Problem of the Week
Problem E and Solutions
Made in the Shade
Problem
Two sides of equilateral 4ABC are each divided into 8 equal
segments, with length 2 cm. Seven line segments, each parallel
to AC, are drawn by connecting division points on BC to
division points on BA, as shown. An altitude is constructed
from A to T on BC, dividing the shaded region into two parts.
In this shaded region, determine the ratio of the area on the
left side of the altitude to the area on the right side of the
altitude.
Solution 1:
Label the shaded region DEF G. Let Q and R be where altitude AT intersects DEF G. We
are required to find the ratio of the area of DRQG to the area of QREF .
We will find the area of DEF G by finding the area of 4DBE and subtracting it from the area
of 4GBF .
But first we will establish two results that will help in our
solution. For some equilateral 4P QS with side length 2, altitude P R right bisects base QS and it follows that QR =
RS√= 1. Using
√ the Pythagorean Theorem, the length of P R
equilateral triangle, the ratio of the
is 22 − 12 = 3. In any √
height to the side length is√ 3 : 2. In other words, the height of
any equilateral triangle is 23 times its side length. The altitude
P R also bisects ∠QP S so ∠QP R = ∠SP R = 30◦ and
√4P QR
◦
◦
◦
is a 30 - 60 - 90 triangle whose sides are in ratio 1 : 3 : 2.
In equilateral 4ABC, we know that ∠ABC = ∠BCA = ∠CAB = 60◦ .
In 4DBE, ∠DBE = ∠ABC = 60◦ since they are the same angle. Since DE k AC,
∠BED = ∠BCA = 60◦ and ∠EDB = ∠CAB = 60◦ . Since all of the angles in 4DBE are
60◦ , it is an equilateral
triangle
with sides of length 12 cm.
√
√ Using√our above result, the height
√
3
of 4DBE = 23 × 12 = 6 3. The area of 4DBE = 12×6
= 36 3 cm2 .
2
In a similar way, we can show
that 4GBF
is an equilateral triangle with
14 cm.
√
√ side length
√
√
3
14×7 3
The height of 4GBF = 2 × 14 = 7 3 and the area of 4GBF = 2 = 49 3 cm2 .
√
√
√
The area of DEF G = the area of 4GBF − the area of 4DBE = 49 3 − 36 3 = 13 3 cm2 .
Next, we will find the area of QREF by finding the area of 4RET and subtracting it from the
area of 4QF T .
In 4RET , ∠RT E = ∠AT C = 90◦ since the altitude AT is perpendicular to base BC and the
two angles represent the same angle. Since DE k AC, ∠RET = ∠ACB = 60◦ . It follows
that
√
∠ERT = 30◦ . Then 4RET is a√30◦ − 60◦ − 90◦ triangle and ET : RT : RE√ = 1 : √ 3 : 2. Since
ET = 4, it follows that RT = 4 3 and RE = 8. The area of 4RET = 4×42 3 = 8 3 cm2 .
In a similar way, we √
can show that 4QF T is a 30◦ − 60◦ − 90◦ triangle
with
√
F T : QT : QF =√1 : 3 √
: 2. Since F T = 6, it follows that QT = 6 3 and QF = 12. The area
6×6 3
of 4QF T = 2 = 18 3 cm2 .
√
√
√
The area of QREF = the area of 4QF T − the area of 4RET = 18 3 − 8 3 = 10 3 cm2 .
√
√
√
The area of DRQG = the area of DEF G − the area of QREF = 13 3 − 10 3 = 3 3 cm2 .
√
√
The ratio of the area of DRQG to the area of QREF is 3 3 : 10 3 = 3 : 10.
Solution 2:
In the diagram, we see that we can tile 4ABC with the bottom left equilateral triangle of side
length 2 cm. (You may wish to justify that this is indeed an equilateral triangle.)
Then, three equilateral triangles fit in the first trapezoid from the left, 5 equilateral triangles fit
in the second trapezoid from the left, 7 equilateral triangles fit in the third trapezoid from the
left, 9 equilateral triangles fit in the fourth trapezoid from the left, 11 equilateral triangles fit
in the fifth trapezoid from the left, and 13 equilateral triangles fit in the sixth trapezoid from
the left.
The shaded trapezoid is the sixth trapezoid from the left and so contains a total of 13
congruent equilateral triangles. To the left of the altitude there are 2 + 12 + 12 = 3 of the
triangles and to the right of the altitude there are 13 − 3 = 10 of the triangles. Each of the 13
triangles in the shaded trapezoid have the same area.
The ratio of the shaded area to the left of the altitude to the shaded area to the right of the
altitude is 3 : 10.
There are many things in this second solution that the solver may wish to justify.