Maths Olympiad Prep

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CEMC Cayley, 2025

Geometry Difficulty 3.2 AMC 10/12 Find the answer Canada

In the diagram, point DD is
on side ABAB of △ABC\triangle ABC and points EE and FF are on BCBC.

Figure 0

For some positive real number tt,
the area of △DBE\triangle DBE is tt, the area of △DEF\triangle DEF is tt, the area of △DFC\triangle DFC is 2t2t, and the area of △DAC\triangle DAC is 4t4t. If the area of △DEC\triangle DEC is 6363, what is the area of △ABC\triangle ABC?

Pick one

Solution

The area of △DEC\triangle DEC is
equal to the sum of the areas of △DEF\triangle DEF and △DFC\triangle DFC, and
so t+2t=63t+2t=63 or 3t=633t=63 or t=21t=21. The area of △ABC\triangle ABC is equal to the sum of the
areas of △ADC\triangle ADC, △DBE\triangle DBE, △DEF\triangle DEF and △DFC\triangle DFC, which is equal to 4t+t+t+2t=8t=8(21)=1684t+t+t+2t=8t=8(21)=168.

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