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Rectangle $ABCD$ contains a point $X$ such that $AX = 3$, $BX = 3$, and $CX = 3$.  Find $DX$.

 Dec 14, 2023
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Since AX=3, BX=3, and CX=3, we can extend lines AX, BX, and CX until they intersect at a point, which we'll call O.

 

Now let U be the foot of the perpendicular from O to line segment BD. Clearly, the sides of quadrilateral AOUX are half the diagonals of rectangle ABCD

 

AO=UO=DX=2AC​=2AD3​​

 

We know that AC=AD3​ from Pythagorean Theorem on right triangle ACD. Also, we know AD=3+3+3=9. Substituting those values, we get

AO=UO=DX=293​​

 

Therefore, DX=293​​​.

 Dec 14, 2023

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