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Geometry Difficulty 4.8 AIME Find the answer

2. As shown in Figure 2, in quadrilateral ABCDABCD, AB=10AB=10, BC=17BC=17, CD=13CD=13, DADA =20=20, AC=21AC=21. Then BD=BD=

A number or a short expression. Spacing and $ signs are ignored.

Solution

2.1052.10 \sqrt{5}.
As shown in Figure 6, draw BEACB E \perp A C at point EE, DFACD F \perp A C at point FF, and draw BG//ACB G / / A C intersecting the extension of DFD F at point GG. Then, quadrilateral BEFGB E F G is a rectangle, and BDG\triangle B D G, ABE\triangle A B E, and CDF\triangle C D F are all right triangles.
In ABC\triangle A B C, we have
p=12(AB+BC+CA)=24 p=\frac{1}{2}(A B+B C+C A)=24 \text{. }

Thus, BE=22124(2410)(2417)(2421)=8B E=\frac{2}{21} \sqrt{24(24-10)(24-17)(24-21)}=8.
Similarly, in ADC\triangle A D C, we have DF=12D F=12.
Therefore, DG=DF+FG=DF+BE=20D G=D F+F G=D F+B E=20.
Also, AE=AB2BE2=6A E=\sqrt{A B^{2}-B E^{2}}=6,
CF=CD2DF2=5 C F=\sqrt{C D^{2}-D F^{2}}=5 \text{, }

Then, BG=EF=ACAECF=10B G=E F=A C-A E-C F=10.
Thus, BD=DG2+BG2=105B D=\sqrt{D G^{2}+B G^{2}}=10 \sqrt{5}.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.