2020 AMC 12A Problem 17

Problem 17 of 25IntermediateAlgebra Geometry

The vertices of a quadrilateral lie on the graph of

y = \ln x,

and the

x

-coordinates of these vertices are consecutive positive integers. The area of the quadrilateral is

\ln\dfrac{91}{90}.

What is the

x

-coordinate of the leftmost vertex?

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Solution

Let the vertices have

x

-coordinates

n, n+1, n+2, n+3

with

y

-coordinates

\ln

of those values. Applying the shoelace formula and simplifying, the area is

\ln\dfrac{(n+1)(n+2)}{n(n+3)}.

Setting

\dfrac{(n+1)(n+2)}{n(n+3)} = \dfrac{91}{90}

gives

1 + \dfrac{2}{n^2 + 3n} = \dfrac{91}{90},

n^2 + 3n = 180.

Then

(n - 12)(n + 15) = 0,

n = 12.

Thus, D is the correct answer.