February Regional
Geometry Individual Solutions
1) B; The area of triangle ABC, according to the formula
, is
. The area of this triangle is also equal to
, where
x is the altitude drawn from vertex A to side BC. Setting these two expressions
equal reveals that
.
2) D; The geometric mean of 11, 1331 is
The square root of 64 is 8.
. The arithmetic mean of 121 and 7 is
3) B; If we let the unknown side of the triangle be x, then according to the triangle inequality
theorem,
. Thus,
, so there are 9 possible integral values for x.
4) C; The complement of the complement of
.
is
. The supplement of
is
5) A; The radius of the circle can be found by dropping a perpendicular to the
base of the equilateral triangle. This will form a 30-60-90 triangle with side
opposite the
angle equal to 4. Since this triangle is a 30-60-90 triangle,
the hypotenuse, which is the radius of the circle, must be
circle is
. The area of the
. The area of the equilateral triangle is given
by the formula
the triangle is
. The area inside the circle but outside
.
6) B; The value of x must be found in order to determine what the largest angle of the triangle is.
Since the sum of the angles of a triangle is equal to
, then:
. Solving
this equation for x, we find that
. Since x is equal to 13, the largest angle of this triangle is 78
degrees, the complement of which is 12 degrees.
7) E; Before the length of segment CE can be found, the
length of segment BE, which we have called x, must be found.
Since segment CE is a secant segment, and segment AE is
tangent to circle D, we know that
.
Plugging in the values for segments AE and BE, we have
. Putting all terms on one side, gives
To find the value of x, we need to use the
quadratic formula,
. Simplifying tells
us that
. The value of x must be positive, so
the length of segment CE is
.
8) A; The formula for the number of diagonals in a convex polygon is
sides, the number of diagonals is
. An undecagon has 11
February Regional
Geometry Individual Solutions
9) B; The formula for the sum of the squares of the lengths of the medians of any triangle is
, where a, b, and c are the sides of the triangle. Plugging in our numbers, we get
. This problem also could have been solved by finding the
actual lengths of the medians.
10) D; This problem involves similar triangles. Since the triangle
made by the tree is similar to the triangle made by Gulete, the
proportion
can be used to find the length of Gulete’s
shadow. Solving for x, we obtain
.
11) C; Since we have been told that this is an isosceles trapezoid,
the legs are equal. If we divide the trapezoid into 3 parts by
drawing perpendiculars from the top base vertices to the bottom
base, we obtain two right triangles. These right triangles have one
leg of length 2 and a hypotenuse of length
. Using the
Pythagorean theorem, it is found that the other leg of both of these triangles is also of length 2. This
means that height of the trapezoid is 2. Now that we know the height, the area can be found using the
formula
, where h is the height, and b1 and b2 are the bases. Substituting our values
in, we get
.
12) D; The sum of the interior angles of any triangle is
any pentagon is
. Using this information, we know
, and the sum of the interior angles of
, as well as
Simplifying these two equations, we obtain
, and
. Solving this system of equations, we find x=40. Substituting
for x, we find that y =-50. The value of x + y = 40 - 50 = -10.
13) B; The area of a kite is given by
, where d is a diagonal of the kite. The diagonals of the
kite are 121 and 16. Thus, the area of the kite is
.
14) E; All of the given choices are properties of Rhombi.
15) E; Vertical angles are congruent, so
. Moving all terms to one side, we
have the quadratic
. Solving for x using the quadratic formula gives us
.
16) B; Triangles ABC and AED are similar. Thus,
. Cross-multiplying gives us
. Solving for x reveals that x= 55/3. The length of segment AE is
.
February Regional
Geometry Individual Solutions
17) A; We are told that the rhombus has diagonals of length 36 and 24. Since the diagonals of a
rhombus bisect each other, and are perpendicular to each other, the diagonals divide the rhombus
into 4 congruent right triangles with legs of length 18 and 12. The perimeter of the rhombus is equal
to 4 times the length of the hypotenuse of one of these triangles, since there are 4 triangles. The
length of the hypotenuse of one of these triangles is
. Thus, the
perimeter of the rhombus is
18) B; Since the area of the square was doubled, the ratio of the area of the larger square to that of
the smaller square is 2. This means that the larger square must have sides that are
times larger
than that of the smaller square. Thus, the ratio of the perimeter of the larger square to that of the
smaller square is
.
19) B; The intersection of the altitudes of a triangle is known as the orthocenter.
20) D; The measure of an interior angle of a regular polygon is given by the formula
, where
n is the number of sides of the polygon. We are told that the interior angle measure is 162. We set
that equal to the above expression, giving us
then isolating the variable n, we get
. Multiplying both sides by n, and
. Dividing both sides by n shows us that n=20.
21) D; This problem can be solved by using the geometric definitions of sine and cosine. Sine of an
angle is defined as
, and cosine of an angle is defined as
. The problem asks
for the value of
. This is equivalent to asking for the value of
, which is equal to
this expression is equal to
The numerator of
by the Pythagorean theorem, so this expressions is really
22) A; Husayn has 6 units of fencing, so the sum of the three fenced sides
of the rectangle must add up to 6. In order to find the maximum area of
this rectangular region, we must first find an expression for the area in
terms of the unknown dimensions. This expression can be found by setting
the two opposite fenced sides equal to x, and the remaining fenced side equal to 6-2x.The area of the
rectangle is then
. Note that the area of the rectangle is a
parabola that opens downward. So, the maximum area of the rectangle is at the vertex of the
parabola. The x-coordinate of the vertex of this parabola is given by – , where the equation of the
parabola is
. Substituting our numbers in, we find that the x-coordinate equals
Now that we know what value of x maximizes the area, we plug this into our
expression for the area to reveal
23) D; The length of the altitude to the hypotenuse of a right triangle can be found
by using the geometric mean. In the figure to the right, the altitude to the hypotenuse
divides the hypotenuse into 45 and 5. The length of the altitude is the geometric
mean of the lengths of these two segments. Thus,
.
February Regional
Geometry Individual Solutions
24) C; Statement I is true. Statement II is false. Statement III is false. Statement IV is true. Only 1 of
the statements is true.
25) D; The distance between these points is found using the distance formula. Distance
=
.
26) E; The only statement that must be true is the contrapositive of the original statement. None of
the choices are the contrapositive of the given statement.
27) A; This problem can be solved by modeling the 13 attendees to the conference as 13 vertices.
This will result in a 13 sides polygon. To find the number of handshakes that occur, first find the
number of diagonals of this polygon. This can be done by using the formula
. Plugging in 13,
we get
. Now we must add the number of sides in this polygon to the number of
diagonals. This results in
. Thus 78 handshakes occur at this meeting.
28) C; It is stated that segment EB is 15 less than the square of segment
CE, so we have called segment CE x, and segment EB
. This
problem can now solved by using power of a point. According to this
theorem, the product of segments CE and ED equals the product of
segments AE and ED. Thus,
. Putting all terms of this
quadratic on one side and dividing by like terms gives us
. This quadratic factors into
Solving for x yields x=
(5, -3), but we know that x must be positive, so x=5. The problem asks for
the geometric mean of the lengths of CD and AB. Since x=5, segment
CD=15 and segment AB=15. The geometric mean of 15 and 15 is 15.
29) D; There are 360 degrees in a circle, so this sector is
the whole circle is
the area of this circle. Since the area of
, the area of this sector is:
30) B; The formula for the angle in degrees between the hour and minute hands of a clock is
, where h is the hour and m is the minute. We are told that the time is 7:36, so the angle is
.