A sinusoid squeezed between two shifted sines — strictness matters

mangoaggro·Jul 13, 2026(edited)

The lower and upper envelopes have corners. How can a single sinusoid lie between them for every x and touch only where the original two sines agree? Problem: Let

a

be a real number. Prove that there exist real numbers

b

and

c

such that the inequalities

\min \{\sin(x), \sin(a+x)\} \le b \sin(x+c) \le \max \{\sin(x), \sin(a+x)\}

hold for all real numbers

x

and the equalities hold only if

\sin(x) = \sin(a+x)

. Source: Baltic Way 2021 Shortlist · MNET-01HD

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parry_this·Jul 13, 2026(edited)

Use their arithmetic mean:

(\sin x+\sin(x+a))/2=\cos(a/2)\sin(x+a/2)

. Take

b=\cos(a/2)

and

c=a/2

. An arithmetic mean lies strictly between two unequal real numbers, giving exactly the required equality condition, including the cases where the two original functions coincide everywhere.

A sinusoid squeezed between two shifted sines — strictness matters

mangoaggro·Jul 13, 2026(edited)

The lower and upper envelopes have corners. How can a single sinusoid lie between them for every x and touch only where the original two sines agree? Problem: Let

a

be a real number. Prove that there exist real numbers

b

and

c

such that the inequalities

\min \{\sin(x), \sin(a+x)\} \le b \sin(x+c) \le \max \{\sin(x), \sin(a+x)\}

hold for all real numbers

x

and the equalities hold only if

\sin(x) = \sin(a+x)

. Source: Baltic Way 2021 Shortlist · MNET-01HD

Open MNET-01HD in the question bank →

1 reply

parry_this·Jul 13, 2026(edited)

Use their arithmetic mean:

(\sin x+\sin(x+a))/2=\cos(a/2)\sin(x+a/2)

. Take

b=\cos(a/2)

and

c=a/2

. An arithmetic mean lies strictly between two unequal real numbers, giving exactly the required equality condition, including the cases where the two original functions coincide everywhere.

A sinusoid squeezed between two shifted sines — strictness matters

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A sinusoid squeezed between two shifted sines — strictness matters

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