Depth · one concept from first principles · step 2 of 4
Rearrange and compare with 1/3
Add 1/z to both sides. Because 1/z > 0, the sum 1/x+1/y is bigger than 1/3.
Seen this way, the equation becomes an inequality with only x and y.
7.EE.B.4Organize Information In More WaysBound y
Since x ≥ y > 0, 1/x ≤ 1/y, so the sum is at most 2/y. Hence 1/3 < 2/y, which means y < 6. With y ≥ 5, y=5.
The larger of two unit fractions must carry at least half of the sum.
7.EE.B.4Convert To AlgebraBecause , the sum is at most , and since it is more than , must be less than .
Why?
A bigger bottom number cuts the whole into more, smaller pieces, so makes no bigger than , and the sum is at most .
Why?
is less than the sum and the sum is at most , so .
Why?
means two pieces of size beat two sixths, which happens only when .
Why?
and are the same amount, because doubling top and bottom only renames it.
Why?
With the same number of pieces on each side, the side with the larger pieces is larger, and larger pieces come from cutting the whole into fewer parts.
Bound x
Put y=5 into the inequality.
With x ≥ 5, the integer x is 5, 6, or 7.
7.EE.B.4Convert To AlgebraFind z in each case
From the equation, 1/z=1/x+1/5-1/3=1/x-2/15. Compute it for each x.
Each case gives a unit fraction, so z=15, 30, 105, all at least 5.
7.NS.A.1Make A Systematic ListAn equation with whole-number answers often becomes easy once an inequality fences the variables into a small pen.
- Rearrange and compare with 1/3
- Bound y
- Bound x
- Find z in each case
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The same shape shows up later on the AMC, as Small-Domain Constrained Search. Why the shapes repeat →