Problems for middle school math enrichment
In triangle ABC, segment AG from vertex A meets opposite side BC at a point G, with
the length of CG 1/3 the length of BC. That is, point G is located 1/3 of the way along
segment BC from C. Similarly, points E and F are located at 1/3 marks along the other
two sides of triangle ABC. The intersections of these segments are the vertices of
another triangle, IJH, in the interior of triangle ABC.
What is the relationship between the areas of triangles ABC and IJH?
Find the difference between the sum of the first 1,000,000 positive even
numbers and the sum of the first 1,000,000 positive odd numbers.
During a baseball game, a pitcher uses the following 4 throws: a fastball, a
curveball, a changeup, and a slider. Of all his pitches, 4 out of every 10 are
fastballs. He throws twice as many curveballs as sliders and more changeups
than curveballs. If he uses each type of throw as least once over 10 pitches,
how many of each pitch does he use for each 10 pitches thrown?
How many different isosceles triangles are possible if the sides must have
whole-number lengths and the perimeter must be ninety-three inches?
Mary's three pet turtles are great racers. In a race, turtle A comes in first,
beating turtle B by 12 cm and turtle C by 15 cm. The race continues for
second place, and turtle B beats turtle C by 5 cm. Assuming that each turtle
moves at a constant rate, how many centimeters long was the race?
Solutions
Nowadays, with so many dynamic geometry applications readily available, it is really tempting to
investigate this problem with dynamic geometry software to get some idea of how the areas compare
by using measuring tools. That will serve to help us make some conjectures, but of course, we would
still have to prove our conjectures!
Consider another approach, together with some initial observations on the problem. Sometimes with
problems like this one, it is helpful to look for relationships among various triangles and polygonal
regions. Just listing things that we do know can sometimes help us to get started. For example, segment
AF is 1/3 the length of segment AC, so we know that triangle ABF has 1/3 the area of triangle ABC,
because both of these triangles also have the same altitude from point B.
The same is true for triangles AGC and CEB, each of which also has 1/3 the area of the original
triangle, ABC. If we shade all three of these 1/3 area triangles, we can see more relationships.
Each of the triangles ABF, CBE, and ACG, has 1/3 the area of triangle ABC, so Area ABF + Area CBE
+ Area ACG = Area ABC. Furthermore, in the figure above, we see that these three triangles overlap in
such a way as to leave the area of triangle IHJ uncovered, and the area of triangle IHJ is exactly what
we are trying to find! Since the areas of ABF, ACG, and CBE do add up to the area of triangle ABC,
the “missing area” in the picture above can be accounted for in the overlap parts, the small triangles
AHF, BIE, and CJG, because those are being ”counted twice.”
So, now we also know that
Area AHF + Area BIE + Area CJG = Area IJH.
Can you determine anything about the areas of these three corner triangles?
That may be a key to the area of triangle IHJ. Stay tuned…
In last month’s Problem to Ponder discussion, we reasoned that the area of triangle HIJ is equal to the
sum of the areas of the three small shaded triangles in figure 2.
Leah Wulfman, a former student of high school teacher Michael Caraco, suggested that if we construct
a line through each of the three vertices of the tangled triangle HIJ so that each line is parallel to the
respective opposite side, we will create “copies” of triangle HIJ, as shown in figure 3.
Construct lines through points H, I, and J respectively parallel to opposite sides IJ, HJ, and HI. This
construction yields parallelogram HIJM (opposite sides parallel), and produces four congruent
triangles all within a similar triangle RMQ that is formed by the intersections of the three constructed
lines. (Note that triangle HIJ is congruent to triangle JMH by SSS, the other two triangles are then
congruent by ASA). We need then to find a way to argue that the small triangle OMP, which is
exterior to our original triangle ABC, is congruent to triangle FHP. Then we could compose at least
four copies of triangle HIJ from regions inside the original triangle ABC.
Because we also know that the three “corner” shaded triangles fill an area equal to a fifth copy of
triangle IHJ, at this point we would know that our tangled triangle, HIJ, has an area less than 1/5 the
area of our original triangle, ABC.
Only three other pieces inside of triangle ABC remain for us to take into account: triangle OJC and two
other unshaded triangles, one with a vertex at A, the other with a vertex at B.
1,000,000. One strategy would be to try a simpler problem. The sum of the first 10 positive odd
numbers (1, 3, 5, 7, 9, 11, 13, 15, 17, 19) is 100, and the sum of the first 10 positive even numbers (2,
4, 6, 8, 10, 12, 14, 16, 18, 20) is 110. The difference is 10, which is the number of odd and even
numbers. We can generalize as follows: The first odd number is 1, and the first even number is 2. The
difference between the sums is 1. If we look at the sums of the first three odd numbers (1 + 3 + 5 = 9)
and the first three even numbers (2 + 4 + 6 = 12), we notice that the difference is 3, the number of
even and odd numbers. Therefore, the even numbers will be 1 greater for each number. Since we are
working with the first 1,000,000 even and the first 1,000,000 odd numbers, the difference is
1,000,000.
4 fastballs, 3 changeups, 2 curveballs, and 1 slider. The problem states that the pitcher throws 4
fastballs for every 10 pitches, which leaves 6 pitches. Since there must be a 2:1 ratio of curveballs to
sliders, if 4 curveballs were thrown, there would be 2 sliders, which account for all 6 pitches and no
changeups. Considering only 2 curveballs and 1 slider leaves 3 changeups, which supports the
condition that there must be more changeups than curveballs.
23 triangles. Use an organized list and look for a pattern. The congruent sides must measure 24 or
greater, and the third side must measure 47 or less. For three segments to form a triangle, the
measure of the longest side must be less than the sum of the measures of the other two. Another
approach is to realize that the sum of the measures of the congruent sides must be greater than half
the triangle's perimeter.