434. Problems with areas II.

by Virgil Nicula, Nov 17, 2015, 9:05 AM

P0. Denote $\triangle ABC$ with the incenter $I$ and $A=60^{\circ}.$ Let $E\in BI\cap AC,$ $F\in CI\cap AB$ and the areas $[BIF]=m,$ $[CIE]=n,$ $[EIF]=p.$ Prove that $\frac 1m+\frac 1n=\frac 1p$ .

Proof. $\left\{\begin{array}{ccccccc} \frac{p}{m} & = &\frac {[EIF]}{[BIF]} & = & \frac{IE}{IB} & = & \frac{b}{a+c}\\\\ \frac{p}{n} & = & \frac{[EIF]}{[CIE]} & = & \frac{IF}{IC} & = & \frac{c}{a+b}\end{array}\right\|\ (*)\ .$ I"ll use the theorem of Cosines $:\ \boxed{b^2+c^2-a^2=2bc\cdot \cos A}\ (1)\ .$ Therefore, $\frac 1m+\frac 1n=\frac 1p\iff$ $\frac{p}{m}+\frac{p}{n}=1\ \stackrel{(*)}{\iff}$

$\frac{b}{a+c}+\frac{c}{a+b}=1$ $\Longleftrightarrow$ $b(a+b)+c(a+c)=(a+b)(a+c)\iff$ $\cancel{a(b+c)}+b^2+c^2=a^2+\cancel{a(b+c)}+bc\iff$ $b^2+c^2-a^2=bc\stackrel{(1)}{\iff}$ $\cos A=\frac 12\iff$ $A=60^{\circ}\ .$

P1. Let $ABC$ be a triangle so that $a$ is constant and $b=2c$ . Find the maximum of the area $S=[ABC]$ .

Proof. $S=S(a,2c,c)$ where $\left\{\begin{array}{ccc} a<2c+c & \implies & \frac a3<c\\\\ 2c<a+c & \implies & c<a\end{array}\right\|$ $\implies$ $\boxed{\frac a3<c<a}$ . Thus, $16S^2=2\sum b^2c^2-\sum a^4=$ $2a^2\left(c^2+4c^2\right)+2c^2\cdot 4c^2-a^4-c^4-16c^4\implies$

$16S^2=-9c^4+10a^2c^2-a^4\implies$ $\boxed{16S^2=\left(9c^2-a^2\right)\left(a^2-c^2\right)}$ . Therefore, $S$ is $\max\iff$ $16S^2$ is $\max\iff$ $\left(9c^2-a^2\right)\left(a^2-c^2\right)$ is $\max\iff$ $\left(9c^2-a^2\right)\left(9a^2-9c^2\right)$ .

Observe that $\left(9c^2-a^2\right)+\left(9a^2-9c^2\right)=8a^2$ (constant). In conclusion, $S$ is $\max\iff$ $9c^2-a^2=$ $9a^2-9c^2=4a^2$ $\iff$ $c^2=$ $\frac {5a^2}{9}$ $\iff$ $\boxed{c_{\max}=\frac {a\sqrt 5}{3}}$ ,

i.e. $\frac a3=$ $\frac {b}{2\sqrt 5}=$ $\frac {c}{\sqrt 5}$ and in this case $16S^2_{\max}=\left(9\cdot \frac {5a^2}{9}-a^2\right)\left(a^2-\frac {5a^2}{9}\right)\iff$ $\boxed{S_{\max}=\frac {a^2}{3}}$ .

Remark. Can use relation $\boxed{\left(b^2-c^2\right)^2+16S^2=4a^2m^2_a}\ (*)$ . Indeed, $4m^2_a=2\left(b^2+c^2\right)-a^2=$ $10c^2-a^2$ and the relation $(*)$ becomes $16S^2=a^2\left(10c^2-a^2\right)-\left(4c^2-c^2\right)^2$ ,

i.e. $\boxed{16S^2=\left(9c^2-a^2\right)\left(a^2-c^2\right)}$ . Can use the Heron's relation. Thus, $\left\{\begin{array}{ccccccc} s & = & \frac {3c+a}{2} & ; & s-a & = & \frac {3c-a}{2}\\\\ s-b & = & \frac {a-c}{2} & ; & s-c & = & \frac {a+c}{2}\end{array}\right\|\implies$ $16S^2=\left(9c^2-a^2\right)\left(a^2-c^2\right)$ where $\frac a3<c<a$ .

Proof 2. Let $D\in (BC)$ , $E\in BC$ so that $B\in (EC)$ and $\frac {DB}{DC}=\frac {EB}{EC}=\frac {AB}{AC}=$ $\frac cb=\frac 12$ . Thus, $DB=\frac a3$ , $EB=a$ and $ED=EB+BD=a+\frac a3$ , i.e. $ED=\frac {4a}{3}$ .

Let the midpoint $M$ of $[ED]$ . Circle $w=C\left(M,\frac {2a}{3}\right)$ is the Appolonius' circle for $\triangle ABC$ and ratio $\frac bc=2$ . Thus, $A\in w$ , $S$ is $\max \iff$ $h_a=\frac {ED}{2}\iff$ $h_a=\frac {2a}{3}\iff$

$S_{\max}=\frac {ah_a}{2}=\frac 12\cdot a\cdot \frac {2a}{3}\iff$ $\boxed{S_{\max}=\frac {a^2}{3}}$ .

Extension. Let $\triangle ABC$ so that $a$ is constant and $b=kc$ . Prove that $\boxed{\ S_{\max}(a,b,c)=\frac {ka^2}{\left|2\left(k^2-1\right)\right|}\ }$ .

Proof. Suppose w.l.o.g. $k>1$ . Denote $D\in (BC)$ and $E\in BC$ so that $B\in (EC)$ and $\frac {DB}{DC}=\frac {EB}{EC}=\frac {AB}{AC}=$ $\frac cb=\frac 1k$ . Thus, $DB=\frac a{k+1}$ and $EB=\frac a{k-1}$

and $ED=EB+BD=$ $\frac a{k-1}+\frac a{k+1}$ , i.e. $ED=\frac {2ak}{k^2-1}$ . Denote the midpoint $M$ of the segment $[ED]$ . The circle $w=C\left(M,\frac {ak}{k^2-1}\right)$ is the Appolonius' circle

for $\triangle ABC$ and the ratio $\frac bc=k$ . In conclusion, $A\in w$ and $S$ is $\max \iff$ $h_a=\frac {ED}{2}$ $\iff$ $h_a=\frac {ak}{k^2-1}$ $\iff$ $S_{\max}=\frac {ah_a}{2}=$ $\frac 12\cdot a\cdot \frac {ka}{k^2-1}$ $\iff$ $\boxed{S_{\max}=\frac {a^2k}{2\left(k^2-1\right)}}$ .

P2. Let $w$ be incircle of $\triangle ABC$ which touch $BC$ , $CA$ , $AB$ at $(D,E,F)$ respectively. Let $\{P,D\}= AD\cap w$ and $\left\{\begin{array}{ccc} BC & = & a\\\\ AP & = & m\\\\ DP & = & n\end{array}\right\|$ . Find area $S=f(m,n,a)$ of $\triangle ABC$ .

Proof. Let $\left\{\begin{array}{c} AE=AF=x\\\\ BF=BD=y\\\\ CD=CE=z\end{array}\right|$ . Thus, $\left\{\begin{array}{c} y+z=a\\\\ z+x=c\\\\ x+y=c\end{array}\right|$ and observe that $AP\cdot AD=AE^2\implies \boxed{x=\sqrt{m(m+n)}}$ and $S=[ABC]=\sqrt{xyz(x+y+z)}\implies$

$\boxed{S=\sqrt {x\cdot yz\cdot (x+a)}}\ (*)$ . Apply the Stewart's relation $zc^2+yb^2=a(m+n)^2+ayz\iff$ $z(x+y)^2+y(x+z)^2=a(m+n)^2+ayz\iff$

$x^2(y+z)+4xyz+yz(y+z)=$ $a(m+n)^2+ayz\iff$ $m(m+n)a+4xyz=a(m+n)^2\iff$ $4xyz=an(m+n)$ . In conclusion,

$\boxed{S\stackrel{(*)}{=}\frac 12\cdot an(m+n)\cdot \left[a+\sqrt {m(m+n)}\right]}$ .

P3. Prove that for any $\{x,y,z\}\subset (0,1)$ exists the inequality $x\left ( 1-y \right )+y\left ( 1-z \right )+z\left ( 1-x \right )< 1$ .

Proof 1 (analytic). Let $f(x)=\sum x-\sum yz-1=(1-y-z)\cdot x+(y+z-yz-1)\ ,\ x\in (0,1)$ . Thus, $f(0)<0$ and $f(1)<0$ . In conclusion, $f(x)<0\ ,\ (\forall ) x\in (0,1)$

Proof 2 (algebraic). $1-\sum_{\mathrm{cyc}}x(1-y)=$ $xyz+\left[1-\sum x+\sum yz-xyz\right]=$ $xyz+\prod (1-x)>0$ . In conclusion, $\sum_{\mathrm{cyc}}x(1-y)< 1$ .

Proof 3 (geometric). Let an equilateral $\triangle ABC$ , where $AB=1$ . Let $\left\{\begin{array}{ccc} A_1\in (BC) & ; & BA_1=x\\\\ B_1\in (CA) & ; & CB_1=y\\\\ C_1\in (AB) & ; & AC_1=z\end{array}\right\|$ . So that $\left\{\begin{array}{c} A_1C=1-x\\\\ B_1A=1-y\\\\ C_1B=1-z\end{array}\right\|$ . Then $\left\{\begin{array}{c} BC_1=1-x\\\\ CA_1=1-z\\\\ AB_1=1-y\end{array}\right\|$ .

From the inequality $[AB_1C_1]+[BC_1A_1]+[CA_1B_1]<[ABC]$ obtain that $\frac{\sin 60^{\circ}}{2}\cdot\left[z(1-y)+x(1-z)+y(1-x)\right]<\frac {\sqrt 3}{4}\iff$

$\sum x-\sum yz<1\iff$ $x(1-y)+y(1-z)+z(1-x)<1$ .

P4. Let $ABC$ be an $A$ -isosceles triangle , let $D\in (BC)$ be a mobile point such that $BD>DC$ and let $w_1$ , $w_2$ be the circumcircles of $\triangle ABD$ , $\triangle ADC$

respectively. Let $BB'$ and $CC'$ be diameters in these two circles and let $M$ be the midpoint of $[B'C']$ . Prove that the area of the triangle $MBC$ is constant.

Proof. Let $w_1=C(O_1,r_1),\ w_2=C(O_2,r_2)$ . Because $AD=2r_1\sin B=2r_2\sin C$ results that the triangles $ABD$ , $ACD$ have the same length $r_1=r_2=\rho$ of the circumradius.

I note $x=m(\widehat {ABB'})$ . Hence $B'D\perp BC,\ C'D\perp BC$ $\Longrightarrow$ $D\in B'C',\ \overline {DB'C'}\perp BC$ and $\widehat {ABB'}\equiv$ $\widehat {ADB'}\equiv$ $\widehat {ADC'}\equiv$ $\widehat {ACC'}\Longrightarrow$ $AB'=2\rho \sin x$ ,

$AC'=2\rho\sin x$ $\Longrightarrow$ $AB'=AC'\Longrightarrow$ $AM\perp B'C'\Longrightarrow$ $AM\parallel BC\Longrightarrow$ $\sigma [BMC]=\sigma [ABC]$ . Remark. $\sigma [XYZ]$ is the area of the triangle $XYZ$ .

P5. Let $w=C(O,R)$ be a circle with two fixed diameters $AE$ and $BF$ so that $AE\perp BF$ . For a mobile point $C\in w$ between

$E$ and $F$ denote $P\in AE \cap BC$ and $Q\in AC\cap BF$ . Prove that the area of the orthogonal quadrilateral $ABPQ$ is constant.

Proof 1 (synthetic - Farenhajt). Denote $\phi =m\left(\widehat{CAO}\right)$ . Then $m\left(\widehat{AQO}\right)=90^\circ-\phi$ , $m\left(\widehat{OBP}\right)=45^\circ-\phi$ and $m\left(\widehat{OPB}\right)=45^\circ+\phi$ . Therefore,

$\left\{\begin{array}{ccc} \triangle ABQ\ : A=45^\circ+\phi \ ;\ B=45^\circ\ ;\ Q=90^\circ-\phi\\\\ \triangle PAB\ :\ P=45^\circ+\phi\ ;\ A=45^\circ\ ;\ B=90^\circ-\phi\end{array}\right\|\implies$ $\triangle ABQ\sim\triangle PAB\implies$ therefore $\frac {AB}{BQ}=\frac {PA}{AB}\implies$ $\boxed{AP\cdot BQ=AB^2}=2R^2$ . So finally $[ABPQ]=R^2$ .

Proof 2. Denote $\left\{\begin{array}{c} x=m\left(\widehat{CBF}\right)\\\\ y=m\left(\widehat{CAE}\right)\end{array}\right\|$ , where $x+y=45^{\circ}\implies$ $\frac {\tan x+\tan y}{1-\tan x\tan y}=1$ $\implies$ $\boxed{(1+\tan x)(1+\tan y)=2}\ (*)$ .

Observe that $\left\{\begin{array}{c} OP=R\tan x\\\\ OQ=R\tan y\end{array}\right\|$ and $AP\cdot BQ=$ $(R+x)(R+y)=$ $R^2(1+\tan x)(1+\tan y)\stackrel{(*)}{\implies}$ $\boxed{AP\cdot BQ=2R^2}$ .

In conclusion, the area of the orthogonal quadrilateral $ABPQ=\frac {AP\cdot BQ}{2}$ is constant, i.e. $[ABPQ]=R^2$ .

Remark. Area $[CPQ]$ is max. $\iff$ $[ABC]$ is max. because $[ABPQ]$ is constant $\iff$ $CA\cdot CB$ is max. because $m\left(\widehat {BCA}\right)$ is constant $\iff$ $2R\cos x\cdot 2R\cos y$ is max, where $x+y=45^{\circ}$ $\iff$ $cos (x-y)+\cos (x+y)$ is max. because $2R^2$ is constant $\iff$ $\cos (x-y)$ is max. because $x+y$ is constant $\iff$ $x-y=0\iff x=y=\frac {\pi}{8}$ .

Extension. Let $NWSE$ (North/West/South/East) be a square with circumcircle $w$ . For a mobile $M\in w$ let $\left\{\begin{array}{c} P\in SM\cap WE\\\ Q\in WM\cap SN\end{array}\right\|$ . Prove that $WP\cdot SQ=2R^2$ (constant).

Proof 1 (metric). Let $\left\{\begin{array}{c} x=m\left(\widehat{MSN}\right)\\\\ y=m\left(\widehat{MWE}\right)\end{array}\right\|$ . Observe that $M\in \left\{\begin{array}{ccccc} \mathrm{Quadrant\ I} & \implies & x+y=45^{\circ} & \implies & (1+\tan x)(1+\tan y)=2\\\\ \mathrm{Quadrant\ II} & \implies & -x+y=45^{\circ} & \implies & (1-\tan x)(1+\tan y)=2\\\\ \mathrm{Quadrant\ III} & \implies & x+y=135^{\circ} & \implies & (\tan x-1)(\tan y-1)=2\\\\ \mathrm{Quadrant\ IV} & \implies & x-y=45^{\circ} & \implies & (1+\tan x)(1-\tan y)=2\end{array}\right\|$ .

Prove easily that $\left\{\begin{array}{cccc} WP=R(1+\tan x) & SQ=R(1+\tan y) & \mathrm{\iff} & M\in \mathrm{Quadrant\ I}\\\\ WP=R(1-\tan x) & SQ=R(1+\tan y) & \mathrm{\iff} & M\in \mathrm{Quadrant\ II}\\\\ WP=R(\tan x-1) & SQ=(\tan y-1) & \mathrm{\iff} & M\in \mathrm{Quadrant\ III}\\\\ WP=R(1+\tan x) & SQ=(1-\tan y) & \mathrm{\iff} & M\in \mathrm{Quadrant\ IV}\end{array}\right\|$ . In conclusion, $WP\cdot SQ=2R^2$ .

Proof 2 (analytic). Suppose w.l.o.g. $\boxed{R=1}$ and denote $N(0,1)$ , $S(0,-1)$ , $W(-1,0)$ and $E(1,0)$ . For a mobile point $M(x,y)$ , where $\boxed{x^2+y^2=1}\ (*)$

obtain easily that $\left\{\begin{array}{c} x_P=\frac {x}{1+y}\\\ y_Q=\frac {y}{1+x}\end{array}\right\|$ . In conclusion, $WP\cdot SQ=\left(\frac {x}{1+y}+1\right)\left(\frac {y}{1+x}+1\right)=$ $\frac {(x+y+1)^2}{(1+x)(1+y)}\stackrel{(*)}{\implies}$ $WP\cdot SQ=2$ .

P6. Find the max. of $f(x)=x\left(1+\sqrt{1-x^2}\right)$ , where $x\in [0,1]$ .

Proof 1. Denote $y\in [0,1]$ so that $x^2+y^2=1$ . Thus the proposed problem becomes : PP. Find the maximum value of $g(x,y)=x(1+y)$ , where $\{x,y\}\subset [0,1]$ , $x^2+y^2=1$ .

Proof 2. Observe that $g(x,y)$ is $\max\iff$ $x^2(1+y)^2$ is $\max\iff$ $(1-y)(1+y)^3$ is $\max\iff$ $(1-y)\left(\frac {1+y}{3}\right)^3$ is $\max$ . Since

$(1-y)+\frac {1+y}{3}+\frac {1+y}{3}+\frac {1+y}{3}=$ $(1-y)+(1+y)=2$ (constant) obtain that $g(x,y)$ is $\max\iff$ $(1-y)=\frac {1+y}{3}=\frac 24=\frac 12$ ,

i.e. $y=\frac 12$ and $x=\frac {\sqrt 3}{2}$ . In conclusion, $\max g$ , where $\left\{\begin{array}{c} \{x,y\}\subset[0,1]\\\\ x^2+y^2=1\end{array}\right|$ is equally to $g\left(\frac {\sqrt 3}{2},\frac 12\right)=\frac {3\sqrt 3}{4}$ .

Application. Find the $A$ -isosceles triangle $ABC$ with the fixed vertex $A$ , which is inscribed in a fixed circle $w=C(O,1)$ and has maximum area.

Proof. Can suppose w.l.o.g. that $\triangle ABC$ is nonobtuse. Denote $D\in AO\cap BC$ and $DB=DC=x$ , $OD=y$ . Thus, the

area of $\triangle ABC$ is $S(x,y)=x(1+y)$ and its maximum is for $x=\frac {\sqrt 3}{2}$ and $y=\frac 12$ , i.e. when it is an equilateral triangle.

A similar proposed problem. Find the dimensions of the cylinder (the lengths of its radius and its generatrix) with constant volum and for which its total surface is minimum.

Proof. Denote $x$ - the length of its radius and $y$ - the length of its generatrix. Then the our problem becomes $\min_{\begin{array}{c} x>0\ ,\ y>0\\\ x^2y=k\ (\mathrm{const})\end{array}}x(x+y)$ . Observe that $x^2\cdot \frac {xy}{2}\cdot\frac {xy}{2}=\frac {k^2}{4}$

is constant and $x(x+y)$ is $\min\iff$ $x^2+\frac {xy}{2}+\frac {xy}{2}$ is $\min\iff$ $x^2=\frac {xy}{2}\iff$ $y=2x=\sqrt[3]{4k}$ , i.e.the axial section of the required cylinder is a square.

P7 - Pythagoras' theorem. Let $ABC$ be a $C$ -right-angled triangle. Prove that $a^2+b^2=c^2$ .

Proof 1. Let $ABCD$ be a rectangle with side lengths $\left\{\begin{array}{c} AB=2b\\\\ AD=a+b\end{array}\right\|$ . Let $\left\{\begin{array}{ccc} \{E,F\}\subset (AB)\ : & AE=BF=a\\\\ G\in (BC)\ ,\ J\in (AD)\ : & BG=AJ=b\\\\ H\in (CD)\ : & HD=HC=b\end{array}\right\|$ .Thus, $m\left(\angle EJH\right)=m\left(\angle FGH\right)=90^\circ$

because if $m\left(\angle AEJ\right)=x$ , then $m\left(\angle EJA\right)=$ $180^{\circ}-m\left(\angle JAE\right)-m\left(\angle AEJ\right)=$ $180^{\circ}-90^{\circ}-x=90^{\circ}-x$ . Also, $\triangle AEJ\cong\triangle DJH$ , so that

$m\left(\widehat{HJD}\right)=\left(\widehat{AJE}\right)=90^{\circ}-x$ . Then $m\left(\angle EJH\right)=90^\circ$ and by symmetry, $m\left(\angle FGH\right)=90^\circ$ . This is useful because if you draw $EH$ and $FH$ , you make two

right $\triangle EJH$ , $\triangle FGH$ . Also, the altitude of $\triangle EFH$ from $H$ is $a+b$ . You can then find area of the rectangle in two different ways $[ABCD]=(2b)(a+b)=$ $4\cdot\dfrac{ab}{2}+2\cdot\dfrac{c^2}{2}+$

$\frac{(2b-2a)(a+b)}{2}\iff$ $2ab+2b^2=$ $2ab+c^2+\frac{2(b-a)(b+a)}{2}\iff$ $2ab+$ $2b^2=2ab+c^2+b^2-a^2$ . And finally, with a little more simplifying, $a^2+b^2=c^2$ .

Proof 2. Let $ABCD$ be a rectangle with $\left\{\begin{array}{c} AB=2(a+b)\\\\ AD=a+b\end{array}\right\|$ . Denote $\left\{\begin{array}{ccc} \{E,F\}\subset (AB)\ : & AE=BF=a\\\\ G\in (BC)\ ,\ J\in (AD)\ : & BG=AJ=b\\\\ \{I,H\}\subset (CD)\ : & ID=HC=b\end{array}\right\|$ . Thus, $JE\perp JI$

and $GF\perp GH$ and $EJI$ , $FGH$ are right isosceles triangles. Now can find the area of the rectangle in two different ways $[ABCD]=AB\cdot AD=$

$2(a+b)\cdot (a+b)=2(a+b)^2$ and $[ABCD]=[AEJ]+[DJI]+$ $[CHG]+BFG]+$ $[EJI]+FGH]$ $+[EFHI]=$ $2ab+c^2+(a+b)^2$ .

In conclusion, $2(a+b)^2= 2ab+c^2+(a+b)^2\iff$ $a^2+b^2=c^2$ . Remark. 97 nice proofs (<== click) of the Pythagoras' theorem.

P8. Let $ABCD$ be a tangential quadrilateral in which $\left\{\begin{array}{ccc} AB\parallel CD & ; & AB\perp AD\\\\ AB=3\cdot CD & ; & S=[ABCD]=4\end{array}\right\|$ . Find the length of its inradius.

Proof 1. Since $ABCD$ is a tangential quadrilateral, then $AB+CD=AD+BC$ , i.e. there are $\{a,b\}\subset\mathbb R^*_+$ so that $\left\{\begin{array}{ccc} AB=3a & ; & CD=a\\\\ AD=b & ; & BC=4a-b\end{array}\right\|$ . Since $AD\perp AB$ , then

$BC^2=AD^2+\left(AB-CD\right)^2$ , i.e. $(4a-b)^2=b^2+4a^2\iff$ $12a^2=8ab\iff$ $\boxed{3a=2b}\ (1)$ . From $S=[ABCD]=\frac 12\cdot AD\cdot (AB+CD)$

obtain that $8=b\cdot 4a\iff$ $\boxed{ab=2}\ (2)$ . From the relations $(1)$ and $(2)$ obtain that $a=\frac {2}{\sqrt 3}$ and $b=\sqrt 3$ . In conclusion, the length of the inradius is $r=\frac b2$ , i.e. $\boxed{\ r=\frac {\sqrt 3}{2}\ }$ .

Proof 2. Since $ABCD$ is a tangential quadrilateral, then there are $\{x,y\}\subset\mathbb R^*_+$ so that $\left\{\begin{array}{ccc} AB=r+y & ; & CD=r+x\\\\ AD=2r & ; & BC=x+y\end{array}\right\|$ such that $\left\{\begin{array}{ccc} r+y & = & 3(r+x)\\\\ r(2r+x+y) & = & 4\\\\ r^2 & = & xy\end{array}\right\|$ .

Therefore, $\left\{\begin{array}{c} x=\frac {1-r^2}{r}\\\\ y=\frac {3-r^2}{r}\end{array}\right\|\implies$ $r^2=xy=\frac {\left(r^2-3\right)\left(r^2-1\right)}{r^2}\implies$ $r^4=\left(r^2-3\right)\left(r^2-1\right)\implies$ $4r^2=3\implies\boxed{r=\frac{\sqrt 3}{2}}$ .

P9. Let $\triangle ABC$ and three points $M\in (BC)$ , $N\in (CA)$ , $P\in (AB)$ . Denote $I\in AM \cap BN$ , $K\in AM \cap CP$ and $J\in BN \cap CP$ .

Suppose that $[APK] = [KIJ] = [CNJ] =[BIM]$ . Prove that $[PKIB] =[AKJN] =[IJCM]$ , where $[XYZ...]$ is the area of the polygon $XYZ...$ .

Proof. Let w.l.o.g. $[ABC]=1$ . Exist $\{m,n,p\}\subset\mathbb R^*_+$ so that $\left\{\begin{array}{c} \frac {MB}{m}=\frac {MC}{1}=\frac {BC}{m+1}\\\\ \frac {NC}{n}=\frac {NA}{1}=\frac {CA}{n+1}\\\\ \frac {PA}{p}=\frac {PB}{1}=\frac {AB}{p+1}\end{array}\right|$ . Apply Menelaus' theorem to $\overline{AKM}/\triangle BCP\ : \frac {AP}{AB}\cdot\frac {MB}{MC}\cdot\frac {KC}{KP}=1\implies$

$\frac {p}{p+1}\cdot \frac m1\cdot \frac {KC}{KP}=1\implies$ $\frac {KC}{1+p}=\frac {KP}{mp}=\frac {CP}{1+p+pm}\implies$ $[APK]=\frac {[APK]}{[APC]}\cdot\frac {[APC]}{[ABC]}=$ $\frac {KP}{CP}\cdot \frac {PA}{BA}$ $\implies$ $\boxed{[APK]=\frac {mp^2}{(1+p)(1+p+pm)}}$ . Obtain analogously

$\boxed{[BMI]=\frac {nm^2}{(1+m)(1+m+mn)}}$ and $\boxed{[CNJ]=\frac {pn^2}{(1+n)(1+n+np)}}$ . It is well-known the area $\boxed{[IJK]=\frac {(1-mnp)^2}{(1+m+mn)(1+n+np)(1+p+pm)}}$ . Observe

that $x=[APK]= [BMI]=[CNJ]=[IJK]$ $\iff$ $\frac {mp^2}{(1+p)(1+p+pm)}=$ $\frac {nm^2}{(1+m)(1+m+mn)}=$ $\frac {pn^2}{(1+n)(1+n+np)}=$ $\frac {(1-mnp)^2}{\prod_{\mathrm{cyc}}(1+m+mn)}\ (*)$ .

Let $[PKIB] =u\ ,\ [AKJN] =v\ ,\ [IJCM]=w$ . Thus, $\left\{\begin{array}{ccccc} 2x+u=[ABM]=\frac {m}{m+1} & \wedge & 2x+v+w=[ACM]=\frac {1}{m+1} & \implies & v+w-u=\frac {1-m}{1+m}\\\\ 2x+w=[BCN]=\frac {n}{n+1} & \wedge & 2x+u+v=[BAN]=\frac {1}{n+1} & \implies & u+v-w=\frac {1-n}{1+n}\\\\ 2x+v=[CAP]=\frac {p}{p+1} & \wedge & 2x+w+u=[CBP]=\frac {1}{p+1} & \implies & w+u-v=\frac {1-p}{1+p} \end{array}\right|$ $\implies$

$u+v+w=\sum\frac {1-m}{1+m}\implies$ $\left\{\begin{array}{c} u=\frac {1-np}{(1+n)(1+p)}\\\\ v=\frac {1-pm}{(1+p)(1+m)}\\\\ w=\frac {1-mn}{(1+m)(1+n)}\end{array}\right|$ and in this case $\boxed{x=\frac {2mnp+mn+np+pm-1}{2(1+m)(1+n)(1+p)}}$ . Can verify easily $[ABC]=1$ , i.e. $u+v+w+4x-1$ .

Indeed, $[ABC]=\frac {(1-mn)(1+p)+(1-np)(1+m)+(1-pm)(1+n)+2\left(2mnp+mn+np+pm-1\right)}{(1+m)(1+n)(1+p)}=$ $\frac {3+s_1-s_2-3s_3+4s_3+2s_2-2}{1+s_1+s_2+s_3}=1$ ,

where $\left\{\begin{array}{c} s_1=m+n+p\\\ s_2=mn+np+pm\\\ s_3=mnp\end{array}\right|$ . Prove easlily that $u=v=w\iff m=n=p$ . Is more difficulty to prove $[APK] = [KIJ] = [CNJ] =[BIM]\iff m=n=p$ .

P10. Let a convex (not rectangle) $ABCD$ . Let $W\in AD\cap BC$ , $X\in (AC)$ , $Y\in (BD)$ so that $\frac {XA}{XC}=\frac {YD}{YB}=m$ . Prove that $[XWY]=\frac {m}{(m+1)^2}\cdot [ABCD]$ .

Proof 1. $[XWY]=[DYW]-[DXW]-[DXY]=$ $\frac {m}{m+1}\cdot ([BDW]-[CDW]-[BDX])=$ $\frac {m}{m+1}\cdot([BCD]-[BDX])=$

$\frac {m}{m+1}\cdot ([CDX]+[BCX])=$ $\frac {m}{m+1}\cdot\frac {1}{m+1}\cdot ([ACD]+[ABC])=$ $\frac {m}{(m+1)^2}\cdot [ABCD]$ . In conclusion, $[XWY]=\frac {m}{(m+1)^2}\cdot [ABCD]$ .

Proof 2 (Delta). Let $M\in (CD)$ so that $\frac {MD}{MC}=m$ . Thus, $\left\{\begin{array}{ccc} XM\parallel AD & \implies & [WMX]=[DMX]\\\\ YM\parallel BC & \implies & [WMY]=[CMY]\end{array}\right|\implies$ $[XYW]=[XYM]+[WMX]+[WMY]=$

$[XYM]+[DMX]+[CMY]=$ $[XYCD]=[XYC]+[XCD]=$ $\frac 1{m+1}\cdot ([ACY]+[ACD])=$ $\frac 1{m+1}\cdot [ADCY]=$ $\frac 1{m+1}\cdot ([CDY]+[ADY])=$

$\frac m{(m+1)^2}\cdot ([CBD]+[BAD])=$ $\frac m{(m+1)^2}\cdot [ABCD]$ .

P11 (Carlos Olivera). Let an acute $\triangle ABC$ with area $S$ , circumcircle $w=\mathbb C(O,R)$ and the orthic $\triangle DEF$

where $\left\{\begin{array}{ccc} D\in BC & E\in CA & F\in AB\\\\ EF=d & FD=e & DE=f\end{array}\right\|$ . Prove that $S=\frac {def\sqrt{d+e+f}}{\sqrt{(e+f-d)(f+d-e)(d+e-f)}}$ .

Proof. Let orthocenter $H$ of $\triangle ABC$ and prove easily that $d=AH\cdot\sin A=2R\cos A\cdot\sin A=a\cdot\cos A\implies$ $d=a\cdot \cos A$ and analogously. Hence $\left\{\begin{array}{ccc} d & = & a\cdot\cos A\\\\ e & = & b\cdot\cos B\\\\ f & =& c\cdot\cos C\end{array}\right\|$ .

I"ll use two identities in $\triangle XYZ$ (standard notations) $:\ \frac {xyz}{4\rho}=[XYZ]=\sqrt {s(s-x)(s-y)(s-z)}\implies$ $\frac {xyz}{4\rho}=\sqrt {s(s-x)(s-y)(s-z)}\ (*)$ , where $\rho$ is the circumcenter

and $s$ is the semiperimeter for $\triangle XYZ\ ,$ i.e. $2s=x+y+z$ . For $\triangle DEF$ obtain that $\left\{\begin{array}{c} x:=d\\\ y:=e\\\ z:=f\end{array}\right\|$ and $s:=\frac {d+e+f}2\ ,\ \rho :=\frac R2$ (Euler's circle). Relation $(*)$ becomes $\frac {def}{2R}=\sqrt {s(s-d)(s-e)(s-f)}\ (1)$ . Since $2S=2([BOC]+[COA]+[AOB])=$ $\sum BC\cdot R\cos A=$ $R\cdot \sum a\cos A=$ $R(d+e+f)=2sR\implies$ $\boxed{S=Rs}$ .

The relation $(1)$ becomes $2S\cdot\sqrt{(s-d)(s-e)(s-f)}=def\cdot\sqrt s\iff$ $\boxed{S=\frac {def\sqrt{d+e+f}}{\sqrt{(e+f-d)(f+d-e)(d+e-f)}}}$ .

Remark. Prove easily that $H$ is the incenter of $\triangle DEF$ , the distance $\delta\equiv\delta_{EF}(H)$ is the length of the inradius of $\triangle DEF$ and the ratio $\boxed{\frac {[DEF]}{[ABC]}=\frac {\delta}R=2\cos A\cos B\cos C}$ .

P12. Let $\triangle ABC$ with the incenter $I$ and $A=60^{\circ}.$ Let $E\in BI\cap AC,$ $F\in CI\cap AB$ and the areas $[BIF]=m,$ $[CIE]=n,$ $[EIF]=p.$ Prove that $\frac 1m+\frac 1n=\frac 1p$ .

Proof. Observe that $A=60^{\circ}\iff$ $\boxed{b^2+c^2=a^2+bc}\ (*)$ and $\left\{\begin{array}{ccccc} \frac pm & = & \frac {IE}{IB} & = & \frac b{a+c}\\\\ \frac pn & = & \frac {IF}{IC} & = & \frac c{a+b}\end{array}\right\|$ $\bigoplus\implies$ $\frac pm+\frac pn=\frac b{a+c}+\frac c{a+b}=$

$\frac {b(a+b)+c(a+c)}{(a+b)(a+c)}=$ $\frac {b^2+c^2+a(b+c)}{(a+b)(a+c)}\ \stackrel{(*)}{=}\ \frac {a^2+bc+a(b+c)}{(a+b)(a+c)}=$ $\frac {(a+b)(a+c)}{(a+b)(a+c)}=1$ $\implies$ $\frac pm+\frac pn=1\implies$ $\frac 1m+\frac 1n=\frac 1p$ .

P13. Let a parallelogram $ABCD$ with $AB=CD=1$ and $P\in (AB)$ , $R\in (CD)$ for which $\left\{\begin{array}{c} R\in PC\cap BR\\\\ S\in PD\cap AR\end{array}\right\|$ . Prove that the area $[PQRS]\le \frac S4$ , where $S=[ABCD]$ .

Proof. I"ll use an well-known property: $"(\forall )$ trapezoid $ABCD$ with $AD\parallel BC$ and $I\in AC\cap BD$ there is the inequality $[AIB]=[CID]\le\frac S4$ , where $S=[ABCD]"$ . Indeed, if denote $[AIB]=[CID]=s$ and $\left\{\begin{array}{ccc} \alpha & = & [AID]\\\\ \beta & = & [BIC]\end{array}\right\|$ , then $\alpha\beta=s^2$ and $S=2s+\alpha +\beta\ge 2s+2\sqrt{\alpha\beta}=2s+2s=4s$ , i.e. $s\le\frac S4$ . Come back to the our

proposed problem. Apply above property to the trapezoids $:\ \odot\begin{array}{cccccc} \nearrow & PBCR: & [PQR] & \le & \frac {[PBCR]}4 & \searrow\\\\ \searrow & PADR: & [PSR] & \le & \frac {[PADR]}4 & \nearrow\end{array}$ $\bigoplus\implies$ $[PQRS]=[PQR]+[PSR]\le \frac S4$ .

P14. Fie $\triangle\ ABC$ si punctele $M\in (AB)$ , $N\in (AC)$ , $P\in BN\cap CM\ .$ Sa se arate ca $\boxed{\begin{array}{cccc} 1. & [MPN]\cdot [BPC] & = & [BPM]\cdot [CPN]\\\\ 2. & [MAN]\cdot [BPC] & = & [ABC]\cdot [MPN]\\\\ 3. & [AMPN]=[BPC] & \Longleftrightarrow & \frac {MA}{MB}\cdot\frac {NA}{NC}=1\end{array}}\ .$

Demonstratie.

$\blacktriangleright\ \frac{[MPN]}{[CPN]}=\frac{MP}{PC}=\frac{[BPM]}{[BPC]}\ \Longrightarrow\ MPN]\cdot [BPC]=[BPM]\cdot [CPN]\ .$

$\blacktriangleright$ Avem $\left\|\begin{array}{ccc} \frac{PM}{PC} & = & \frac{AM}{AC}\ \cdot\ \frac{\sin(\angle MAP)}{\sin(\angle PAC)}\\\\ \frac{PN}{PB} & = & \frac{AN}{AB}\ \cdot\ \frac{\sin(\angle PAN)}{\sin(\angle PAB)} \end{array}\right| \bigodot\ \Longrightarrow\ \frac{PM\cdot PN}{PC\cdot PB}=\frac{AM\cdot AN}{AB\cdot AC}\ \Longleftrightarrow\ \frac{[MPN]}{[BPC]}=\frac{[MAN]}{[ABC]}\ .$

$\blacktriangleright$ Avem $[ABC]=[AMPN]+[BMC]+[BNC]-[BPC]$ $\Longleftrightarrow$ $[AMPN]-[BPC]=[ABC]-[BMC]-[BNC]$ $\Longleftrightarrow$

$[AMPN]-[BPC]=[ABC]$ $\left(1-\frac{MB}{AB}-\frac{NC}{AC}\right)$ $\Longrightarrow$ $[AMPN]=[BPC]$ $\Longleftrightarrow$ $\frac{MB}{AB}+\frac{NC}{AC}=1$ $\Longleftrightarrow$ $\frac{MB}{MA}=\frac{NA}{NC}$ $\Longleftrightarrow$ $\frac{MA}{MB}\cdot \frac{NA}{NC}=1\ .$

PP15. Let $\triangle ABC\ ,$ $D\in (BC)\ ,$ $\left\{\begin{array}{cccc} F\in (AB) & ; & DF\parallel AC & (1)\\\\ E\in (AC) & ; & DE\parallel AB & (2) \end{array}\right\|$ and $P\in BE\cap CF\ .$ Prove that $[AEPF]=[BPC]\ .$

Proof. $\left\{\begin{array}{ccccc} \frac {[FBC]}{[ABC]} & = & \frac {FB}{AB} & \stackrel{(1)}{=} & \frac {DB}{CB}\\\\ \frac {[EAB]}{[CAB]} & = & \frac {EA}{CA} & \stackrel{(2)}{=} & \frac {DB}{CB}\end{array}\right\|\implies$ $[FBC]=EAB]\iff$ $[BPC]+[BPF]=[AEPF]+[BPF]\iff$ $[BPC]=[AEPF]\ .$

P16. Let an acute $\triangle ABC$ with circumcenter $O$ and orthocenter $H\ .$ Prove that the area of one of $\triangle AOH\ ,$ $\triangle BOH$ and $\triangle COH$ is equal to the sum of the areas of the other two.

Proof 1. Suppose w.l.o.g. $a<b<c\ .$ Denote the midpoint $M$ of $[BC]$ and the projection $D$ of $A$ on $BC\ .$ Is well-known that $MD=\frac {|b^2-c^2|}{2a}\ .$ Since $G\in (OH)\ ,$ $OH=3\cdot OG$

and $AG=2\cdot GM$ obtain that $[AOH]=3\cdot [AOG]=2\cdot [AOM]=2\cdot [DOM]=$ $MD\cdot MO=R\cdot \frac {|b^2-c^2|}{2a}\cdot \frac {b^2+c^2-a^2}{2bc}=$ $\frac {R\left(c^2-b^2\right)\left(b^2+c^2-a^2\right)}{16RS}$ $\implies$

$\boxed{16S\cdot [AOH]=\left(c^2-b^2\right)\left(b^2+c^2-a^2\right)}=$ $\frac {c^2-b^2}{4\tan A}$ $\implies$ $[AOH]=\frac {c^2-b^2}{4\tan A}\ .$ Obtain analogously $\left\{\begin{array}{ccc} 16S\cdot [BOH] & = & \left(c^2-a^2\right)\left(a^2+c^2-b^2\right)\\\\ 16S\cdot [COH] & = & \left(b^2-a^2\right)\left(a^2+b^2-c^2\right)\end{array}\right\|\ .$ In conclusion,

$16S\cdot\left([AOH]+[COH]\right)=$ $\left(c^2-b^2\right)\left(b^2+c^2-a^2\right)+$ $\left(b^2-a^2\right)\left(a^2+b^2-c^2\right)=$ $\left(c^2-a^2\right)\left(a^2+c^2-b^2\right)=16S\cdot [BOH]\implies$ $[AOH]+[COH]=[BOH]\ .$

P17. Let $ABCD$ be a convex quadrilateral with $P\in AC\cap BD.$ Denote the distancies $\left\{\begin{array}{ccc} \delta_{AB}(P)=h_a & ; & \delta_{BC}(P)=h_b\\\\ \delta_{CD}(P)=h_c & ; & \delta_{DA}(P)=h_d\end{array}\right\|.$ Prove that $\frac {AB\cdot CD}{BC\cdot DA}=\frac {h_bh_d}{h_ah_c} .$

Proof. Let $\left\{\begin{array}{c} AB=a\ ;\ BC=b\\\\ CD=c\ ;\ DA=d\end{array}\right\|.$ Apply (well known) $\boxed{[APB]\cdot [CPD]=[BPC]\cdot [DPA]}\ (*).$ Thus, $ah_a\cdot ch_c=bh_b\cdot dh_d\iff$ $\frac {ac}{bd}=\frac {h_bh_d}{h_ah_c},$ i.e. $\frac {AB\cdot CD}{BC\cdot DA}=\frac {h_bh_d}{h_ah_c}$

P18. Let $\triangle ABC$ and an interior point $G\left(\alpha ,\beta ,\gamma\right)$ with its barycentrical coordinates, where $\alpha +\beta +\gamma =1.$ Denote $D\in AG\cap BC$ and choose

$P\in (AB)$ and $Q\in (AC)$ so that $G\in (PQ).$ Prove that $\beta\cdot \frac mn+\gamma \cdot\frac qp=\frac 1{\beta +\gamma}-\left(\beta +\gamma\right),$ where $\left\{\begin{array}{ccc} m=[BPGD] & ; & q=[CQGD]\\\\ n=[APG] & ; & p=[AQG]\end{array}\right\|\ .$

Proof. I"ll use $\left\{\begin{array}{ccccc} \frac {GA}{\beta +\gamma} & = & \frac {GD}{\alpha} & = & AD\\\\ \frac {DB}{\gamma} & = & \frac {DC}{\beta} & = & \frac {BC}{\beta +\gamma}\end{array}\right\|$ and apply Cristea's theorem $:\ \frac {PB}{PA}\cdot DC+\frac {QC}{QA}\cdot DB=\frac {GD}{GA}\cdot BC\iff$ $\frac {AB}{AP}\cdot DC+\frac {AC}{AQ}\cdot DB=\frac {AD}{AG}\cdot BC\iff$

$\frac {AB}{AP}\cdot \beta +\frac {AC}{AQ}\cdot \gamma =\frac {AD}{AG}\cdot \left(\beta +\gamma\right)\iff$ $\boxed{\beta\cdot \frac {AB}{AP} +\gamma\cdot \frac {AC}{AQ} =1}\iff$ $\beta\cdot \frac {AB\cdot AD}{AP\cdot AG} +\gamma\cdot \frac {AC\cdot AD}{AQ\cdot AG} =\frac {AD}{AG}\iff$ $\beta\cdot\frac {[ABD]}{[APG]}+$ $\gamma\cdot\frac {[ACD]}{AQG]}=\frac {AD}{AG}$ $\iff$

$\beta\cdot\frac {n+m}n+\gamma\cdot\frac {p+q}p=$ $\frac 1{\beta +\gamma}$ $\iff$ $\boxed{\beta\cdot\frac mn+\gamma\cdot\frac qp=\frac 1{\beta +\gamma}- (\beta+\gamma )}\ (*)\ .$ If $G$ is centroid, i.e. $G\left(\frac 13,\frac 13,\frac 13\right),$ then required relation $(*)$ becomes $\frac mn+\frac qp=\frac 52.$

P19 (TST USA 2003). Let $ABC$ be a triangle and let $P$ be a point in its interior. Lines $PA$ , $PB$ , $PC$ intersect sides $BC$ , $CA$ , $AB$

at $D$ , $E$ , $F$ respectively. Prove that $[PAF]+[PBD]+[PCE]=\frac S2\iff P$ lies on at least one of the medians of triangle $ABC$ .

Proof. Denote the area $[XYZ]$ of the triangle $\triangle XYZ$ . I will suppose w.l.o.g. that $[ABC]=2$ and I note:

$\bullet\ M,N,P$ - the midpoints of the sides $[BC],[CA],[AB]$ respectively;

$\bullet\ x_1=[PAF]\ ,\ x_2=[PAE]\ ;\ y_1=[PBD]\ ,\ y_2=$ $[PBF]\ ;\ z_1=[PCE]\ ,\ z_2=[PCD]\ ;$

$\bullet\ x=y_2-z_1\ ,\ y=z_2-x_1\ ,\ z=x_2-y_1\ .$ Thus, $\boxed{x_1+y_1+z_1=x_2+y_2+z_2=1\ ;\ x+y+z=0}\ .$

From the Ceva's theorem results the relation $\frac{DB}{DC}\cdot \frac{EC}{EA}\cdot \frac{FA}{FB}=1$ , i.e. $\frac{(x_1+y_1)+y_2}{(x_2+z_2)+z_1}\cdot \frac{(y_1+z_1)+z_2}{(x_2+y_2)+x_1}\cdot \frac{(z_1+x_1)+x_2}{(y_2+z_2)+y_1}=1\Longleftrightarrow$

$\frac{1+(y_2-z_1)}{1-(y_2-z_1)}\cdot \frac{1+(z_2-x_1)}{1-(z_2-x_1)}\cdot \frac{1+(x_2-y_1)}{1-(x_2-y_1)}=1\Longleftrightarrow$ $(1+x)(1+y)(1+z)=(1-x)(1-y)(1-z)\Longleftrightarrow$ $xyz=0\Longleftrightarrow$

$y_2=z_1\ \vee\ z_2=x_1\ \vee\ x_2=y_1\Longleftrightarrow$ $EF\parallel BC\ \vee\ FD\parallel CA\ \vee\ DE\parallel AB\Longleftrightarrow$ $P\in [AM]\cup [BN]\cup [CP]\ .$

PP. Let $ABC$ be a triangle and let $P$ be a point in its interior. Lines $PA$ , $PB$ , $PC$ intersect sides $BC$ , $CA$ , $AB$ at $D$ , $E$ , $F$
respectively. Prove that $[PAF]+[PBD]+[PCE]=\frac S2\iff P$ lies on at least one of the medians of triangle $ABC$ .

Proof. Denote the area $[XYZ]$ of the triangle $\triangle XYZ$ . I will suppose w.l.o.g. that $[ABC]=2$ and I note:

$\bullet\ M,N,P$ - the midpoints of the sides $[BC],[CA],[AB]$ respectively;

$\bullet\ x_1=[PAF]\ ,\ x_2=[PAE]\ ;\ y_1=[PBD]\ ,\ y_2=$ $[PBF]\ ;\ z_1=[PCE]\ ,\ z_2=[PCD]\ ;$

$\bullet\ x=y_2-z_1\ ,\ y=z_2-x_1\ ,\ z=x_2-y_1\ .$ Thus, $\boxed{x_1+y_1+z_1=x_2+y_2+z_2=1\ ;\ x+y+z=0}\ .$

From the Ceva's theorem results the relation $\frac{DB}{DC}\cdot \frac{EC}{EA}\cdot \frac{FA}{FB}=1$ , i.e. $\frac{(x_1+y_1)+y_2}{(x_2+z_2)+z_1}\cdot \frac{(y_1+z_1)+z_2}{(x_2+y_2)+x_1}\cdot \frac{(z_1+x_1)+x_2}{(y_2+z_2)+y_1}=1\Longleftrightarrow$

$\frac{1+(y_2-z_1)}{1-(y_2-z_1)}\cdot \frac{1+(z_2-x_1)}{1-(z_2-x_1)}\cdot \frac{1+(x_2-y_1)}{1-(x_2-y_1)}=1\Longleftrightarrow$ $(1+x)(1+y)(1+z)=(1-x)(1-y)(1-z)\Longleftrightarrow$ $xyz=0\Longleftrightarrow$

$y_2=z_1\ \vee\ z_2=x_1\ \vee\ x_2=y_1\Longleftrightarrow$ $EF\parallel BC\ \vee\ FD\parallel CA\ \vee\ DE\parallel AB\Longleftrightarrow$ $P\in [AM]\cup [BN]\cup [CP]\ .$

This post has been edited 185 times. Last edited by Virgil Nicula, Apr 10, 2018, 7:21 AM

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434. Problems with areas II.

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