2022 USAJMO Problems/Problem 4

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 $\triangle AKY \cong \triangle CYL$ (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)

2022 USAJMO (Problems • Resources)
Preceded by Problem 3	Followed by Problem 5
1 • 2 • 3 • 4 • 5 • 6
All USAJMO Problems and Solutions

Page

Toolbox

Search

2022 USAJMO Problems/Problem 4

Problem

Solution

See Also