287. Some metrical problems with or without trigonometry.

by Virgil Nicula, Jun 18, 2011, 1:33 PM

PP1 (Skytin). Let $ABC$ be a triangle with $B=90^{\circ}$ and the incircle $w=C(I,r)$ . Denote the midpoint $M$ of the side

$[AC]$ and the point $X$ for which $X\in MI$ and $AX\perp AI$ . Ascertain the length of the segment $IX=f(b,r)$ .

Proof 1. Supppose w.l.o.g. $a>c$ . Observe that $\left\{\begin{array}{ccc} s=b+r & \iff & a+c=b+2r\\\ ac=2sr & \iff & ac=2r(b+r)\end{array}\right\|\implies \boxed{a-c=\sqrt {b^2-4br-4r^2}}$ . Remark

that $b\ge 2r\left(1+\sqrt 2\right)$ (existence condition). Denote $\theta=m\left(\widehat{BMI}\right)$ . Thus, $\boxed{\tan\theta=\frac {2r}{a-c}}\ ,\ AI=r\sqrt 2\ ,\ m\left(\widehat{AIX}\right)=45^{\circ}+C-\theta$ .

Thus, $\cos\widehat{AIX}=\cos\left(45^{\circ}+C-\theta\right)=$ $\frac {1}{\sqrt 2}\cdot\left[\cos (C-\theta )-\sin (C-\theta )\right]=$ $\frac {1}{\sqrt 2}\cdot\left(\cos C\cos\theta +\sin C\sin\theta -\sin C\cos\theta +\cos C\sin\theta \right)=$

$\frac {1}{b\sqrt 2}\cdot \left(a\cdot \cos\theta +c\cdot \sin\theta -c\cdot \cos\theta +a\cdot\sin\theta\right)=$ $\frac {1}{b\sqrt 2}\cdot \left[(a+c)\cdot\sin\theta +(a-c)\cdot\cos\theta\right]$ . Therefore, $IX=\frac {AI}{\cos \widehat{AIX}}=$

$\frac {2br}{(a+c)\cdot\sin\theta +(a-c)\cdot\cos\theta}=$ $\frac {2br\cdot\sqrt {(a-c)^2+4r^2}}{2r(a+c) +(a-c)^2}=$ $\frac {2br\sqrt{b(b-4r)}}{2r(b+2r)+\left(b^2-4br-4r^2\right)}\implies$ $\boxed{IX=\frac {2r\sqrt{b(b-4r)}}{b-2r}}$ .

Proof 2. Let $E$ , $D$ be the midpoints of the arcs $AC$ and $AB$ of the circumcircle of $\triangle ABC$ . Incircle $(I,r)$ touches $AC$ at $K$ and $P$ is the orthogonal projection

of $B$ onto $AC$ . If $IM$ cuts $BP$ at $U$ , then we know that $BIKU$ is a parallelogram (which is true for any scalene triangle) . Thus $BU=IK=r$ . From

$\triangle XUB \sim \triangle XMD$ we get $\frac{BX}{BX+BD}=\frac{BU}{DM}=\frac{2r}{b} \Longrightarrow \ BX= BD \cdot \frac{2r}{b-2r}$ . Hence, by the Pythagoras' theorem for $\triangle BIX$ with hypotenuse $IX$ we

get $IX^2=BD^2 \cdot \frac{4r^2}{(b-2r)^2}+2r^2=2r^2 \cdot \frac{2BD^2+(AC-2r)^2}{(b-2r)^2} \ (\star)$ . Since $EA=EC=EI= \frac{\sqrt{2}}{2}b$ , then $EB=EI+IB= \sqrt{2} \left (r+\frac{b}{2} \right)$ .

Hence, by the Pythagoras' theorem for $\triangle DEB$ with hypotenuse $ED=b$ we get $BD^2b^2-2 \left ( r+ \frac{b}{2} \right)^2=\frac{b^2}{2}-2r^2-2r \cdot b$ . Substituting $BD^2$ from

the latter equation into $(\star)$ yields $IX^2=2r^2 \cdot \frac{b^2-4r^2-4r \cdot b+b^2+4r^2-4r \cdot b}{(b-2r)^2} \ \Longrightarrow$ $IX= 2r \cdot \frac{\sqrt{b(b-4r)}}{b-2r}$ .

PP2. In $\triangle ABC$ consider the points $\left\{\begin{array}{cc} D\in (BC)\ ;\ \widehat{DAB}\equiv\widehat {DAC}\\\ M\in (AB)\ ;\ \widehat{ADM}\equiv\widehat{ABC}\\\ N\in (AC)\ ;\ \widehat{ADN}\equiv\widehat{ACB}\end{array}\right\|$ . Denote $P\in AD\cap MN$ . Prove that $AD^3 = AB\cdot AC\cdot AP$ .

Proof. $\left\{\begin{array}{ccc} \triangle ADM\sim\triangle ABD & \implies & AD^2=AM\cdot AB\\\ \triangle ADN\sim\triangle ACD & \implies & AD^2=AN\cdot AC\end{array}\right\|\implies$ $AD^4=bc\cdot AM\cdot AN\ (1)$ . Since $AM\cdot AB=AN\cdot AC\iff$ $AMDN$

is cyclically. Observe that $\triangle AMD\sim\triangle APN$ $\implies$ $AM\cdot AN=AD\cdot AP\ (2)$ . From the relations $(1)$ and $(2)$ obtain that $AD^3 = AB\cdot AC\cdot AP$ .

PP3. Let $H$ be the orthocenter of an acute triangle $ABC$ . The circle centered at the midpoint of $BC$ and passing through $H$ intersects

the line $BC$ at $A_{1}$ , $A_{2}$ . Similarly define the pairs $B_{1}$ , $B_{2}$ and $C_{1}$ , $C_{2}$ . Prove that $A_{1}$ , $A_{2}$ , $B_{1}$ , $B_{2}$ , $C_{1}$ , $C_{2}$ are concyclically.

Proof 1. Denote the midpoints $D$ , $E$ , $F$ of the sides $[BC]$ , $[CA]$ , $[AB]$ respectively. Prove easily or is well-known that

$\boxed {\ HA^2 + a^2 = HB^2 + b^2 = HC^2 + c^2 = 4R^2\ }\ (*)$ . Thus, $4\cdot\overline {BA_1}\cdot\overline {BA_2} = 4\cdot\left(BD^2 - HD^2\right) =$ $a^2 - 4\cdot HD^2 =$

$a^2 - 2\cdot\left(HB^2 + HC^2\right) + a^2$ $\stackrel {(*)}{\ \ \implies\ \ }$ $\overline {BA_1}\cdot\overline {BA_2} = \frac 12\cdot \left(a^2 + b^2 + c^2\right) - 4R^2$ $\implies$ $\boxed {\ \overline {BA_1}\cdot\overline {BA_2} = 4R^2\prod\cos A\ }$

because prove easily or is well-known that $\boxed {\ \left(a^2 + b^2 + c^2\right) - 8R^2 = 8 R^2\prod\cos A\ } > 0$ (remark that $\{A_1,A_2\}\subset (BC)$ a.s.o.).

Thus, $\overline {BA_1}\cdot \overline {BA_2} = - \overline {A_1B}\cdot\overline {A_1C} = R^2 - OA_1^2\ \implies\ 4R^2\prod\cos A = R^2 - OA_1^2$ $\implies$ $OA_1^2 = R^2\left(1 - 4\prod\cos A\right)$

(symmetrically in $a$ , $b$ , $c$ ). In conclusion, $\boxed {\ \rho = OA_1 = OA_2 = OB_1 = OB_2 = OC_1 = OC_2 = R\sqrt {1 - 4\cos A\cos B\cos C}\ }$ .

Remark. Since $1 - 8\prod\cos A\ge 0$ obtain easily that $\rho\ge 2R\sqrt {\prod\cos A}$ .

Proof 2. Denote the circumcenter $O$ and the midpoint $E$ of $[HO]$ (center of the Euler's circle). Observe

that $OA_1^2=OD^2+DA_1^2=OD^2+DH^2=2\cdot DN^2+\frac 12\cdot HO^2$ (is symmetrically in $a$ , $b$ , $c$ ).

Proof 3. Denote diameter $[AA']$ of circumcircle $w=C(O,R)$ . Apply theorem of median to $[HO]$ in $\triangle A'HA\ :\ 4\cdot HO^2=$ $2\cdot\left(HA^2+A'H^2\right)-A'A^2=$

$2\cdot\left(4\cdot DO^2+4\cdot DH^2\right)-R^2$ . Thus, $HO^2=2\cdot\left(DO^2+DA_1^2\right)-R^2=$ $2\cdot OA_1^2-R^2$ . In conclusion, $\boxed{2\cdot OA_1^2=HO^2+R^2}$ .

I used the well-known property : the quadrilateral $HBA'C$ is a parallelogram, i.e. $D\in HA'$ and $DH=DA'$ .

Proof 4. $4\cdot OA_1^2=4\cdot \left(DA_1^2+DO^2\right)=$ $4\cdot DH^2+HA^2=$ $2\cdot\left(HB^2+HC^2\right)-a^2+HA^2=$

$2\cdot\left(HA^2+HB^2+HC^2\right)-\left(a^2+HA^2\right)\implies$ $\boxed {2\cdot OA_1^2=\left(HA^2+HB^2+HC^2\right)-2R^2}$ .

Proof 5. Denote $S\in BC\cap AH$ and the midpoint $E$ of $[HO]$ - the center of the Euler's circle, i.e. $ED=ES=\frac R2$ . Observe that $OA_1^2=DA_1^2+DO^2=$

$DH^2+DO^2=$ $HS^2+SD^2+DO^2=$ $HS^2+OS^2$ . Apply the theorem of median $SE$ in $\triangle HSO\ :\ 2\cdot SE^2=$ $\left(SH^2+SO^2\right)-\frac 12\cdot HO^2$ .

Therefore, $OA_1^2=2\cdot SE^2+\frac 12\cdot HO^2=$ $2\cdot\frac {R^2}{4}+\frac 12\cdot HO^2$ , i.e. $2\cdot OA_1^2=HO^2+R^2$ a.s.o.

Proof 6. Fix $O$ as center of a coordinate vector system, so that $H = A + B + C$ . If $M = \frac{A + B}2$ we have to prove that $MC_1^2 + OM^2$ is symmetric in $A$ , $B$ and $C$

(it would be the square of the radius of the circle). But $MC_1^2 + OM^2 = HM^2 + OM^2 =$ $\left(A + B + C - \frac{A+B}2\right)\cdot \left(A + B + C - \frac{A+B}2\right) +$

$\left(\frac{A+B}2\right)\cdot \left(\frac{A+B}2\right) =$ $\frac12A\cdot A + \frac12B\cdot B + C\cdot C + A\cdot B + B\cdot C + C\cdot A =$ $2R^2 + A\cdot B + B\cdot C + C\cdot A$ and we are done.

Remark. Observe that $HA^2+HB^2+HC^2=HO^2+3R^2$ . Prove easily that for any point $X$ we have $\overrightarrow{XA}+\overrightarrow{XB}+\overrightarrow{XC}=\overrightarrow{XH}+2\cdot\overrightarrow{XO}$ .

PP4. Let $P$ be an interior point of the square $ABCD$ so that $PA=1$ , $PB=2$ , $PC=3$ . Find the length of $[AB]$ .

Proof. Let $l=AB$ and $\phi =m\left(\widehat{ABP}\right)$ . Thus, $m\left(\widehat{CBP}\right)=90^{\circ}-\phi$ and $l\sqrt 2>3\ ,\ l<1+2$ , i.e. $l\in\left(\frac {3\sqrt 2}{2},3\right)\ (*)$ .

Apply the generalized Pytagoras' theorem to $:\ \left\{\begin{array}{ccc} PA/\triangle ABP & \implies & 4l\cdot\cos\phi =l^2+3\\\ PC/\triangle CPB & \implies & 4l\cdot\sin \phi =l^2-5\end{array}\right\|\implies$

$\left(l^2+3\right)^2+\left(l^2-5\right)^2=16l^2\iff$ $l^4-10l^2+17=0\iff$ $l^2\in\{5\pm2\sqrt 2\}\stackrel{(*)}{\implies}$ $\boxed{\ l=\sqrt{5+2\sqrt 2}\ }$ .

PP5. Prove that in any triangle $ABC$ exists the implication $A=2B\implies a^2=b(b+c)$ .

Proof 1. Denote the points $\left\{\begin{array}{ccc} D\in (BC) & ; & \widehat{DAB}\equiv\widehat{DAC}\\\\ E\in AB\ ,\ A\in (BE) & ; & AB=AE\end{array}\right\|$ . Observe that $A=2B\implies$ $\widehat{EBC}\equiv$

$\widehat{DAB}\equiv$ $\widehat{DAC}\equiv$ $\widehat{BEC}\equiv$ $\widehat{ECA}$ $\implies$ $AE=AC$ and $\triangle EAC\sim\triangle ECB\implies$ $\frac {EA}{EC}=\frac {EC}{EB}\implies$ $a^2=b(b+c)$ .

Proof 2. $A=2B\implies$ $\sin A=\sin 2B\iff$ $\sin A=2\sin B\cos B\iff$ $a=2b\cos B$ ,

i.e. $a^2c=b\left(a^2+c^2-b^2\right)\implies$ $a^2(b-c)=b\left(b^2-c^2\right)\iff$ $b=c\ \ \vee\ \ a^2=b(b+c)$ .

Proof 3. Consider the circumcircle $w$ of $\triangle ABC$ and denote the second intersection $E$ of the $A$ -angle bisector

with $w$ . Prove easily that $A=2B\implies$ $ABEC$ is isosceles triangle , where $BE=EC=b$ and $AE=a$ .

Using the Ptolemy's theorem obtain that $AE\cdot BC=$ $EB\cdot AC+EC\cdot AB$ , i.e. $a^2=b(b+c)$ .

Remark. In this case $AE=a$ and $\triangle ABE\sim\triangle ADC\implies$ $AD\cdot AE=AB\cdot AC\implies$ $AD\cdot a=bc\implies$ $AD=\frac {bc}{a}$ .

Proof 4. $a^2=b(b+c)\iff$ $\sin^2A=\sin B(\sin B+\sin C)\iff$ $\sin^2A-\sin^2B=\sin B\sin C\iff$

$\sin C\sin (A-B)=$ $\sin B\sin C\iff$ $\sin (A-B)=\sin B\iff$ $A-B=B\ \vee\ (A-B)+B=\pi\iff$ $A=2B$ .

PP6. Let $ABC$ be a triangle. Prove that $\boxed{3A+2B=180^{\circ}\iff c^2=a^2+bc}$ .

Proof. $c^2=a^2+bc\iff$ $\sin^2C=\sin^2A+\sin B\sin C\iff$ $\sin (C-A)\sin (C+A)=\sin B\sin C\iff$

$\sin (C-A)=$ $\sin C\iff$ $(C-A)+C=\pi\iff$ $2C=\pi +A\iff$ $2(\pi -A-B)=\pi +A\iff$ $3A+2B=\pi$ .

PP7. Let $ABC$ be a triangle. Prove that $\boxed{A=3B\ \implies\ c^2=b(a+b)\ \vee\ (a+b)(a-b)^2=bc^2}$ .

Proof 1. Suppose $A=3B$ . Denote circumcircle $w$ of $\triangle ABC$ , points $\{M,N\}\subset (BC)$ so that $M\in (BN)$ and $m\left(\widehat{MAB}\right)=$ $m\left(\widehat{MAN}\right)=$

$m\left(\widehat{NAC}\right)=B$ , second intersections $E$ , $F$ of $AM$ , $AN$ with $w$ respectively and $AF=BF=EC=x$ . So $AE=a$ and $BE=EF=FG=b$ .

Apply Ptolemy's theorem to : $\left\{\begin{array}{cccc} ABEC\ : & cx+b^2=a^2 & \implies & x=\frac {a^2-b^2}{c}\\\\ AEFC\ : & & ab+b^2=x^2 & \\\\ ABEF\ : & bx+bc=ax & \implies & x=\frac {bc}{a-b}\end{array}\right\|$ $\implies$ $\frac {a^2-b^2}{c}=\frac {bc}{a-b}\iff$ $(a+b)(a-b)^2=bc^2$ .

Proof 2. Observe that $m(\angle ADC)=2B$ , $m(\angle AEC)=3B$ . Hence $\triangle ACD$ is isosceles $\implies CD=b\implies$ $\boxed{BD=a-b}\implies$

$\boxed{AD=a-b}$ . By the Angle Bisector Theorem in the $\triangle ACD$ we get $\boxed{CE={b^2\over a}}$ , $\boxed{DE={b(a-b)\over a}}$ . Since $\triangle ACE\sim\triangle BCA$ we have

$\boxed{AE={bc\over a}}$ . Since $\triangle ADE\sim BAE$ we have $EA^2=EB\cdot ED\iff {b(a-b)\over a}\cdot{a^2-b^2\over a}={b^2c^2\over a^2}\iff (a-b)^2(a+b)=bc^2$ .

Proof 3. $A=3B\implies$ $\sin A=\sin B\left(3-4\sin^2B\right)\implies$ $\sin A=\sin B\left(-1+4\cos^2B\right)\iff$ $\frac {a+b}{4b}=\left(\frac {a^2+c^2-b^2}{2ac}\right)^2\iff$

$a^2c^2(a+b)=b\left(a^2+c^2-b^2\right)^2\iff$ $bc^4+\left[2b\left(a^2-b^2\right)-a^2(a+b)\right]c^2+b\left(a^2-b^2\right)^2=0\iff$

$b\cdot c^4-(a+b)\left[(a-b)^2+b^2\right]\cdot c^2+b(a-b)^2(a+b)^2=0\iff$ $(a+b)(a-b)^2c^2+b^2(a+b)c^2=bc^4+b(a-b)^2(a+b)^2\iff$

$(a+b)((a-b)^2\left[b(a+b)-c^2\right]=bc^2\left[b(a+b)-c^2\right]=0\iff$ $c^2=b(a+b)\ \vee\ (a+b)(a-b)^2=bc^2$ .

Proof 4. $(a+b)(a-b)^2=bc^2\iff$ $(a-b)\left(a^2-b^2\right)=bc^2\iff$ $(\sin A-\sin B)\sin (A-B)\sin (A+B)=\sin B\sin ^2C\iff$

$\sin A\sin (A-B)=\sin B\left[\sin (A-B)+\sin C\right]\iff$ $\cos B-\cos (2A-B)=\cos (2B-A)-\cos A+\cos (B-C)+\cos A\iff$

$\cos B-\cos (B-C)=\cos (2A-B)+\cos (2B-A)\iff$ $-\sin\frac {2B-C}{2}\sin \frac C2=\cos\frac {A+B}{2}\cos\frac {3(A-B)}{2}\iff$

$\cos\frac {\pi +2B-C}{2}=\cos\frac {3A-3B}{2}\iff$ $\cos\frac {A+3B}{2}=\cos\frac {3A-3B}{2}\iff$ $A+3B=3A-3B\iff$ $A=3B$ .

PP8. Let $ABC$ be a triangle with the circumcircle $w=C(O,R)$ and the orthocenter $H$ . Show that there exist the points

$D\in BC$ , $E\in CA$ and $F\in AB$ such that $OD+DH=OE+EH=OF+FH$ and $AD\cap BE\cap CF\ne\emptyset$ .

Proof. Denote the reflection $X$ , $Y$ , $Z$ of $H$ w.r.t. $BC$ , $CA$ , $AB$ respectively. Can choose $D\in OX\cap BC$ , $E\in OY\cap CA$ and $F\in OZ\cap AB$ .

Prove easily that $OD+DH=OE+EH=OF+FH=R$ and $\boxed{\frac {DB}{DC}=\frac {\sin 2B}{\sin 2C}}$ , i.e. $D$ is the symetrical point of $U\in AO\cap BC$

w.r.t. the midpoint of $[BC]$ . Indeed, $\frac {DB}{DC}=\frac{XB}{XC}\cdot\frac {\sin\widehat{DXB}}{\sin\widehat{DXC}}=$ $\frac {HB}{HC}\cdot\frac {\sin [C+(B-C)]}{\sin [B-(B-C)]}\implies$ $\frac {DB}{DC}=\frac {\cos B\sin B}{\cos C\sin C}=\frac {\sin 2B}{\sin 2C}$ .

Show analogously that $\frac {EC}{EA}=\frac {\sin 2C}{\sin 2A}$ and $\frac {FA}{FB}=\frac {\sin 2A}{\sin 2B}$ . In conclusion, using the Ceva's theorem obtain that $AD\cap BE\cap CF\ne\emptyset$ .

Lemma. A point $D$ lies inside of $\triangle ABC$ . Let $A_1$ , $B_1$ , $C_1$ be the second intersections of $AD$ , $BD$ , $CD$

with the circumcircles of $BDC$ , $CDA$ , $ADB$ respectively. Prove that $\frac {AD}{AA_1} + \frac {BD}{BB_1} + \frac {CD}{CC_1} = 1$ .

Proof. Let $A_1B, A_1C$ cut the circumcircles of the triangles $ABD$ and $ACD$ again at $C_2, B_2$ . By the Miquel's theorem obtain that $A\in B_2C_2$ .

From the circumircles of the triangles $ACD$ and $BCD$ we have that $\angle B_1B_2C = \angle BA_1C$ , which means $B_1B_2 \parallel A_1B$ and similarly,

$C_1C_2 \parallel A_1C$ . Let $B_2D$ , $C_2D$ cut $A_1C_2$ , $A_1B_2$ at $B_0$ , $C_0,$ respectively. Then $\frac {BD}{BB_1} = \frac {B_0D}{B_0B_2}$ and $\frac {CD}{CC_1} = \frac {C_0D}{C_0C_2}$ . Since $A_1A$ , $B_2B_0$ ,

$C_2C_0$ are concurrent cevians at $D$ in the $\triangle A_1B_2C_2$ , it follows that $\frac {AD}{AA_1} + \frac {BD}{BB_1} + \frac {CD}{CC_1} =$ $\frac {AD}{AA_1} + \frac {B_0D}{B_0B_2} + \frac {C_0D}{C_0C_2} = 1$ .

PP9. Let $M$ and $N$ be points inside triangle $ABC$ such that $m(\angle MAB)=m(\angle NAC)$ and

$m(\angle MBA)=m(\angle NBC)$ . Prove that $\frac {AM \cdot AN}{AB \cdot AC} + \frac {BM \cdot BN}{BA \cdot BC} + \frac {CM \cdot CN}{CA \cdot CB} = 1$ .

Proof. It is quite well-known that $\angle MCA = \angle NCB$ , which can be proven by taking the sine version of Ceva's Theorem on $M$ and $N$ on dividing (I won't do

that here.) Denote the second intersections $P$ , $Q$ and $R$ of $AM$ , $BM$ , and $CM$ respectively with the circumcircles of the triangles $BMC$ , $CMA$ and $AMB$

respectively. Observe that $\triangle ANB\sim \triangle ACP$ , $\triangle BAQ\sim \triangle BNC$ , and $\triangle CNA\sim \triangle CBR$ . By the Miquel's theorem obtain that $\left\{\begin{array}{c} A\in QR\\\ B\in RP\\\ C\in PQ\end{array}\right\|$ .

By upper lemma have $\frac {BM}{BQ} + \frac {AM}{AP} + \frac {CM}{CR} = 1$ . Since $\triangle ANB\sim \triangle ACP$ $\implies$ $\frac {AB}{AN} = \frac {AP}{AC}$ $\implies$ $\frac {AB\cdot AC}{AN} = AP$ $\implies$ $\frac {AM}{AP} = \frac {AM\cdot AN}{AB\cdot AC}$ .

Similarly, we have that $\frac {BM}{BQ} = \frac {BM\cdot BN}{AB\cdot BC}$ and $\frac {CM}{CR} = \frac {CM\cdot CN}{AC\cdot BC}$ . Hence, $\frac {BM\cdot BN}{AB\cdot BC} + \frac {CM\cdot CN}{AC\cdot BC} + \frac {AM\cdot AN}{AB\cdot AC} =$ $\frac {AM}{AP} + \frac {BM}{BQ} + \frac {CM}{CR} = 1$ .

PP10. Consider a right triangle $ABC$ with $m(\angle BAC)=90^{\circ}$ and $AB=AC$ . Let $M$ and $N$ be two

points on hypotenuse $(BC)$ . Prove that $BM^2+CN^2=MN^2\ \iff\ m(\angle MAN)=45^{\circ}$ .

Proof 1 (metric). Denote the midpoint $D$ of $[BC]$ and suppose that $AD=1$ , $M\in (BD)$ . Then $AB=AC=\sqrt 2$ ,

$DB=DC=1$ and $N\in (DC)$ . Denote $\left\{\begin{array}{c} m(\angle DAB)=x\\\ m(\angle DAC)=y\end{array}\right|$ . Then $\left\{\begin{array}{c} DM=\tan x\\\ DN=\tan y\end{array}\right|$ and $MN=\tan x+\tan y$ .

Therefore, $\left\{\begin{array}{ccc} \frac {MB}{\sin (45-x)}=\frac {\sqrt 2}{\cos x} & \implies & MB=1-\tan x\\\\ \frac {MC}{\sin (45-y)}=\frac {\sqrt 2}{\cos y} & \implies & MB=1-\tan y\end{array}\right|$ .

In conclusion, $\boxed{\ MN^2=BM^2+CN^2\ }\iff$ $\left(\tan x+\tan y\right)^2=(1-\tan x)^2+(1-\tan y)^2\iff$

$2\tan x\tan y=2-2(\tan x+\tan y)\iff$ $\frac {\tan x+\tan y}{1-\tan x\tan y}=1\iff$ $\tan (x+y)=1\iff$ $\boxed{\ x+y=45^{\circ}\ }$ .

Proof 2 (analytical). Chose a system of coordinates for which $A(0,0)\ ,\ B(1,0)\ ,\ C(0,1)$ . Let $M(m,1-m)\ ,\ N(n,1-n)$ so that

$0<n<m<1$ . Denote $\phi =m\left(\widehat{MAN}\right)\in \left[0,\frac {\pi}{2}\right]$ . The slopes of the lines $AM$ and $AN$ are $s_m=\frac {1-m}{m}$ and $s_n=\frac {1-n}{n}$ respectively.

Thus, $\left\{\begin{array}{cccccc} MB^2+NC^2=MN^2 & \iff & (m-1)^2+n^2=(m-n)^2 & \iff & 1+2mn=2m & (1)\\\\ \tan\phi =\frac {\frac {1-n}{n}-\frac {1-m}{m}}{1+\frac {(1-n)(1-m)}{nm}} & \iff & \tan\phi =\frac {m-n}{1+2mn-(m+n)} & \iff & \phi =\arctan \frac {m-n}{1+2mn-(m+n)} & (2)\end{array}\right|$ . From the relations

$(1)$ and $(2)$ obtain that $\phi =45^{\circ}\iff$ $\tan\phi =1\iff$ $\frac {m-n}{1+2mn-(m+n)}=1\iff$ $1+2mn=2m\iff$ $MB^2+NC^2=MN^2$ .

Proof 3 (synthetical). Rotate point $M$ clockwise by $90^\circ$ about $A$ to $M'$ . Then, $M'C = BM$ and $M'C \perp NC \implies NM'^2 = BM^2 + CN^2 = MN^2$ .

Because $AM = AM'$ , $AN = AN$ , $MN = M'N$ , it follows that $\triangle MAN \cong \triangle M'AN$ . But $\angle MAM'$ is a right angle. Hence, $\angle MAN = 45^\circ$ .

PP11. Fie paralelogramul $ABCD\ ,\ B>90^{\circ}$ . Notam proiectiile $E$ si $F$ ale lui $C$ pe $AB$ , $AD$ respectiv. Demonstrati relatia $AE\cdot AB+AD\cdot AF=AC^2$ .

Metoda 1 (Stefan Spataru). Notam mijloacele $M$ si $N$ ale laturilor $BC$ si $CD$ respectiv. Din teorema medianei aplicata $A$ -medianelor din $\triangle BAC$ si $\triangle DAC$

se obtin relatiile $\left\{\begin{array}{c} 4\cdot AM^2=2\cdot\left(AC^2+AB^2\right)-BC^2\\\ 4\cdot AN^2=2\cdot\left(AC^2+AD^2\right)-CD^2\end{array}\right|$ . Din puterea punctului $A$ fata de fiecare dintre cercurile cu diametre $[BC]$ , $[CD]$ se obtin relatiile

$\left\{\begin{array}{cccc} 4\cdot AB\cdot AE=4\cdot AM^2-BC^2\\\ 4\cdot AD\cdot AF=4\cdot AN^2-CD^2\end{array}\right|\implies$ $4\cdot\left(AB\cdot AE+AD\cdot AF\right)=$ $2\cdot\left(AC^2+AB^2-BC^2\right)+$ $2\cdot\left(AC^2+AD^2-CD^2\right)=4\cdot AC^2$ .

Metoda 2. Aplicam teorema Pitagora generalizata laturilor $[CB]$ si $[CD]$ in $\triangle ABC$ si $\triangle ADC\ :\ \left\{\begin{array}{c} CB^2=AC^2+AB^2-2\cdot AB\cdot AE\\\ CD^2=AC^2+AD^2- 2\cdot AD\cdot AF\end{array}\right|\ \bigoplus\ \implies$

$AE\cdot AB+AD\cdot AF=AC^2$ deoarece $\left\{\begin{array}{c} CB=AD\\\ CD=AB\end{array}\right|$ , iar unghiurile $\angle BAC$ si $\angle DAC$ sunt ascutite.

PP12. Let $ABC$ be an acute triangle with incenter $I$ , circumcircle $w=C(O,R)$ and

orthocenter $H$ . Let $\{A,A'\}=AI\cap w$ . Prove that $A=60^{\circ}\implies$ $AA'=HB+HC$ .

Proof 1. $HB+HC=AA'\iff$ $2R(\cos B+\cos C)=2R\sin\left(B+\frac A2\right)\iff$ $2\cos\frac {B-C}{2}\sin\frac A2=\cos\frac {C-B}{2}\iff$ $A=60^{\circ}$ .

Proof 2. Denote $\{B,L\}=BH\cap w$ . Therefore, $A=60^{\circ}\iff$ $m(\angle ABH)=m(\angle BAA')=30^{\circ}\iff$ $A'L\parallel AB\iff$ $AA'=BL\ (1)$ . On

other hand, observe that $A=60^{\circ}\iff$ $\triangle CHL$ is equilateral. In conclusion, $A=60^{\circ}\iff$ $AA'=BL=BH+HL\iff$ $AA'=HB+HC$ .

Proof 3 (Mateescu Constantin). It is well-known that $\boxed{AH+BH+CH=2\left(R+r\right)}$ (in acute-angled triangles) and that

$\boxed{60^{\circ}\in\{A,B,C\}\iff s=\sqrt 3\cdot\left(R+r\right)}$ (see here a proof for the latter). Thus, the conclusion reduces to proving that $A=60^{\circ}\implies$

$AA^{\prime}+AH=\frac {2s}{\sqrt 3}$ . Let $D$ be the foot of the internal bisector coresponding to vertex $A$ . Then $l_a=AD=\frac {2bc}{b+c}\cdot\cos 30^{\circ}=\frac {bc\sqrt 3}{b+c}$ and

since $AD\cdot DA^{\prime}=BD\cdot DC=\frac {a^2bc}{(b+c)^2}$ one obtains that: $DA^{\prime}=\frac {a^2}{\sqrt 3\left(b+c\right)}$ . In addition, by using the law of cosine in $\triangle\, ABC$ get

$a^2=b^2+c^2-bc$ so $AA^{\prime}=\frac {a^2+3bc}{\sqrt 3\left(b+c\right)}=\frac {b+c}{\sqrt 3}$ . Also, $AH=2R\cos A=R=\frac a{\sqrt 3}$ (the last equality following from the law of sines).

In conclusion, $AA^{\prime}+AH=\frac {b+c}{\sqrt 3}+\frac a{\sqrt 3}=\frac {2s}{\sqrt 3}$ , so our proof is now completed.

PP13. Let $ABC$ be a triangle. Choose two points $\{D,E\}\subset (BC)$ so that $D$ $\widehat{BAD}\equiv\widehat{CAE}$ . The incircles of $\triangle ABD$

and $\triangle ACE$ are tangent to sideline $BC$ at $M$ and $N$ respectively. Prove that $\frac{1}{MB}+\frac{1}{MD}=\frac{1}{NC}+\frac{1}{NE}$ .

Proof. $\left\{\begin{array}{ccccccc} 2\cdot MB & = & AB+BD-AD & ; & 2\cdot NC & = & AC+CE-AE\\\ 2\cdot MD & = & -AB+BD+AD & ; & 2\cdot NE & = & -AC+CE+AE\end{array}\right|\ \ (1)\ .$

$\frac {[ABD]}{[ACE]}=\frac {BD}{CE}\ \implies\ \frac {AB\cdot AD}{AC\cdot AE}=$ $\frac {BD}{CE}\ \implies\ BD\cdot AE\cdot AC=CE\cdot AD\cdot AB\ \ (2)\ .$

Denote $\phi =m\left(\widehat{BAD}\right)=$ $m\left(\widehat{CAE}\right)\ \implies\ \left\{\begin{array}{c} 2\cdot AE\cdot AC\cdot\cos\phi =AE^2+AC^2-CE^2\\\ 2\cdot AD\cdot AB\cdot\cos\phi=AD^2+AB^2-BD^2\end{array}\right|\ \ (3)\ .$

$\frac{1}{MB}+\frac{1}{MD}=$ $\frac{1}{NC}+\frac{1}{NE}\ \iff\ BD\cdot NC\cdot NE=$ $CE\cdot MB\cdot MD\ \stackrel{(1)}{\iff}\ BD\cdot\left[CE^2-(AE-AC)^2\right]=$

$CE\cdot\left[BD^2-(AD-AB)^2\right]\ \stackrel {(2)}{\iff}\ BD\left[AE^2+AC^2-CE^2\right]=$ $CE\cdot\left[AD^2+AB^2-BD^2\right]\ \stackrel{(3)}{\iff}$

$2\cdot BD\cdot AE\cdot AC\cdot\cos\phi =2\cdot CE\cdot AD\cdot AB\cdot\cos\phi$ , what is truly again from same relation $(2)$ .

PP14. Let $ABCD$ be a parallelogram. Let $X\in (DA)$ , $Y\in (DB)$ and $Z\in (DC)$

so that $DXYZ$ is cyclically. Prove that $\boxed{DY\cdot DB=DX\cdot DA+DZ\cdot DC}$ .

Proof. Apply the Ptolemy's relation to $DXYZ\ :\ \ DY\cdot XZ=DX\cdot YZ+DZ\cdot XY\ (*)$ .

Observe that $\left\{\begin{array}{ccc} \widehat{YXZ}\equiv\widehat{YDZ}\equiv\widehat{BDC}\equiv\widehat{DBA} & \implies & \widehat{YXZ}\equiv\widehat{DBA}\\\\ \widehat{YZX}\equiv\widehat{YDX}\equiv\widehat{BDA} & \implies & \widehat{YZX}\equiv\widehat{BDA}\end{array}\right|\ \implies$

$\triangle XYZ$ $\sim\triangle BAD\ \implies\ \frac {XY}{AB}=$ $\frac {YZ}{DA}=$ $\frac {XZ}{DB}\ \stackrel{(*)}{\implies}\ DY\cdot DB=DX\cdot DA+DZ\cdot DC$ .

Particular case.. Let $ABCD$ be a parallelogram. The circle with diameter $[BD]$ meet $(BC$ , $(BA$ at $E$ , $F$ respectively $\implies \boxed{BD^2=BC\cdot BE+BA\cdot BF}$ .

PP15. Let $ABCD$ be a cyclical quadrilateral. Let projections $E$ , $F$ of $C$ on $AB$ , $AD$ respectively. Prove that $\boxed{AC\cdot BD=AE\cdot DC+AF\cdot BC}\ \ (*)$ .

Proof 1. Apply the Ptolemy's theorem to $AECF\ :\ \ AC\cdot EF=AF\cdot CE+AE\cdot CF\ \ (1)$ . Observe that $\widehat{FEC}\equiv\widehat{CAD}\equiv\widehat{DBC}\Longrightarrow \widehat{FEC}\equiv\widehat{DBC}$ .

Prove analogously that $\widehat{EFC}\equiv\widehat{BAC}\equiv\widehat{BDC}$ . In conclusion, using the relation $(1)$ obtain that $\triangle{ECF}\sim\triangle{BCD}$ $\Longrightarrow$ $\frac{EF}{BD}=$ $\frac{CF}{CD}=\frac{CE}{BC}\ \implies\ (*)$ .

Proof 2. Observe that $\widehat{BCE}\equiv\widehat {DCF}$ . Thus, $\triangle BCE\sim\triangle DCF\iff$ $\frac {BE}{BC}=\frac {DF}{DC}\ \ (1)$ . Apply Ptolemy's theorem to $ABCD\ :$

$AC\cdot BD=AB\cdot DC+AD\cdot BC\ \ (2)$ . In conclusion, the required relation $AC\cdot BD=AE\cdot DC+AF\cdot BC$ is equivalently

with $AC\cdot BD=(AB+BE)\cdot DC+(AD-FD)\cdot BC\stackrel{(2)}{\iff}$ $BE\cdot DC=FD\cdot BC\iff (1)$ .

Proof 3. Denote $\left\{\begin{array}{c} m\left(\widehat{CBD}\right)=x\\\\ m\left(\widehat{CDB}\right)=y\end{array}\right|$ . Since $BD=CB\cdot\cos x+CD\cdot\cos y$ and $\left\{\begin{array}{c} AE=AC\cdot\cos y\\\\ AF=AC\cdot\cos x\end{array}\right|$ obtain that

$AC\cdot BD=AE\cdot DC+AF\cdot BC\iff$ $AC\cdot \left(CB\cdot\cos x+CD\cdot\cos y\right)=AC\cdot \cos y\cdot DC+AC\cdot \cos x\cdot BC$ , what is truly.

PP16. Consider in $\triangle ABC$ a point $D\in (BC)$ . The bisectors of the angles $\widehat{ADB}$ and $\widehat{ADC}$ cut the sides $[AB]$

and $[AC]$ at $E$ and $F$ respectively. Prove that $AB\cdot AE+AC\cdot AF=AD\cdot BC\ (*)\iff$ $m(<A)=90^\circ$ .

Proof. I"ll use the Stewart's relation $a\cdot AD^2+a\cdot DB\cdot DC=b^2\cdot DB+c^2\cdot DC\ (1)$ . Since $\left\{\begin{array}{c} AE=\frac {c\cdot AD}{DA+DB}\\\\ AF=\frac {b\cdot AD}{DA+DC}\end{array}\right|$ ,

the relation $(*)$ becomes $c\cdot \frac {c\cdot AD}{DA+DB}+$ $b\cdot \frac {b\cdot AD}{DA+DC}=a\cdot AD\iff$ $\left(b^2+c^2\right)\cdot AD+\underline{c^2\cdot DC+b^2\cdot DB}=$

$a^2\cdot AD+\underline{a\cdot AD^2+a\cdot DB\cdot DC}\stackrel{(1)}{\iff}$ $\left(b^2+c^2\right)\cdot AD=a^2\cdot AD\iff$ $b^2+c^2=a^2\iff$ $A=90^{\circ}$ .

PP17. Let $D\in [BC]$ of $\triangle ABC$ such that $AD = \frac {BD^2}{AB + AD} = \frac {CD^2}{AC + AD}$ . Let $E$ be a point such that $D\in [AE]$ and $CD = \frac {DE^2}{CD + CE}$ . Prove that $AE = AB + AC$ .

Proof. The relations from hypothesis suggest me to apply the well-known lemma:

Lemma. Let $\triangle ABC$ and $D\in BC$ for which $AD\perp BC$ . Suppose $D\in (BC)$ . Then $b^2 = c(a + c)\Longleftrightarrow B = 2C\Longleftrightarrow$ $DC = DB + BA$ .

$\boxed {\ b^2 = c(a + c)\ }$ $\Longleftrightarrow b^2 - c^2 = ac$ $\Longleftrightarrow$ $DC^2 - DB^2 = ac$ $\Longleftrightarrow$ $\boxed {\ DC = DB + BA\ }$ $\Longleftrightarrow$ $b\cos C = c + c\cdot \cos B$ $\Longleftrightarrow$ $\sin B\cos C =$

$\sin C(1 + \cos B)$ $\Longleftrightarrow$ $\tan\frac B2 = \tan C$ $\Longleftrightarrow$ $\boxed {\ B = 2C\ }$ . Return to proof of the proposed problem. Let $\left\|\begin{array}{c} m \left(\widehat {ABC}\right) = x \\ \\ m\left(\widehat {ACB}\right) = y \\ \\ m\left(\widehat {CED}\right) = z\end{array}\right\|$ . Apply the upper

lemma to $:\ \left\|\begin{array}{cccc} \triangle\ ABD\ : & BD^2=AD\cdot (AB+AD) & \implies & m(\widehat {BAD}) = 2\cdot\left(\widehat {ABD}\right) = 2x \\ \\ \triangle\ ACD\ : & CD^2= AD\cdot (AC+AD) & \implies & m(\widehat {CAD}) = 2\cdot m\left(\widehat {ACD}\right) = 2y \\ \\ \triangle\ CDE\ : & DE^2=.CD\cdot (CE+CD) & \implies & m(\widehat {DCE}) = 2\cdot m\left(\widehat {CED}\right) = 2z\end{array}\right\|$ . Hence $A + B + C = 180^{\circ}$ $\Longleftrightarrow$

$(2x + 2y) + x + y = 180^{\circ}$ $\Longleftrightarrow$ $\boxed {\ x + y = 60^{\circ}\ }$ , $m(\widehat {ABD}) + m(\widehat {BAD}) =$ $m(\widehat {CED}) + m(\widehat {DCE})$ $\Longleftrightarrow$ $x + 2x = z + 2z$ $\Longleftrightarrow$ $\boxed {\ z = x\ }$

and $ABEC$ is cyclic. Thus, $AE = AB + AC$ $\Longleftrightarrow$ $\sin \widehat {ACE}=\sin\widehat {ACB}+\sin\widehat {ABC}$ $\Longleftrightarrow$ $\sin (2x + y) = \sin y + \sin x$ $\Longleftrightarrow$

$\sin (2x + y) - \sin y = \sin x$ $\Longleftrightarrow$ $2\sin x\cos (x + y) = \sin x$ $\Longleftrightarrow$ $\cos (x + y) = \frac 12$ $\Longleftrightarrow$ $x + y = 60^{\circ}$ , what is truly.

PP18. Prove that in any $\triangle ABC$ there are the equivalencies $\left\{\begin{array}{cccc} (1) & a^2=b(b+c) & \iff & A=2B\ .\\\\ (2) & a^2=b(b-c) & \iff & A=2B-180^{\circ}\ .\end{array}\right\|$

Proof 1 (synthetic). Denote $M\in (AC)$ so that $AN=c$ . Thus, $MC=b-c$ and $a^2=b(b-c)\iff$ $CB^2=CA\cdot CM\iff$ $\triangle CBM\sim\triangle CAB$

$\iff$ $m\left(\widehat{ABM}\right)=m\left(\widehat{AMB}\right)=90^{\circ}-\frac A2$ . In conclusion, $B=m\left(\widehat{ABM}\right)+m\left(\widehat{CBM}\right)=$ $\left(90^{\circ}-\frac A2\right)+A$ $\iff$ $A=2B-180^{\circ}$ .

Proof 2 (trigonometric). $a^2=b(b-c)\iff$ $\sin^2A=\sin B(\sin B-\sin C)\iff$ $2\sin A\sin\frac A2\cos\frac A2=2\sin B\sin\frac {B-C}{2}\sin \frac A2\iff$

$\sin A\cos\frac A2=\sin B\sin\frac {B-C}{2}\iff$ $\sin\frac {3A}{2}+\sin\frac A2=\cos\frac {B+C}{2}-\cos\frac {3B-C}{2}\iff$ $\sin\left(-\frac {3A}{2}\right)=\cos\frac {3B-C}{2}\iff$

$\cos\left(90^{\circ}+\frac {3A}{2}\right)=\cos\frac{3B-C}{2}\iff$ $180^{\circ}+3A=3B-C\iff$ $2A+C=B\iff$ $A+180^{\circ}=2B\iff$ $A=2B-180^{\circ}$ .

Proof 3 (trigonometric). $a^2=b(b-c)\iff$ $a^2+bc=a^2+c^2-2ac\cdot\cos B\iff$ $c-b=2a\cdot\cos B\iff$ $\sin C-\sin B=2\sin A\cos B\iff$

$\sin\frac {C-B}{2}\sin \frac A2=2\sin\frac A2\cos\frac A2\cos B\iff$ $\sin\frac {C-B}2=\cos\left(\frac A2+B\right)+\cos\left(\frac A2-B\right)\iff$ $\cos\left(\frac A2-B\right)=0\iff$ $B-\frac A2=90^{\circ}\iff$ $A=2B-180^{\circ}$ .

Remark. For the first one see the lemma from PP17.

PP19. Let $\triangle ABC$ with $b\not =c$ and the incircle $w=C(I,r)$ which touches $AB\ ,\ AC$ in $E\ ,\ F$ . Denote

$M\in (AB)\ ,\ N\in (AC)$ so that $I\in MN$ and $MN\parallel BC$ . Prove that $\frac {ME}{NF}=\left(\frac cb\right)^2\iff$ $AB\perp AC$ .

Proof. $\left\{\begin{array}{ccccc} m\left(\widehat{AMN}\right)=B & \implies & \tan B=\frac {EI}{ME}=\frac r{ME} & \implies & ME=r\cot B\\\\ m\left(\widehat{ANM}\right)=C & \implies & \tan C=\frac {FI}{NF}=\frac r{NF} & \implies & NF=r\cot C\end{array}\right\|$ $\implies \boxed{\frac {ME}{NF}=\frac {\tan C}{\tan B}}\ (1)$ . Using relation obtain that $\left(a^2+c^2-b^2\right)\tan B=4S=$

$\left(a^2+b^2-c^2\right)\tan C\implies$ $\boxed{\frac {\tan C}{\tan B}=\frac {a^2+c^2-b^2}{a^2+b^2-c^2}}\ (2)$ . In conclusion, $\frac {ME}{NF}=\left(\frac cb\right)^2\ \stackrel{(1\wedge 2)}{\iff}\ \frac {a^2+c^2-b^2}{a^2+b^2-c^2}$ $=\frac {c^2}{b^2}\stackrel{(b\not =c)}{\iff} a^2=b^2+c^2\iff AB\perp AC$ .

This post has been edited 159 times. Last edited by Virgil Nicula, Nov 21, 2015, 3:46 PM

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El_Ectric:
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Dear Virgil !

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287. Some metrical problems with or without trigonometry.

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