\circ}$ and the line $AC$ separates $B$ and $D$ . Prove that $BC+CD=AB$ .

Proof 1. $AB=BC+CD\iff$ $\frac {BC}{AB}+\frac {CD}{AB}=1\iff$ $\frac {\sin 20^\circ}{\sin 80^\circ}+\frac {\sin 40^\circ}{\sin 100^\circ}=1\iff$ $\sin 20^\circ+\sin 40^\circ=\cos 10^\circ\iff$ $2\sin 30^\circ\cos 10^\circ=\cos 10^\circ$ , what is true.

Proof 2. Let $E\in (AB)$ so that $AE=AD$ . Prove easily that $AD=AE=ED=DC$ and $BE=BC$ . In conclusion, $AB=BE+EA=BC+CD$ .

Proof 3. Let $E\in (AB)\ ,\ BE=BC$ . Prove easily that $BCDE$ is a kite. So $DE=CD\ (\ 1\ )\ ,\ m(\angle CDE)=40^\circ\ ,\ m(\angle ADE)=60^\circ$ and $(1)\implies \Delta ADE$ is equilateral.

PP15. Let $\triangle ABC$ and an interior $D$ for which $\left\{\begin{array}{ccc} m\left(\widehat{DBA}\right)=12^{\circ} & ; & m\left(\widehat{DBC}\right)=6^{\circ}\\\ m\left(\widehat{DCA}\right)=24^{\circ} & ; & m\left(\widehat{DCB}\right)=6^{\circ}\end{array}\right|$ . Find $m\left(\widehat{ADC}\right)$ .

Proof 1. Denote $E\in CD\cap AB$ and $m\left(\widehat{DAB}\right)=y$ . Observe that $\left\{\begin{array}{c} DB=DC\\\ EB=ED\\\ AE=AC\end{array}\right|$ . Thus, $\frac {\sin \widehat{BDE}}{\sin\widehat{BED}}=$ $\frac {BE}{BD}=$ $\frac {DE}{DC}=$ $\frac {\sin\widehat{DAE}}{\sin\widehat{DAC}}$ $\implies$ $\frac {\sin 12^{\circ}}{\sin 24^{\circ}}=\frac {\sin y}{\sin \left(48^{\circ}+y\right)}$

$\iff$ $\sin \left(48^{\circ}+y\right)=$ $2\sin y\cos 12^{\circ}\iff$ $\sin \left(48^{\circ}+y\right)=$ $\sin \left(y+12^{\circ}\right)+\sin\left(y-12^{\circ}\right)\iff$ $\sin \left(48^{\circ}+y\right)-\sin \left(y-12^{\circ}\right)=\sin\left(y+12^{\circ}\right)\iff$

$2\sin 30^{\circ}\cos\left(y+18^{\circ}\right)=\sin\left(y+12^{\circ}\right)\iff$ $\cos\left(y+18^{\circ}\right)=\cos\left(78^{\circ}-y\right)\iff$ $y+18^{\circ}=78^{\circ}-y\iff$ $y=30^{\circ}\iff$ $\boxed{m\left(\widehat{ADC}\right)=54^{\circ}}$ .

Proof 2 (synthetic - Sunken Rock). Let $O$ be the circumcenter of $\triangle ABC$ . Oviously, $\Delta ABO$ is equilateral and $m\left(\widehat{BOD}\right)= 48^\circ$ . Therefore,

$\Delta BOD$ is $D-$ isosceles, i.e. $AD\bot BO\implies$ $m\left(\widehat{BAD}\right)=30^\circ$ and $m\left(\widehat{ADB}\right)=138^\circ$ , consequently $m\left(\widehat{ADC}\right)=54^\circ$ .

An easy extension. Let $\triangle ABC$ and an interior point $D$ for which $\left\{\begin{array}{ccc} m\left(\widehat{DBA}\right)=2x & ; & m\left(\widehat{DBC}\right)=x\\\ m\left(\widehat{DCA}\right)=4x & ; & m\left(\widehat{DCB}\right)=x\end{array}\right|$ .

Prove that $m\left(\widehat{DAB}\right)=30^{\circ}\iff x\in\left\{6^{\circ},10^{\circ}\right\}\iff m\left(\widehat{ADC}\right)=30^{\circ}+4x$ .

Proof (trigonometric). Denote $E\in CD\cap AB$ and $m\left(\widehat{DAB}\right)=y$ . Observe that $\left\{\begin{array}{c} DB=DC\\\ EB=ED\\\ AE=AC\end{array}\right|$ . Thus, $\frac {\sin \widehat{BDE}}{\sin\widehat{BED}}=$ $\frac {BE}{BD}=$ $\frac {DE}{DC}=$ $\frac {\sin\widehat{DAE}}{\sin\widehat{DAC}}$ $\implies$ $\frac {\sin 2x}{\sin 4x}=\frac {\sin y}{\sin (y+8x)}$

$\iff$ $\boxed{\sin (y+8x)=2\sin y\cos 2x}\ (*)$ . In conclusion, $y=30^{\circ}\stackrel{(*)}{\iff}$ $\cos \left(60^{\circ}-8x\right)=\cos 2x\iff$ $\left\{\begin{array}{ccc} 60^{\circ}-8x=2x & \iff & x=6^{\circ}\\\\ 60^{\circ}-8x=-2x & \iff & x=10^{\circ}\end{array}\right|\iff$ $x\in\left\{6^{\circ},10^{\circ}\right\}$ .

Remark. $\sin (y+8x)=2\sin y\cos 2x\iff$ $\sin (y+8x)=\sin (y+2x)+\sin (y-2x)\iff$ $\sin (y+8x)-\sin (y-2x)=\sin (y+2x)\iff$

$2\sin 5x\cos (y+3x)=\sin (y+2x)$ . So $2\sin 5x\left(\cos 3x-\sin 3x\tan y\right)=\cos 2x\tan y+\sin 2x\iff$ $\boxed{\tan y=\frac {\sin 8x}{2\cos 2x-\cos 8x}}$ , where $2x<45^{\circ}$ .

PP16. Let $\triangle ABC\ ,\ A=50^{\circ}$ . Suppose that exists an interior $P$ of $\triangle ABC$ so that $PB=PC\ ,\ \left\{\begin{array}{c} E\in PB\cap AC\\\\ F\in PC\cap AB\end{array}\right|$ and $\left\{\begin{array}{c} m\left(\widehat{PEF}\right)=50^{\circ}\\\\ m\left(\widehat{PFE}\right)=30^{\circ}\end{array}\right|$ . Find $m\left(\widehat{ACB}\right)$ .

Proof. Let a fixed $P$ -isosceles $\triangle BPC$ with $m\left(\widehat{BPC}\right)=100^{\circ}$ . Suppose w.l.o.g. $PB=PC=1$ . For two mobile points $E\in (BP\ ,\ F\in (CF$ so that $P\in (BE)\cap (CF)$ and

$\left\{\begin{array}{c} m\left(\widehat{PEF}\right)=50^{\circ}\\\\ m\left(\widehat{PFE}\right)=30^{\circ}\end{array}\right|$ (these angles are constant, i.e. the line $EF$ has a constant slope) denote $A\in BF\cap CE$ and $\left\{\begin{array}{c} m\left(\widehat{BAC}\right)=\alpha\\\\ m\left(\widehat{ABP}\right)=x\\\\ m\left(\widehat{ACP}\right)=y\end{array}\right|$ . Therefore, $A$ is a mobil point for which

$0<\alpha <100^{\circ}$ and $x+y+\alpha =100^{\circ}$ . Thus, if $x$ increases, then $y$ , $x+y$ increase and $\alpha$ decreases. Hence exists and is unique the position of the mobile point $A$ so that $\alpha =50^{\circ}$ .

I"ll show that this position is feasiblely when $x=30^{\circ}$ and $y=20^{\circ}$ , i.e. $\alpha =50^{\circ}\implies$ $x=30^{\circ}\ \wedge\ y=20^{\circ}$ . Indeed, $\alpha =50^{\circ}\implies x+y=50^{\circ}$ and $\sin 30^{\circ}\sin x\sin (100^{\circ}-y)=$

$\sin 50^{\circ}\sin (100^{\circ}-x)\sin y$ $\implies$ $\sin 30^{\circ}\sin x\cos (40^{\circ}-x)=$ $\cos 40^{\circ}\cos (10^{\circ}-x)\cos (40^{\circ}+x)$ $\implies$ $\sin x\cos (40^{\circ} -x)=$ $\cos 40^{\circ}[\cos 50^{\circ}+\cos (30^{\circ}+2x)]$ $\implies$

$\sin 40^{\circ}+\sin (2x-40^{\circ})-\sin 80^{\circ}=$ $2\cos 40^{\circ}\cos (30^{\circ}+2x)$ $\implies$ $\sin (2x-40^{\circ})-2\sin 20^{\circ}\cos 60^{\circ}=$ $2\cos 40^{\circ}\cos (30^{\circ}+2x)$ $\implies$ $\sin (2x-40^{\circ})-\sin 20^{\circ}=$

$2\cos 40^{\circ}\sin (60^{\circ}-2x)$ $\implies$ $\sin (x-30^{\circ})\cos (x-10^{\circ})=$ $\cos 40^{\circ}\sin (60^{\circ}-2x)$ $\implies$ $\sin (x-30^{\circ})\cos (x-10^{\circ})+$ $2\cos 40^{\circ}\sin (x-30^{\circ})\cos (x-30^{\circ})=0$ $\implies$

$\boxed{x=30^{\circ}\ \wedge\ y=20^{\circ}}$ because $x\ne 30^{\circ}$ $\implies$ $\cos (x-10^{\circ})+2\cos 40^{\circ}\cos (x-30^{\circ})=0\iff$ $\cos (x-10^{\circ})+\cos (x+10^{\circ})+$ $\cos (70^{\circ}-x)=0\iff$

$\sin (x+20^{\circ})+2\cos x\cos 10^{\circ}=0$ , what is falsely because $x\in\left(0,50^{\circ}\right)$ .

PP17. Let $\triangle ABC\ ,\ B=30^{\circ}\ ,\ C=15^{\circ}$ . Find $m\left(\widehat{MAC}\right)$ , where $M$ is the midpoint of $[BC]$ .

Proof 1 (metric). Apply theorem of Sine : $\frac {AC}{\sin B}=\frac {BC}{\sin A}\iff$ $\frac b{\frac 12}=\frac {a}{\frac {\sqrt 2}{2}}\iff$ $\boxed{a=b\sqrt 2}\iff$ $b^2=\frac a2\cdot a\iff$

$CA^2=CM\cdot CB\iff$ $\triangle CAM\sim\triangle CBA\iff$ $m\left(\widehat{CAM}\right)=m\left(\widehat{CBA}\right)\iff$ $\boxed{m\left(\widehat{CAM}\right)=30^{\circ}}$ .

Proof 2 (synthetic). Let $w=C(O,R)$ be the circumcircle of $\triangle ABC$ . Prove easily that the $\triangle AOC$ is echilateral and the ray $[AM$ is the bisector $\widehat{OAC}$ . Thus, $\boxed{m\left(\widehat{CAM}\right)=30^{\circ}}$ .

Proof 3 (metric). Apply completly theorem of Sine : $\frac a{\sin 135^{\circ}}=\frac b{\sin 30^\circ}=\frac c{\sin 15^{\circ}}\iff$ $\frac a{\sqrt 2+\sqrt 6}=\frac b{1+\sqrt 3}=\frac c{\sqrt 2}$ . Suppose w.l.o.g. that $\left\{\begin{array}{c} a=\sqrt 2+\sqrt 6\\\\ b=1+\sqrt 3\\\\ c=\sqrt 2\end{array}\right\|\implies$

$4m_a^2=2\left(b^2+c^2\right)-a^2\iff$ $m_a=1\implies$ $\cos\widehat{CAM}=\frac {m_a^2+b^2-\frac {a^2}{4}}{2bm_a}=$ $\frac {\sqrt 3}{2}\implies$ $\boxed{m\left(\widehat{CAM}\right)=30^{\circ}}$ . Otherwise. Denote $m\left(\widehat{CAM}\right)=x$ .

Thus, $\frac {m_a}{\sin\widehat{ACM}}=\frac {MC}{\sin x}\iff$ $\frac 1{\sin 15^{\circ}}=\frac {\sqrt 2+\sqrt 6}{\sin x}\iff$ $\frac 4{\sqrt 6-\sqrt 2}=\frac {\sqrt 2+\sqrt 6}{\sin x}\iff$ $\sin x=\frac 12\iff$ $\boxed{m\left(\widehat{CAM}\right)=30^{\circ}}$ .

Proof 4 (metric). Denote $D\in BC$ so that $AD\perp BC$ and suppose w.l.o.g. $AB=2$ . Prove easily that $AD=1$ , $BD=\sqrt 3$ ,

$DC=\tan 75^{\circ}=2+\sqrt 3$ , $MC=1+\sqrt 3$ and $DM=AD=1\implies$ $m\left(\widehat{DAM}\right)=45^{\circ}\implies$ $\boxed{m\left(\widehat{CAM}\right)=30^{\circ}}$ .

Proof 5 (synthetic - Sayan). Construct an equilateral $\triangle DMC$ such that $\{A,D\}\subset [BC]$ . Since $BMD$ is $M$ -isosceles with $\angle BMD=120^\circ$ obtain $B,A,D$ collinear. Hence

$\angle BDC=90^\circ$ and $\triangle ADC$ is right isosceles. Hence $DM=DC=AD$ implying $ADM$ is $D$ -isosceles. Hence $\angle AMD=\angle MAD=75^\circ$ implying $\angle MAC=30^\circ$ .

Remarks. There is generally the angular relation $\cot \widehat{MAC}=\cot C+2\cot A$ , where $M$ is the midpoint of $[BC]$ .

Easy extension. $\left\{\begin{array}{c} \{M,D\}\subset (BC)\\\ MB=MC\\\ AD\perp BC\end{array}\right|\implies$ $\boxed{AD=DM\iff\cot C=2+\cot B\iff m\left(\widehat{MAC}\right)=45^{\circ}-C}$ .

Proof. Suppose w.l.o.g. that $AB=1$ . Thus, $\left\{\begin{array}{c} BD=\cos B\\\\ AD=\sin B\\\\ DC=\sin B\cot C\end{array}\right|$ and $\boxed{AD=DM}\iff$ $DM=\sin B\iff$ $MC=\sin B\cot C-\sin B\stackrel{MB=MC}{\iff}$

$\cos B+\sin B= \sin B\cot C-\sin B\iff$ $\sin C(\cos B+2\sin B)=\sin B\cos C$ $\iff$ $\tan B-\tan C=2\tan B\tan C\iff$

$\boxed{\cot C=2+\cot B}\iff$ $m\left(\widehat{MAC}\right)=A-\left(90^{\circ}-B\right)-45^{\circ}=$ $A+B-135^{\circ}=$ $\left(180^{\circ}-C\right)-135^{\circ}\iff$ $\boxed{m\left(\widehat{MAC}\right)=45^{\circ}-C}$ .

PP18. Let $ABC$ be a triangle and $D\in (BC)$ be a point such that $AB=CD$ . Suppose

that $m\left(\widehat{CAD}\right)=\alpha\in \left\{12^{\circ},132^{\circ}\right\}$ and $m\left(\widehat{ABD}\right)=144^{\circ}-\alpha$ . Find $m\left(\widehat{ACB}\right)$ .

Proof. Remark that $\boxed{\cos\left(2\alpha +36^{\circ}\right)=\frac 12}\ (1)$ and $\boxed{\sin\frac {72^{\circ}-\alpha}2=\begin{array}{cccc} \nearrow & \frac 12 & \mathrm{if} & \alpha =12^{\circ}\\\\ \searrow & -\frac 12 & \mathrm{if} & \alpha =132^{\circ}\end{array}}\ (2)$ . I"ll show the well-known identity $\boxed{\cos 36^{\circ}=\sin 18^{\circ}+\frac 12 }\ (*)$ . Indeed,

$\cos 36^{\circ}-\cos 72^{\circ}=\frac 12$ $\iff$ $4\sin 54^{\circ}\sin 18^{\circ}=1\ \|\odot \cos 18^{\circ}$ $\iff$ $2\sin 54^{\circ}\sin 36^{\circ}=\cos 18^{\circ}\iff$ $2\cos 36^{\circ}\sin 36^{\circ}=\cos 18^{\circ}\iff$ $\sin 72^{\circ}=\cos 18^{\circ}$ , what is truly.

Now I"ll apply the theorem of Sines: $\left\{\begin{array}{cccc} \triangle ABD\ : & \frac {AD}{AB}=\frac {\sin\widehat {ABD}}{\sin\widehat {ADB}} & \implies & \frac {AD}{AB}=\frac {\sin \left(36^{\circ}+\alpha\right)}{\sin (x+\alpha )}\\\\ \triangle ACD\ : & \frac {AD}{CD}=\frac {\sin\widehat {ACD}}{\sin\widehat {CAD}} & \implies & \frac {AD}{CD}=\frac {\sin x}{\sin\alpha}\end{array}\right|$ $\stackrel{AB=CD}{\implies}$ $\sin x\sin (x+\alpha )=\sin\alpha\sin\left(\alpha +36^{\circ}\right)\iff$ $\cos\alpha -\cos (2x+\alpha )=$

$\cos 36^{\circ}-\cos \left(2\alpha +36^{\circ}\right)\stackrel{(1)}{\iff}$ $\cos (2x+\alpha )=\cos \alpha -\cos 36^{\circ}+\frac 12$ $\stackrel{(*)}{\iff}$ $\cos (2x+\alpha )=\cos \alpha -\sin 18^{\circ}=$ $\cos \alpha -\cos 72^{\circ}=$ $2\sin\frac {\alpha +72^{\circ}}{2}\sin\frac {72^{\circ}-\alpha }{2}\stackrel{(2)}{\implies}$

$\cos (2x+\alpha )=\left\{\begin{array}{ccc} \cos 48^{\circ} & \mathrm{if} & \alpha =12^{\circ}\\\\ \cos \begin{array}{cc} \nearrow & 168^{\circ}\\\\ \searrow & 192^{\circ}\end{array} & \mathrm{if} & \alpha =132^{\circ}\end{array}\right\|\implies$ $\left\{\begin{array}{ccc} 2x+12^{\circ}=48^{\circ} & \mathrm{if} & \alpha =12^{\circ}\\\\ 2x+132^{\circ}\in\left\{168^{\circ},192^{\circ}\right\} & \mathrm{if} & \alpha =132^{\circ}\end{array}\right\|\iff$ $x=\left\{\begin{array}{ccc} 18^{\circ} & \mathrm{if} & \alpha =12^{\circ}\\\\ \begin{array}{cc} 18^{\circ} & \searrow\\\\ 30^{\circ} & \nearrow\end{array} & \mathrm{if} & \alpha =132^{\circ}\end{array}\right\|$ .

This post has been edited 207 times. Last edited by Virgil Nicula, Nov 18, 2015, 12:55 PM

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341. Some nice and interesting "slicing" problems.

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