Construct a triangle

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Construct a triangle

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Source: Middle European Mathematical Olympiad 2013 T-5

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#1 h May 17, 2014, 8:57 AM • 1 Y

Y by Adventure10

Let $ABC$ be and acute triangle. Construct a triangle $PQR$ such that $AB = 2PQ$ , $BC = 2QR$ , $CA = 2 RP$ , and the lines $PQ, QR,$ and $RP$ pass through the points $A, B ,$ and $C$ , respectively. (All six points $A, B, C, P, Q,$ and $R$ are distinct.)

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#2 h Jul 16, 2014, 8:09 AM • 2 Y

Y by Adventure10, Mango247

solution:

make high $CD \perp AB$ ,since it is acute triangle, $D$ will fall on $AB$ . $EC \perp CB,$ circle C @ $r=CB$ , cross $EC$ at $E$ , make $EF \perp EC$ , extend $CA$ , cross $FE$ at $F$ , circle B @ $r=BA$ , cross line $CB$ at $G$ , connect $BF$ , make $GH$ // $BF$ , cross $EF$ at $H$ ,connect $CH$ , make it's midpoint $M$ , circle M@ $r=MH$ , cross $CG$ at $N$ , connect $HN$ , make it's midpoint $S$ ,,circle H @r=HS, cross circle $M$ at $T$ , connect $CT$ , circle C@ $r=CB$ , cross $CT$ at $B'$ ,circle C @ $r=CA$ , cross $CB$ at $H$ , cross $CT$ at $U$ ,make $\angle B'CA$ bisector $CV$ , cross circle $CA$ at $V$ , circle V@r=VH, cross circle CA at $A'$ , connect $A'B'$ , make $RB$ // $A'B'$ , cross $CT$ at $R$ , connect $A'C$ ,make $AQ$ // $A'C$ , cross $CT$ at $Q$ , cross $BR$ at $P, \triangle PQR$ is the solution.

Proof:

put $CB$ on $x$ , $CE$ on y, $C$ is on $O$ $\implies C(0,0),B(a,0),A(b\cos{C},b\sin{C})$ , $\angle B'CB=\alpha,B'(a\cos{\alpha},a\sin{\alpha)},A'(b\cos{(\alpha+C)},b\sin{(\alpha+C)})$ , $k_{A'B'}=\dfrac{b\sin{(\alpha+C)}-a\sin{\alpha}}{b\cos{(\alpha+C)}-a\sin{\alpha}}$ $=\dfrac{b(\sin{C}+\cos{C}\tan{\alpha})-a\tan{\alpha}}{b(\cos{C}-\sin{C}\tan{\alpha})-a},k_{A'C}=\tan{(\alpha+C)}$

line $CB' : y=\tan{\alpha}*x$
line $BR: y=k_{A'B'}(x-a)$
line $AQ: y-b\sin{C}=k_{A'C}(x-b\cos{C})$

let $R(x_1,y_1),Q(x_2,y-2) \implies x_1-x_2=\dfrac{a\cos{\alpha}}{2}$ ,

$x_1=\dfrac{a\cos{\alpha}(b\sin{C}\cos{\alpha}+(b\ \cos{C}-a)\sin{\alpha})}{b\sin{C}} \\ x_2=\dfrac{b\sin{\alpha}\cos{\alpha}}{\sin{C}} \implies \\ (ab\sin{C})\cos{\alpha}+(ab\cos{C}-(a^2+b^2))\sin{\alpha}=\dfrac{ab\sin{C}}{2}$

note:

${S=\dfrac{ab\sin{C}}{2}=\dfrac{ch_c}{2},c^2=a^2+b^2-2ab\cos{C} \implies \cos{\alpha}}-p\sin{\alpha}=\dfrac{1}{2},p=\dfrac{c+\dfrac{h}{\tan{C}}}{h}$

define: $\sin{\theta}=\dfrac{1}{\sqrt{p^2+1}}=\dfrac{h}{(\sqrt{c+\dfrac{h}{\tan{C}})^2+h^2}},\implies \sin{(\theta-\alpha)}=\dfrac{1}{2\sqrt{p^2+1}}$

define : $\sin{D}=\dfrac{1}{2\sqrt{p^2+1}}\implies \alpha=\theta-D$

it is trivial $\triangle ABC = \triangle A'B'C$

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#3 h Jul 16, 2014, 8:29 AM • 2 Y

Y by Adventure10, Mango247

in fact, it is always have the solution for any triangle, because we can always find a side which the high foot fall on the side.that make the construction method works.

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#5 h Jan 18, 2022, 9:50 AM

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bir tinga qimmat masala bu

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