403. Some nice geometry problems for the high school. II

by Virgil Nicula, Nov 10, 2014, 6:31 AM

PP15. Let $\alpha =\mathbb C(I,a)$ and $\beta =\mathbb C(J,b)$ be two circles so that $IJ=a+b\ .$ The line $AB$ is a common tangent for $\alpha$ and $\beta$ in $A\in\alpha$ and $B\in\beta$.

Let $w=\mathbb C(O,x)$ be the semicircle with $O\in (AB)$ so that the circles $\alpha\ ,$ $\beta$ are interior tangent to $w\ .$ Prove that $x\equiv g(a,b)=\frac {2ab}{2\sqrt{2ab}-(a+b)}\ .$


Proof. Apply the Pythagoras's theorem to $:\ \left\{\begin{array}{ccccccc}
\triangle AOI\ : & OI^2=AO^2+AI^2 & \implies & (x-a)^2=AO^2+a^2 & \implies & AO^2=x^2-2ax\\\\
\triangle BOJ\ : & OJ^2=BO^2+BJ^2 & \implies & (x-b)^2=BO^2+b^2 & \implies & BO^2=x^2-2bx\end{array}\right\|\ .$

Therefore, $OA+OB=AB\iff$ $\sqrt{x^2-2ax}+\sqrt{x^2-2bx}=2\sqrt{ab}\iff$ $2x^2-2x(a+b)+2\sqrt{\left(x^2-2ax\right)\left(x^2-2bx\right)}=4ab\iff$

$x^2-x(a+b)+\sqrt{\left(x^2-2ax\right)\left(x^2-2bx\right)}=2ab\iff$ $\left(x^2-2ax\right)\left(x^2-2bx\right)=2ab-x^2+x(a+b)$ , where $x\ge\max \{2a,2b\}\ .$ Thus, $x^4-2(a+b)x^3+4abx^2=$

$x^4+(a+b)^2x^2+4a^2b^2-4abx^2+4ab(a+b)x-2(a+b)x^3\iff$ $\left(a^2-6ab+b^2\right)\cdot x^2+4ab(a+b)\cdot x+4a^2b^2=0\implies$ $\boxed{x=\frac {2ab}{2\sqrt{2ab}-(a+b)}}\ (*)\ .$

$\blacktriangleright$ Let $U\in AO\cap w$ and $V\in BO\cap w$ so that $AU=u$ and $BV=v\ .$ Observe that $AB=\boxed{2x-(u+v)=2\sqrt{ab}}\ (3)\ .$ Apply the Pythagoras's

theorem
to $:\ \left\{\begin{array}{ccccccc}
\triangle AOI\ : & OI^2=AO^2+AI^2 & \implies & (x-a)^2=(x-u)^2+a^2 & \implies & u^2-2xu+2xa=0 & (1)\\\\
\triangle BOJ\ : & OJ^2=BO^2+BJ^2 & \implies & (x-b)^2=(x.-v)^2+b^2 & \implies & v^2-2xv+2xb=0 & (2)\end{array}\right\|\ .$ From the relations $(1)$

and the relation $(2)$ obtain that $u^2-v^2-2x(u-v)+2x(a-b)=0\iff$ $(u-v)(u+v-2x)+2x(a-b)=0\ \stackrel{(3)}{=}\ -2
(u-v)\sqrt{ab}+2x(a-b)=0\iff$

$\boxed{u-v=\frac {x(a-b)}{\sqrt{ab}}}\ (4)\ .$ From the relations $(3)$ and $(4)$ obtain that $\left\{\begin{array}{ccccc}
u & = & \left(x-\sqrt{ab}\right) & + & \frac {x(a-b)}{2\sqrt{ab}}\\\\
v & = & \left(x-\sqrt{ab}\right) & - & \frac {x(a-b)}{2\sqrt{ab}}\end{array}\right\|\ \stackrel{(*)}{\implies}\ \boxed{\begin{array}{ccccc}
u & = & \frac {2a\sqrt {ab}-2\left(\sqrt 2-1\right)ab}{2\sqrt{2ab}-(a+b)} & (5)\\\\
v & = & \frac {2b\sqrt {ab}-2\left(\sqrt 2-1\right)ab}{2\sqrt{2ab}-(a+b)} & (6)\end{array}}$. Drawing of PP15 & PP16.

PP16. Let $\alpha =\mathbb C(I,a)$ and $\beta =\mathbb C(J,b)$ be two circles so that $IJ=a+b\ .$ The line $AB$ is a common tangent for $\alpha$ and $\beta$ in $A\in\alpha$ and $B\in\beta$.

Let $w=\mathbb C(O,x)$ be the semicircle with $O\in (AB)$ and what is exterior tangent to the circles $\alpha\ ,$ $\beta\ .$ Prove that $x\equiv f(a,b)=\frac {2ab}{a+b+2\sqrt{2ab}}\ .$


Proof. Apply the Pythagoras's theorem to $:\ \left\{\begin{array}{ccccccc}
\triangle AOI\ : & OI^2=AO^2+AI^2 & \implies & (x+a)^2=AO^2+a^2 & \implies & AO^2=x^2+2ax\\\\
\triangle BOJ\ : & OJ^2=BO^2+BJ^2 & \implies & (x+b)^2=BO^2+b^2 & \implies & BO^2=x^2+2bx\end{array}\right\|\ .$

Therefore, $OA+OB=AB\iff$ $\sqrt{x^2+2ax}+\sqrt{x^2+2bx}=2\sqrt{ab}\iff$ $2x^2+2x(a+b)+2\sqrt{\left(x^2+2ax\right)\left(x^2+2bx\right)}=4ab\iff$

$x^2+x(a+b)+\sqrt{\left(x^2+2ax\right)\left(x^2+2bx\right)}=2ab\iff$ $\left(x^2+2ax\right)\left(x^2+2bx\right)=\underline{\underline{2ab-x^2-x(a+b)\ge 0}}\iff$ $x^4+2(a+b)x^3+4abx^2=$

$x^4+(a+b)^2x^2+4a^2b^2-4abx^2-4ab(a+b)x+2(a+b)x^3\iff$ $\left(a^2-6ab+b^2\right)\cdot x^2-4ab(a+b)\cdot x+4a^2b^2=0\implies$ $\boxed{x=\frac {2ab}{a+b+2\sqrt{2ab}}}\ (*)\ .$

$\blacktriangleright$ Let $U\in (AO)\cap w$ and $V\in (BO)\cap w$ so that $AU=u$ and $BV=v\ .$ Observe that $AB=\boxed{u+v+2x=2\sqrt{ab}}\ (3)\ .$ Apply the Pythagoras's

theorem
to $:\ \left\{\begin{array}{ccccccc}
\triangle AOI\ : & OI^2=AO^2+AI^2 & \implies & (x+a)^2=(x+u)^2+a^2 & \implies & u^2+2xu=2xa & (1)\\\\
\triangle BOJ\ : & OJ^2=BO^2+BJ^2 & \implies & (x+b)^2=(x+v)^2+b^2 & \implies & v^2+2xv=2xb & (2)\end{array}\right\|\ .$ From the relations $(1)$

and the relation $(2)$ obtain that $u^2-v^2+2x(u-v)=2x(a-b)\iff$ $(u-v)(u+v+2x)=2x(a-b)\ \stackrel{(3)}{=}\ 2(u-v)\sqrt{ab}=2x(a-b)\iff$

$\boxed{u-v=\frac {x(a-b)}{\sqrt{ab}}}\ (4)\ .$ From the relations $(3)$ and $(4)$ obtain that $\left\{\begin{array}{ccccc}
u & = & \left(\sqrt{ab}-x\right) & + & \frac {x(a-b)}{2\sqrt{ab}}\\\\
v & = & \left(\sqrt{ab}-x\right) & - & \frac {x(a-b)}{2\sqrt{ab}}\end{array}\right\|\ \stackrel{(*)}{\implies}\ \boxed{\begin{array}{ccccc}
u & = & \frac {2a\sqrt {ab}+2\left(\sqrt 2-1\right)ab}{a+b+2\sqrt{2ab}} & (5)\\\\
v & = & \frac {2b\sqrt {ab}+2\left(\sqrt 2-1\right)ab}{a+b+2\sqrt{2ab}} & (6)\end{array}}$



PP17. Let incircle $w=C(I,r)$ of $\triangle ABC$ and $M\in (AB)$ , $N\in (AC)$ so that $MN$ is tangent to $w$ . Prove that there is the relation $\boxed{\frac {MA}{MB}\cdot\cot\frac B2+\frac {NA}{NC}\cdot\cot\frac C2=\cot\frac A2}\ .$

Proof. Denote $E\in w\cap AC$ , $F\in w\cap AB$ and $T\in w\cap MN$ . Thus, $[AMN]=2\cdot [AIM]+2\cdot [AIN]-[AEIF]$ , i.e. $xy\sin A=2(x+y)r-2(s-a)r$ because

$AE=AF=s-a$ and $IE=IF=IT=r$ . Hence $bc\sin A=2sr$ , i.e. $\sin A=\frac {2sr}{bc}$ . Thus, $xy\cdot\frac {2sr}{bc}=2(x+y)r-2(s-a)r\iff$ $\boxed{\frac {xys}{bc}=x+y-(s-a)}\ (*)$ .

The required relation is equivalently with $\frac {x(s-b)}{r(c-x)}+\frac {y(s-c)}{r(b-y)}=$ $\frac {s-a}r\iff$ $x(b-y)(s-b)+y(c-x)(s-c)=$ $(s-a)(c-x)(b-y)$ $\iff$

$xb(s-b)-xy(s-b)+yc(s-c)-xy(s-c)=$ $bc(s-a)-(s-a)(bx+cy)+xy(s-a)$ $\iff$ $bcx+bcy-axy=$ $bc(s-a)+xy(s-a)$ $\iff$

$bc(x+y)-bc(s-a)+sxy$ $\iff$ $x+y-(s-a)=\frac {xys}{bc}$ , i.e. the relation $(*)$ .



PP18. Let the incircle $w=C(I,r)$ of $\triangle ABC$ , $E\in w\cap AC$ , $F\in w\cap AB$ and for $P\in \overarc[]{EF}$ let the lengths $(x,y,z)$

of the distancies from $P$ to $BC$ , $CA$ , $AB$ respectively. Prove that $\boxed{\sqrt x\cdot\cos\frac A2=\sqrt y\cdot\cos\frac B2+\sqrt z\cdot\cos\frac C2}$ .


Proof. Denote $V\in AC$ so that $PV\perp AC$ and $R\in IE$ so that $PR\perp IE$ . Thus, $PR\perp IE\iff$ $PE^2-PI^2=RE^2-RI^2\iff$ $PE^2-r^2=y^2-(r-y)^2\iff$

$PE^2=2ry$ . Obtain analogously $\frac {PD^2}x=\frac {PE^2}y=\frac {PF^2}z=2r$ . Prove easily that $\frac {EF}{\cos \frac A2}=\frac {DE}{\cos \frac C2}=\frac {FD}{\cos \frac B2}=2r$ . Apply the Ptolemy theorem to the cyclical quadrilateral

$DEPF\ :\ PD\cdot EF=PF\cdot DE+PE\cdot DF\iff$ $\sqrt{2rx}\cdot 2r\cos\frac A2=$ $\sqrt {2ry}\cdot 2r\cos\frac B2+\sqrt{2rz}\cdot 2r\cos\frac C2\iff$ $\sqrt x\cdot\cos\frac A2=\sqrt y\cdot\cos\frac B2+\sqrt z\cdot\cos\frac C2$.



PP19. Let $\triangle ABC$ and $M\in (BC)$ so that $m\left(\widehat{AMB}\right)=\frac B2$ . Prove that $MB=c\ \ \vee\ \ a\cdot MC+c(a+c)=b^2\ .$ Particularly $MC=\frac a4\ \iff\ a=2(b-c)\ .$

Proof 1. Apply Stewart's relation to cevian $[AM$ in $\triangle ABC\ :\  AM^2\cdot a+MB\cdot MC\cdot a=$ $c^2\cdot MC+b^2\cdot MB\ (*)\ .$ Using an well-known

property in $\triangle ABM$ obtain that $m\left(\widehat{AMB}\right)=\frac B2\iff$ $MB=$ $c\ \vee\ AM^2=c(c+MB)\ .$ If $AM^2=c(c+MB)\ ,$ then from $(*)$ obtain that

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

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



PP20. Let $\triangle ABC$ with the orthocenter $H$ and the circumcircle $w=C(O,R)$ . Denote

$\left\{\begin{array}{ccc} 
E\in BH\cap AC & ; & F\in CH\cap AB\\\\
X\in EF\cap AH & ; & Y\in AO\cap BC\end{array}\right\|$ and $\left\|\begin{array}{ccc}
M\in BC & , & MB = MC\\\\
N\in XY & , & NX = NY\end{array}\right\|$ . Prove that $A\in MN$ .


Proof. Suppose w.l.o.g. that $c>b$ . Let $A'\in w$ so that $AA'$ is a diameter of $w\ ,\  P\in EF\cap BC$ and $D\in BC$ so that $AD\perp BC$ . Is well-known

that $A'CHB$ is a parallelogram. Thus, $M$ is the intersection of its diagonals. Prove easily that $\frac {XA} {XH}=$ $\frac {YA} {YA'}\implies$ $A'H\parallel YX\implies A\in MN$ .



PP21. $ABCD$ is an inscribed quadrilateral . Through $A$ and $B$ had a circle tangent to $CD$ at $E$; through $B$ and $C$ had a circle tangent to

$DA$ at $F$; through $C$ and $D$ had a circle tangent to $AB$ at $G$; through $D$ and $A$ had a circle tangent to $BC$ at $H$. Prove that $EG\perp FH$.


Proof. Let $ X =AB \cap CD\ ,\ Y= BC \cap DA $ . Easy to see the bisector of $ \angle BXC $ and the bisector of $ \angle AYB $ are perpendicular $ (*) $ . Since

$\left\{\begin{array}{ccc}
XE^2=XA*XB=XC*XD=XG^2 & \implies & XE=XG\\\\
YF^2=YB*YC=YA*YD=YH^2 & \implies & YF=YH\end{array}\right\|\implies$ the triangles $XGE $ and $YHF $ are isosceles $\stackrel{(*)}{\implies} EG \perp FH $ .

PP22. Let $\triangle ABC$ with incircle $w=C(I,r)$ which touches $\triangle ABC$ at $D\in BC$ , $E\in CA$ and $F\in AB$. Let $\left\{\begin{array}{ccccc}
L\in (AE) & , & S\in (AF) & ; & LS\parallel BC\\\\
P\in (BF) & , & Q\in (BD) & ; & PQ\parallel CA\\\\
M\in (CD) & , & N\in (CE) & ; & MN\parallel AB\end{array}\right\|$ . Prove that

$SP\cdot QM\cdot LN=\sqrt {AS\cdot AL\cdot BP\cdot BQ\cdot CM\cdot CN}$ and $r_1+r_2+r_3=r$ , where $r\ ,\ r_1\ ,\ ,r_2\ ,\ r_3$ are the lengths of the inradii of $ABC\ ,\ ALS\ ,\ BPQ\ ,\ CMN$.


Proof. $\triangle ALS\sim \triangle BPQ\sim \triangle CMN\sim \triangle ABC\implies$ $\left\{\begin{array}{c}
\frac {SL}a=\frac {AL}b=\frac {AS}c=\frac {s-a}{s}=\frac {r_1}r\\\\
\frac {BQ}a=\frac {PQ}b=\frac {BP}c=\frac {s-b}{s}=\frac {r_2}r\\\\
\frac {CM}a=\frac {CN}b=\frac {MN}c=\frac {s-c}{s}=\frac {r_3}r\end{array}\right\|$ $\implies\sum_{k=1}^3 \frac {r_k}r=\sum\frac {s-a}s=1\implies \boxed{r_1+r_2+r_3=r}$ . Thus,

$\left\{\begin{array}{ccccc}
QM=BC-CM-BQ & = & a-\frac {a(s-c)}s-\frac {a(s-b)}s & \implies & QM=\frac {a(s-a)}s=SL\\\\
LN=AC-AL-CN & = & b-\frac {b(s-a)}s-\frac {b(s-c)}s & \implies & LN=\frac {b(s-b)}s=PQ\\\\
SP=AB-AS-BP & = & c-\frac {c(s-a)}s-\frac {c(s-b)}s & \implies & SP=\frac {c(s-c)}s=MN\end{array}\right\|$ . In conclusion, $SP\cdot QM\cdot LN=\sqrt {AS\cdot AL\cdot BP\cdot BQ\cdot CM\cdot CN}\iff$

$\frac {c(s-c)}s\cdot\frac {a(s-a)}s\cdot\frac {b(s-b)}s=$ $\sqrt {\frac {cb(s-a)^2}{s^2}\cdot \frac {ca(s-b)^2}{s^2}\cdot \frac {ab(s-c)^2}{s^2}}\iff$ $\frac {abc(s-a)(s-b)(s-c)}{s^3}=\sqrt{\frac {a^2b^2c^2(s-a)^2(s-b)^2(s-c)^2}{s^6}}$ , what is truly.

Remark. Denote $\left\{\begin{array}{c}
X\in LS\cap w\\\\
Y\in PQ\cap w\\\\
Z\in MN\cap w\end{array}\right\|$ . Thus, $I\in XD\cap YE\cap ZF$ . Using the Brianchon's theorem obtain that $I\in LQ\cap NP\cap MS$ .



PP23. Let $\triangle ABC$ and the midpoint $M$ of $[BC]$, Nagel's point $N$, Gergonne's point $\Gamma$ and Feuerbach's point $F$. Prove that $N\Gamma\parallel BC$ $\iff$ $F\in (AM)$ $\iff$ $a(b+c)=b^2+c^2$.

Proof. (See here). Let the second intersection $L$ of the Euler's circle $e$ with the $A$-median $AM$ , the orthocenter $H$ , $D\in AH\cap BC$ and the midpoint $E$ of $[AH]$ . Is well-known

$\{M,D,E\}\subset w$ and from the power of $A$ w.r.t. $e$ obtain that $:\ AE\cdot AD=AL\cdot AM\iff $ $R\cos A\cdot h_a=AL\cdot m_a\iff$ $4Rh_a\cdot\cos A=4m_a\cdot AL\iff$

$2bc\cdot\cos A=4m_a\cdot AL\iff$ $\boxed{AL=\frac {b^2+c^2-a^2}{4m_a}}$ . Thus, $LM=AM-AL=\frac {4m_a^2-\left(b^2+c^2-a^2\right)}{4m_a}=$ $\frac {2\left(b^2+c^2\right)-a^2-\left(b^2+c^2\right)+a^2}{4m_a}$, i.e. $\boxed{LM=\frac {b^2+c^2}{4m_a}}$.

$\blacktriangleright\ \boxed{N\Gamma\parallel BC}$ $\Longleftrightarrow$ $\frac {p-a}{p}=\frac {(p-b)(p-c)}{\sum (p-b)(p-c)}\iff$ $\frac {p}{(p-a)}=1+\frac {a(p-a)}{(p-b)(p-c)}\iff$ $\frac {p}{(p-a)}-1=\frac {a(p-a)}{(p-b)(p-c)}\iff$ $\boxed{a(b+c)=b^2+c^2}\ .$

$\blacktriangleright\ \boxed{F\in AM}\iff$ $L\in AM\cap w$ , where $w=C(I,r)\ .$ Let $\{L,S\}=AM\cap w$ , where $L\in (AS)\ .$ From $AL=\frac {b^2+c^2-a^2}{4m_a}$ and $LM=\frac {b^2+c^2}{4m_a}$ obtain that

$p_w(A)=(p-a)^2=AL\cdot AS$ $\iff$ $AS=\frac {4m_a(p-a)^2}{b^2+c^2-a^2}\ .$ Thus, $MS=MA-AS=$ $m_a-\frac {4m_a(p-a)^2}{b^2+c^2-a^2}$ , i.e. $MS=\frac {2m_a\left[a(b+c)-a^2-bc\right]}{b^2+c^2-a^2}\ .$ Also,

$p_w(M)=\frac {(b-c)^2}{4}=MS\cdot ML=$ $\frac {2m_a\left[a(b+c)-a^2-bc\right]}{b^2+c^2-a^2}\cdot \frac {b^2+c^2}{4m_a}$ , i.e. $\left(b^2+c^2-a^2\right)(b-c)^2=$ $2\left(b^2+c^2\right)\left[a(b+c)-a^2-bc\right]\iff$ $\boxed{a(b+c)=b^2+c^2}\ .$

Observatie. Apropo de subiectul topicului "Puncte importante intr-un triunghi". Pentru cei care nu cunosc semnificatiile punctelor remarcabile mentionate in enuntul problemei propuse

le recomand sa consulte Google, unde, spre surprinderea lor, vor intalni cel putin 100 de asemenea puncte ceea ce inseamna ca triunghiul a fost si ramane o preocupare de secole

a multor matematicieni, multi dintre ei cu rezultate remarcabile la nivel inalt. In demonstratie am folosit urmatoarele relatii metrice dintr-un triunghi care se pot dovedi fara dificultate $:$

$ab+bc+ca=p^2+\sum(p-b)(p-c)=p^2+r(4R+r)$ and $IA^2=\frac {bc(p-a)}{p}=bc-4Rr\ .$



PP24. Let a square $ABCD$ and its interior point $I$ for which $\left\{\begin{array}{ccc}
m\left(\widehat{AIB}\right) & = & b\\\\
 m\left(\widehat{CID}\right) & = & a\end{array}\right|$ , where $a+b=180^{\circ}$ , i.e. $\sin a=\sin b$ and $a\ne b$ . Prove that $m\left(\widehat{ICD}\right)\in\left\{45^{\circ},135^{\circ}-a\right\}$ .

Proof. Let $\left\{\begin{array}{ccc}
BI=u & ; & m\left(\widehat{IBA}\right)=y\\\
CI=v & ; & m\left(\widehat{ICD}\right)=x\end{array}\right\|$ and suppose w.l.o.g. $AB=1$ . Apply theorem of Sines in $:\ \left\{\begin{array}{cccc}
\triangle IBA\ : & \frac u{\sin (b+y)}=\frac 1{\sin b}\\\\
\triangle BIC\ : & \frac u{\cos x}=\frac v{\cos y}=\frac 1{\sin (x+y)}\\\\
\triangle ICD\ : & \frac v{\sin (a+x)}=\frac 1{\sin a}\end{array}\right\|$ $\implies$

$\odot\begin{array}{ccc}
\nearrow & u=\frac {\sin (b+y)}{\sin b}=\frac {\cos x}{\sin (x+y)} & (1)\\\\
\searrow & v=\frac {\sin (a+x)}{\sin a}=\frac {\cos y}{\sin (x+y)} & (2)\end{array}\odot\ \implies$ $\frac uv$ $=\frac {\sin(b+y)}{\sin (a+x)}=$ $\frac {\cos x}{\cos y}\ \stackrel{(a+b=\pi )}{\implies}\ \frac {\sin (a-y)}{\sin (a+x)}$ $=\frac {\cos x}{\cos y}\implies$ $\sin (a-y)\cos y=\sin (a+x)\cos x\implies$

$\sin a+\sin (a-2y)=\sin (a+2x)+\sin a\implies$ $\sin (a-2y)=\sin (a+2x)\iff$ $(a-2y)+(a+2x)=\pi\implies y-x=\frac {\pi}2-b=a-\frac {\pi}2\implies$

$\boxed{y=x+a-\frac {\pi}2}\ \stackrel{(2)}{\implies}\ \frac {\sin (a+x)}{\sin a}=$ $\frac {\sin (x+a)}{\sin \left(2x+a-\frac {\pi}2\right)}\implies$ $\sin a=\sin\left(2x+a-90^{\circ}\right)\implies$ $\boxed{x\in \left\{45^{\circ},135^{\circ}-a\right\}}$ .
This post has been edited 175 times. Last edited by Virgil Nicula, Jun 4, 2016, 3:21 PM

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