Convex hexagon ABCDEF is drawn in the plane such that ACDF and ABDE are parallelograms with area 168. AC and BD intersect at G. Given that the area of AGB is 10 more than the area of CGB, find the smallest possible area of hexagon ABCDEF.
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Official solution
1. Given that ACDF and ABDE are parallelograms with area 168, we know that [ACDF]=[ABDE]=168. 2. Since ACDF and ABDE are parallelograms, AC∥DF and AB∥DE. 3. The diagonals AC and BD intersect at G. 4. Let the area of △AGB be 10+a and the area of △CGB be a. 5. Since [ACDF]=168, we have [ACD]=[ACF]=84. 6. Similarly, since [ABDE]=168, we have [ABD]=[ADE]=84. 7. Note that [ABD]=[ACD]=84, so AD∥BC. 8. This implies that ABCD is a trapezoid. 9. Let [CGB]=a. Then, [AGB]=10+a and [CGD]=10+a. 10. Since [AGD]=74−a, we have △AGD∼△CGB. 11. Using the similarity ratios and area ratios, we get: 74−aa=(a+10a)2 12. Solving the equation: 74−aa=(a+10a)2 a(74−a)=(a+10)2 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 74a−a2=a2+20a+100 \[ 74a - a
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