SW-MO-ARML
Practice 9/14/14 – “A Dirty Dozen”
The first six problems come from a puzzle blog of the New York Times.
1. Diameter ACE is divided at C in the ratio of 2:3.
The two semicircles, ABC and CDE, divide the
circular region into an upper (shaded) region and a
lower region. Find the ration of the area of the
upper region to that of the lower region.
2. Line AB, which is 1” long, is tangent to the inner of two
concentric circles at A and intersects the outer circle at B. What
is the area of the annulus formed between the two circles?
3. In the figure, the sides of the smaller square are 3”
and those of the larger square are 4”. Point D is a
vertex of the larger square and is located at the
center of the smaller square. The length of AB is 1”
and BC is 2”. Calculate the area of BCED, the
overlap of the two squares.
4. Divide the rectilinear right-angled figure
shown into two regions of equal area with
one straight line PQ.
5. Using only a straightedge and a compass, construct 5
smaller squares whose area will total that of the larger
square. (You can describe your process and/or sketch
it.)
6. Find the area of the shaded region as a fraction
of the entire regular pentagram. Note: Figure is
not drawn to scale.
The following problems come from the Stanford Math Tournament website – the SMT China
2014 Geometry test and the SMT 2014 Geometry test:
7. Consider a square of side length 1 and erect equilateral triangles of side length 1 on all
four sides of the square such that one triangle lies inside the square and the remaining
three lie outside. Going clockwise around the square, let A, B, C, D be the circumcenters
of the four equilateral triangles. Compute the area of ABCD.
8. Let ABC be a triangle with sides AB = 19, BC = 21 and AC = 20. Let ω be the incircle of
ABC with center I. Extend BI so that it intersects AC at E. If ω is tangent to AC at the
point D, then compute the length of DE.
9. Compute the perimeter of the triangle that has area 3  3 and angles 45◦, 60◦, and 75◦.
10. Consider a square ABCD with side length 4 and label the midpoint of the side BC as M.
Let X be the point along AM obtained by dropping a perpendicular from D onto AM.
Compute the product of the lengths XC and MD.
11. Consider a triangle ABC with AB = 4, BC = 3, and AC = 2. Let D be the midpoint of line
BC. Find the length of AD.
12. Let E be an ellipse with major axis length 4 and minor axis length 2. Inscribe an
equilateral triangle ABC in E such that A lies on the minor axis and BC is parallel to the
major axis. Compute the area of ABC .
For fun: The Goldbach Conjecture says that every integer greater than 2 can be written as the
sum of three primes (allowing 1 as a prime number). Euler reasserted the “Strong Goldbach
Conjecture” which says that any integer greater than 4 can be written as the sum of two primes.
For today’s date: 91414, see if you can find at least one sum for either version of the conjecture.
SW-MO-ARML
Practice 9/14/14 – Solutions
1. Without loss of generality, let AC = 2 and CE = 3. The areas can then be calculated:
1
1
2
5 1
 3  15
Shaded:          1       
2
2
4
2 2
2
2
2
1
10
2
5 1
 3 1
Unshaded:                1 
2
4
2 2
2 2
2
2
Ratio: 3:2
2. Let r be the radius of the smaller circle and R be the
radius of the larger one. By the Pythagorean
Theorem, we know:
r 2  1  R2
So,  r 2     R 2
Or:    R 2   r 2 SO, the area is 
3. Draw DG  AC and DF  FC . Since D is at the
center of the smaller square, we know that DF =
DG = 1.5 and FE = GB = 0.5. So,
2
2
5
 3 1
m2       
2
2 2
Also,
A  BCED   A  BDE   A  BCE  
9
1 2 1
m  2 
4
2
2
4. Choose Q a distance of p from A. Since the area of
the entire region is 3, we need to find p such that:
1  2  p  3 , so p =
2
2
3
2
5. Construct the midpoint of each side of the
square. Draw lines connecting midpoints to
vertices as shown. Construct perpendiculars to
complete the shape. The areas outside the
square can be shown to be equal to the unshaded
areas inside the square, making the areas equal.
6. Reflect the triangle at the top of the
pentagram over its base, forming SAR , as
shown in the figure. Now, SUZ  TAU .
Then the darker shaded trapezoid has the
same area as three of the outside triangles
and ZRS . It can also be shown that
ZRS  ZST , so the area of the lightly
shaded portion that is in the pentagram but
not in the darker region is also the same as
three of the outside triangles and ZRS .
Thus, the shaded region is ½ the total area.
7. Let A denote the circumcenter of the triangle which lies inside the square. We first note
that in an equilateral triangle of side length 1, its altitudes goes through the circumcenter
1
1
and the circumcenter divides the altitude into two segments of lengths
and
.
2 3
3
Thus, ABCD is convex and its area is the sum of the area of triangle ABD and the area of
triangle CBD. The length of BD is
corresponding to base BD is
1
2 3
1
1
2 3

3 1
. In triangle ABD, the altitude
3
1
1
1
3 1
. Thus, the area of ABD is . Next, in


6
2 2 3
2 3
triangle CBD, the altitude corresponding to base BD is
1
1
3 1


Thus, the area
2 2 3
2 3
of CBD is
2 3
3 3
. Therefore, the area of ABCD is
. (I used a kite and thought it
6
6
easier.)
8. Since I is the incenter, we know that BE is the angle
bisector of ∠ABC. By the angle bisector theorem,
AE AB 19

 . Also, we know that CE = 20 −
CE BC 21
19
AE. Thus, AE  . Because D is the point of
2
tangency, we also know that AD = s − BC, where s
is the semiperimeter, which is 30. This implies that
AD = 9. Finally, DE = AE − AD = ½
9. Let the triangle be denoted ABC, with ∠A = 45◦, ∠B = 60◦, and ∠C = 75◦. We first find
side AB. Drop an altitude D onto side AB and note that ACD is a 45-45-90 triangle and
CD
BCD is a 30-60-90 triangle. Therefore, AD = CD and BD 
. Thus, AB   1  1  CD
3
3

1
1
3
Since CD is an altitude, the area of ABC is  AB  CD  
 AB 2 . Since the area of
2
2 1 3
ABC is 3  3 , it follows that AB 2  2 
CD 
3
1 3
3 1
 3  3 and hence AB = 2. Now,
3


AB  3  3 . Since ACD is a 45-45-90 triangle, AC  2  CD  3 2  6 .
Since BCD is a 30-60-90 triangle, BC 
2
 CD  2 3  2 . Thus, the perimeter of ABC
3
is 3 2  2 3  6 .
10. First, by Pythagoras, AM  DM  2 5 . Next, we solve for XD by computing the area of
triangle AMD in two different ways. Using AD as the base, we find the area to be
1
1
 4  4  8 . Using AM as the base, we find the area to be  XD  2 5  5  XD . Thus,
2
2
8
5  XD  8 and hence XD 
.
5
6
. Finally, note that since
5
∠DXM = ∠DCM = 90◦, the quadrilateral DCMX is cyclic. Therefore, we may compute
the product XC · MD via Ptolemy’s theorem: XC · MD = XM · DC + XD · MC and
Next, applying Pythagoras to triangle XMD, we get XM 
XC  MD  8 5 .
11. Let   ACB . Then, applying the Law of Cosines, we get cos  
Law of Cosines to triangle ADC and we get AD 
1
. Then apply the
4
31
2
12. Option 1: Consider a transformation that scales along the major axis by a factor of ½ so
that the ellipse becomes a circle of radius 1 and the equilateral triangle becomes an
BC 
isosceles triangle. Let the transformed triangle be denoted ABC . Now, let x 
.
2
Then the original length |BC| = 4x and since ABC is equilateral, |AB| = |BC| = 4x. Next,
we calculate the altitude dropped from A as 1  1  x 2 . Applying Pythagoras’s theorem,

we get that AB  1  1  x 2
2
that x 2 
  4 x . Substitute |AB| = 4x in and solve for x . We get
2
2
2
48
.
169
Finally, we calculate the area of ABC . Since
ABC is equilateral, its area is given by
3
192 3
2
 BC  4 3  x 2 . Substituting in for x 2 , we get
.
4
169
Option 2: Write the equation of the ellipse:
x 2  4 y 2  4 . Let the side of the equilateral triangle
have length 2m. Then a vertex of the triangle would


fall on m, m 3  1 . Put this ordered pair into the
equation of the ellipse and you get that m 
2
3  16 3  192 3
Then the area is

 
4  13 
169
For fun: I got: 91411 + 1 + 2 or 91411 + 3
8 3
.
13