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Difference between revisions of "2022 USAJMO Problems/Problem 4"

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(Image of the solution is here [https://wiki-images.artofproblemsolving.com//d/de/Usajmo2022-4.png])
  
 
Let's draw (<math>\ell</math>) perpendicular bisector of <math>\overline{KL}</math>. Let <math>X, Y</math> be intersections of <math>\ell</math> with <math>AC</math> and <math>BD</math>, respectively. <math>KXLY</math> is a kite. Let <math>O</math> mid-point of <math>\overline{KL}</math>. Let <math>M</math> mid-point of <math>\overline{BD}</math> (and also <math>M</math> is mid-point of <math>\overline{AC}</math>). <math>X, O, Y</math> are on the line <math>\ell</math>.
 
Let's draw (<math>\ell</math>) perpendicular bisector of <math>\overline{KL}</math>. Let <math>X, Y</math> be intersections of <math>\ell</math> with <math>AC</math> and <math>BD</math>, respectively. <math>KXLY</math> is a kite. Let <math>O</math> mid-point of <math>\overline{KL}</math>. Let <math>M</math> mid-point of <math>\overline{BD}</math> (and also <math>M</math> is mid-point of <math>\overline{AC}</math>). <math>X, O, Y</math> are on the line <math>\ell</math>.

Revision as of 05:49, 15 May 2022

Problem

Let $ABCD$ be a rhombus, and let $K$ and $L$ be points such that $K$ lies inside the rhombus, $L$ lies outside the rhombus, and $KA=KB=LC=LD$. Prove that there exist points $X$ and $Y$ on lines $AC$ and $BD$ such that $KXLY$ is also a rhombus.

Solution

(Image of the solution is here [1])

Let's draw ($\ell$) perpendicular bisector of $\overline{KL}$. Let $X, Y$ be intersections of $\ell$ with $AC$ and $BD$, respectively. $KXLY$ is a kite. Let $O$ mid-point of $\overline{KL}$. Let $M$ mid-point of $\overline{BD}$ (and also $M$ is mid-point of $\overline{AC}$). $X, O, Y$ are on the line $\ell$.

$BK=DL$, $BX=XD$, $XK=XL$ and so $\triangle BXK \cong \triangle DXL$ (side-side-side). By spiral similarity, $\triangle BXD \sim\triangle KXL$. Hence, we get \[\angle XBD = \angle XDB = \angle XKL = \angle XLK = b .\]

Similarly, $AK =CL$, $YK = YL$, $YA=YC$ and so $\triangle BXK \cong \triangle DXL$ (side-side-side). From spiral similarity, $\triangle YKL\sim \triangle YAC$. Thus,

\[\angle YAC = \angle YCA = \angle YKL = \angle YLK = a .\]

If we can show that $a=b$, then the kite $KXLY$ will be a rhombus.

By spiral similarities, $\dfrac{BX}{BD} = \dfrac{XK}{KL}$ and $\dfrac{YA}{AC} = \dfrac{YK}{KL}$. Then, $KL = \dfrac{BD \cdot XK}{BX} = \dfrac{AC \cdot YK}{YA}$.

$\dfrac{XK}{YK} = \dfrac{AC \cdot BX}{AY \cdot BD}$. Then, $\dfrac{YK}{XK} = \dfrac{(BD/2) \cdot AY}{(AC/2) \cdot BX} = \dfrac{\sin b}{\sin a}$. Also, in the right triangles $\triangle KXO$ and $\triangle KYO$, $\dfrac{YK}{XK} = \dfrac{OK/\cos a}{OK/\cos b} = \dfrac{\cos b}{\cos a}$. Therefore, \[\dfrac{\sin b}{\sin a} = \dfrac{\cos b}{\cos a}.\]

$\sin a \cos b = \sin b \cos a \implies \sin a \cos b - \sin b \cos a = 0 \implies \sin(a-b) = 0$ and we get $a=b$.

(Lokman GÖKÇE)

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