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If a square Mercator map shows 1000 miles at latitude 30°, how many miles does it show at latitude 60°?

As far as I know, in a Mercator map, every horizontal strip is stretched by $\cos x$ so that the distance from the equator to $x$ north is $\int_0^xR\sec x\,dx$. The distance to latitude 30° should be $\int_0^{\pi/6}R\sec x\,dx=R\ln\sqrt3=1000$. Thus, $R=\frac{1000}{\ln\sqrt3}$. At latitude 60°, $\int_0^{\pi/3}R\sec x\,dx=R\ln(2+\sqrt3)=1000\frac{\ln(2+\sqrt3)}{\ln\sqrt3}$. But the answer key says $\frac{1000}{\sqrt3}$. Where was I wrong?

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No need to integrate $R\sec{x}$.

All you need to do is find the scale factor $k$.

$$k=\frac{R\sec{30^o}}{R\sec{60^o}}$$ $$k=\frac{\sec{30^o}}{\sec{60^o}}$$ $$k=\frac{\cos{60^o}}{\cos{30^o}}$$ $$k=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}$$ $$k=\frac{1}{\sqrt{3}}$$ Now multiply this by your radius $R$. Hence, the Mercator map will show $\frac{1000}{\sqrt{3}}$ miles at latitude $60^o$.

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    The resource you provided says $k=\sec\phi$. Why did you take the ratios? – 2017-01-06
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    The value of $k$ is relative to $\phi=0$ (the equator). Hence, to get the **relative** ratio between $\phi=30^o$ to $\phi=60^o$, we divide both. – 2017-01-06
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    We want to find the distance at 60°. Shouldn't the ratio be $\frac{\sec60°}{\sec30°}$? – 2017-01-06
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    Yes, it would if you take the definition given by Wikipedia. However, I found it more convenient to define our scale factor relative from $\phi=30^o$ over $\phi=60^o$, since we can multiply the ratios rather than divide them. If you prefer to use the other definition, I can edit this post. – 2017-01-06
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    WIth $\frac{\sec60°}{\sec30°}$, we would get $1000\sqrt3$, which doesn't seem right because it should get shorter as it goes away from the equator. The mistake might be in $d\sec30°=1000$ because $1000$ is really a distance in the real world but not on the map. So I think it should be $d\cos30°=1000$. That gives $d=1000\sec30°$. Then, $d\cos60°=1000\sec30°\cos60°=\frac{1000}{\sqrt3}$. We do not really need to know the radius of the earth, so I used $d$. – 2017-01-06
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    The methodology I used is exactly the same as yours, however I defined the scale factor $k$ differently. If we denote your definition (also wikipedia's definition) as $k_1$, then the definition I used would be $k=\frac{1}{k_1}$. Which is why the ratio I used was $\frac{\sec{30^o}}{\sec{60^o}}$ instead of the reciprocal $\frac{\sec{60^o}}{\sec{30^o}}$. Therefore, to account for the reciprocal ratio I used, I multiplied $R$ by the ratio $k$ instead of dividing it. – 2017-01-06