SAT1000 - P901

Geometry Level 3

For a closed region, the maximum distance of the two points in the region is called the diameter of the region, and the ratio of the perimeter to the diameter is denoted by τ \tau .

As shown above, τ 1 , τ 2 , τ 3 , τ 4 \tau_1, \tau_2, \tau_3, \tau_4 denotes the ratio of the perimeter to the diameter of the four regions, from left to right. Then compare them from smallest to largest .

For example , if τ 2 < τ 1 < τ 3 < τ 4 \tau_2<\tau_1<\tau_3<\tau_4 , submit 2134 2134 .

Have a look at my problem set: SAT 1000 problems


The answer is 1324.

Alice Smith by Alice Smith , 20, China

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1 solution

τ 1 = 16 4 2 ≈ 2.828 \tau_1=\dfrac {16}{4\sqrt 2}\approx 2.828

τ 2 = 4 π 4 ≈ 3.142 \tau_2=\dfrac {4π}{4}\approx 3.142

τ 3 = 3 ( 2 − n ) 2 − n = 3 \tau_3=\dfrac {3(2-n)}{2-n}=3

τ 4 = 24 4 3 ≈ 3.464 \tau_4=\dfrac {24}{4\sqrt 3}\approx 3.464

Hence, τ 1 < τ 3 < τ 2 < τ 4 \tau_1<\tau_3<\tau_2<\tau_4 , and the answer is 1324 \boxed {1324} .

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