When you don’t know how to approach a problem, you won’t get anywhere unless you try *something, anything*. In this case, find the digit sum of the product when k=1, k=2, k=3 and see if a pattern becomes evident.

(Source: AMC)

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# Just do something

# Multiples of 11

# Ovelapping Triangle Inequalities

# Right triangles inside right triangles

# Trapezoid existence

# Operation

# Areas of Unusual Shape (AOUS)

When you don’t know how to approach a problem, you won’t get anywhere unless you try *something, anything*. In this case, find the digit sum of the product when k=1, k=2, k=3 and see if a pattern becomes evident.

(Source: AMC)

Every once in a while you’ll need to test a number for divisibility by 11. Multiples of 11 can be quickly identified if the alternating sum of their digits is a multiple of 11. This works because the consecutive powers of 10 are either one more or one less than a multiple of 11. For example, 10 is one less than a multiple of 11, and 100 is one more than a multiple of 11, and 1000 is one less than a multiple of 11.

If you pull out these multiples of 11 from a number, then you are left with an alternating sum. Since the sum of 2 multiples of 11 is a multiple of 11, then the alternating sum is also a multiple of 11.

This is clearer in my notes, where a, b, c, and d are the digits of a 3 and 4 digit number:

I’m not a big fan of inequality problems because I like to get an *answer* not a range of possibilities. But in this case, when the length of a segment must be an integer, and satisfy 2 overlapping triangle inequalities, there is only one possible value.

When you see a right triangle containing other right triangles (as when you drop an altitude to the hypotenuse) you should reflexively check that those triangles are similar. Label the congruent angles to avoid confusion, and even redraw the triangles with corresponding vertices oriented the same way.

(Source: AMC)

Not all 3 line segments will form a triangle. If you don’t believe me, try forming a triangle using line segments of lengths 2, 2 and 20. The two sides of length 2 aren’t long enough to span the 3rd side. A quick test for this is to see that each side is smaller than the combined lengths of the other 2 sides. And that each side is greater than the difference between the other 2 sides.

This week I learned that there is also a fun test to determine whether 4 given side lengths, when assigned to be either the parallel or non-parallel sides can actually form a trapezoid. Not all 4 side lengths in a particular orientation can form a trapezoid. The formula seems weird, but it’s actually an extension of the triangle inequality. Here a and b are the lengths of the parallel sides and c and d are the non-parallel sides.

To form the triangle, draw a line from a vertex parallel to a non-parallel side to the opposite parallel side (see below). The triangle has side lengths and .

An old AMC gives 4 side lengths of possible trapezoids, and the student must test each orientation to determine whether a trapezoid can be formed.

In the old game of Operation, the surgeon/player could earn $100 by removing Butterflies in the Stomach, an actual butterfly from the patient’s torso. But there’s no need to feel butterflies when you see a strange operation defined for the symbol: ¶.

I like to tell my students they can define any symbol to represent any operation they can dream up.

This AMC problem adds a bit of complexity as it’s an operation that is defined for 3 inputs.

I sometimes think the AMC exam writers take special pleasure in defining strange new shapes so students can find their areas.