Question 1
Find the sum of the answers to the four questions below.
a) Find the area of a square with side length 2.
Solution: The area is 22 = 4.
b) How many sides does a pentagon have?
Answer: 5.
c) What is the maximum number of distinct points in which a line and a circle can intersect?
Answer: 2
d) How many diagonals does an octagon have?
Solution: There are n(n-3)/2 = 8(8-3)/2 = 20 diagonals in an octagon.
Question 2
Find the sum of the answers to the two questions below.
a) A hexagonal pyramid has a base with a side length 2 and a height of 5. Find its volume.
Answer: 10 3
Solution: The area of the hexagonal base is
!! ! !
!
= 6 3. Then the volume is
!
!
6 3 5 =
10 3.
b) The greatest distance between two vertices of a rectangular prism is 5. The greatest distance
between two vertices which is less than that is 3. Find the greatest possible volume of the prism.
Answer: 162
Solution: Let the dimensions be l, w, h, with l > w > h. The greatest distance is the length of the
space diagonal: l2 + w2 + h2 = 25. The next greatest distance is the length of the longest face
diagonal: l2 + w2 = 9. Subtraction gives h2 = 16, so h = 4. The largest volume is attained when the
largest area of the lw-face is attained, which is when it is a square, i.e., l = w = 9/ 2. Then the
largest volume is lwh = (81/2)(4) = 162.
Question 3
Find the number of true statements below.
a) The reflections of the altitudes of a triangle about the angle bisectors from their respective
vertices intersect at the triangle’s circumcenter.
Solution: This is true. See diagram. To show this, let D be the foot of the altitude from A, and let
O be the circumcenter. It suffices to show that ∠BAD = ∠OAC. Since BAD is a right triangle,
∠BAD = 90 – ∠B. We now want to show that ∠OAC = 90 – ∠B. The triangles ∆OBC, ∆OAC,
∆OAB are isosceles since the circumcenter is equidistant from the vertices; let their base angles
be be x, y, z respectively. Then ∠A = y+z, ∠B = x+z, ∠C = x+y. Manipulating these equations
gives y = 90 – ∠B, and we are done. (Another way to show that ∠OAC = 90 – ∠B is to use the
inscribed angle theorem to see that ∠AOC = 2∠B since ∠AOC is a central angle in the
circumcircle and subtends the same arc as the inscribed angle ∠B.) Remark: The isogonal
conjugate of a point P with respect to a triangle ABC is the point of intersection of the reflections
of the lines PA, PB, PC about their respective angle bisectors. We have shown here that the
isogonal conjugate of the orthocenter is the circumcenter.
b) The circumcenter of a triangle can lie outside the triangle.
Solution: This is true. The circumcenter lies outside the triangle iff the triangle is obtuse. By the
Inscribed Angle Theorem, the obtuse angle of the triangle is an inscribed angle spanning an arc of
the circumcircle which is greater than 180 degrees.
c) If two distinct planes intersect, they intersect in a line.
Solution: This is true. They cannot intersect in a point.
d) In the plane, one circle is centered at A with radius x, and another circle is centered at B with
radius y. If they intersect at the distinct points M and N, for every point P on the line MN we have
PA2 – x2 = PB2 – y2.
Solution: This is true. We have three cases.
Case 1. If P=M or P=N, both sides of the given equation are 0.
Case 2. If P is in the interior of the circles, consider chord MN and the chord of circle A
containing M and A. By the Power of a Point Theorem for intersecting chords, we have (x –
PA)(x + PA) = PM*PN. A similar argument with circle B shows that (y – PB)(y + PB) = PM*PN.
Combining the two equations gives (x – PA)(x + PA) = (y – PB)(y + PB), which is a factored
form of what we want.
Case 3. If P is outside the circles, consider circle A and the lines PA and PMN. By the Power of a
Point Theorem for intersecting secants, (PA – x)(PA + x) = PM*PN. Then consider the lines PB
and PMN. By the Power of a Point Theorem for intersecting secants, (PB – y)(PB +y) = PM*PN.
Combining the two equations gives (PA – x)(PA + x) = (PB – y)(PB +y), which is a factored
form of what we want.
Remark: It can be shown that the points on line MN are the only points for which the given
equation holds. For this reason, line MN is known as the radical axis of the circles A and B.
Question 4
Find the sum of the answers to the four questions below.
a) Find the sum of measures of the interior angles of an octagon in degrees.
Answer: 1080
Solution: Pick a vertex and draw all the diagonals which come from it. These split the interior of
the octagon into those of 6 triangles whose interior angles comprise those of the octagon. Then
the sum of the interior angles of the octagon is 180(6) = 1080.
b) Find the measure of an interior angle of a regular hexagon in degrees.
Answer: 120
Solution: The measure of an exterior angle is 360/6 = 60. Then a measure of an interior angle is
180 – 60 = 120.
c) Find the measure of the acute angle whose sine is 1/2, in degrees.
Answer: 30
Solution: Use the 30-60-90 right triangle.
d) A circle is divided into 90 congruent sectors. Find the degree measure of the central angle
spanned by one of these sectors.
Answer: 4
Solution: 360/90 = 4.
Question 5
Find the sum of the answers to the three questions below.
a) Triangle ABC has AB = 13, BC = 14, and CA = 15. Let D and E be the points of tangency of the
incircle of ABC with BC and CA, respectively. Find DE 5.
Answer: 16
Outline of solution: Let I be the incenter of ABC. Using congruent triangles, it can be shown that
CI is perpendicular to DE. Let their intersection be G. Since IDC = 90 and ID and DC can be
found, DG = DE/2 is the altitude to the hypotenuse of right triangle IDC.
Detailed solution: Let I be the incenter of ABC, and let CI intersect DE at G. Let s be the
semiperimeter. Then we have CD = CE = s – AB = 8. Using the formula A = rs, where A is the
area and r is the inradius, we can find r, which is ID. We have s = (13+14+15)/2 = 21. By Heron’s
formula, A = 21 8 7 (6) = 84. Then r = ID = 84/21 = 4. We also have that CGD and CGE are
congruent by SAS. Then angle DGC = EGC, and since DGC + EGC = 180, DGC = EGC = 90.
Then DG is an altitude of CID. The area of CID is 4(8)/2 = 16 = (CI)(DG)/2 = 4 5(DG)/2, so DG
= 8/ 5 = DE/2, so DE = 16/ 5, so DE 5 = 16.
b) The shortest two sides of a right triangle have lengths 7 and 24. Find the length of its hypotenuse.
Solution: 7! + 24! = 625 = 25.
c) The medial triangle of a triangle ABC is the triangle with the midpoints of the sides of ABC as its
vertices. If a triangle has side lengths 9, 10, and 11, find the perimeter of its medial triangle.
Answer: 15
General solution: Using similar triangles, each side of the medial triangle is half the length of the
side of ABC is it parallel to. Then the perimeter of the medial triangle is half the perimeter of
ABC.
Detailed solution: Let M and N be the midpoints of AB and AC, respectively. Then AMN is
similar to ABC by SAS similarity. Since AM/AB = 1/2, MN/BC = 1/2, so MN = BC/2. Similar
arguments for the other two sides of the medial triangle show that each side of the medial triangle
is half the length of the side of ABC it is parallel to. Then the perimeter of the medial triangle is
(9+10+11)/2 = 15.
Question 6
Find the sum of the answers to the two questions below.
a) Tony and Mitchell are about to eat the interior of an isosceles triangle with side lengths of 10, 10,
and 16. They have to cut it into two pieces of equal area, one piece for each of them. The Mu
Alpha Theta gods require them to cut it along a straight line parallel to the base. How far from the
base should they cut?
Answer: 6 – 3 2
General solution: Use the fact that the smaller triangle is similar to the larger one.
Detailed solution: Let the triangle be ABC, with AB = AC = 10 and BC = 16. Let M and N be on
AB and AC respectively such that MN || BC and MN splits the triangle into two pieces of equal
area. Let AD be the altitude from A to BC, and let it intersect MN at E. ABD is congruent to
ACD, so they are both right triangles and BD = 16/2 = 8. Then AD = 10! − 8! = 6. Let DE =
x, so AE = 6–x. AME = ABD since they are corresponding, so AME is similar to ABD to AngleAngle. AD/BD = 6/8, so AE/ME = (6–x)/ME = 6/8, ME = (4/3)(6–x). The area of ABC is 48, so
the area of AMN is 24. The area of AMN is AE*ME = (4/3)(6–x)2 = 24. This gives (6–x)2 = 18.
Taking the square root, 6–x = ±3 2. Solving for x and taking the positive value gives x = 6 –
3 2.
b) Determine the area of the largest square which can be inscribed in a semicircle of diameter 10.
Answer: 20
Solution: Let the square be ABCD and the center of the semicircle be O. Then if ABCD is the
largest square, it has A and B on the diameter and C and D on the arc. Let the side length be x. By
symmetry AO = x/2. DO = 5 since this the radius is half the diameter, and DA = x. By
Pythagorean Theorem on right triangle DAO, x2 + (x/2)2 = 25, which gives x2 = 20.
Question 7
Find the sum of the answers to the two questions below.
a) ABCDE…R is a regular octadecagon (18-sided polygon). Find the measure of angle ADE in
degrees. (13) Answer: 140, 140 degrees, or 140°
b) In isosceles triangle ABC, AB = AC, and BC is the shortest side. There exists a point D on
segment AC (strictly between A and C) such that AD = DB = BC. Find the measure of angle
BAC in degrees. (10)
Answer: 36
Solution: See diagram. Let the measure of angle BAC be x. Then also ABD = x because ABD is
isosceles. Since ABC is isosceles, ACB = (180-x)/2 = 90 – x/2. But BCD is also isosceles, so
BDC = 90 – x/2. Then DBC = 180 – 2(90 – x/2) = x. So ABC = ABD + DBC = x+x = 2x. Then
we also know from isosceles triangle ABC that ABC = 90 – x/2, so 2x = 90 – x/2. Solving, we get
x = 36
Question 8
All four of the vertices of quadrilateral JOHN lie on the same circle.
JH bisects angle NJO and intersects ON at X.
If JO = 8, JN = 10, and OX = 2, find OH ! .
Answer: 16/3
Solution: By the Angle Bisector Theorem, JO/OX = JN/NX, so 8/2 = 10/NX, so NX = 5/2. By the
Inscribed Angle Theorem, ∠HON = ∠HJN since both angles subtend the same arc. We also have ∠HXO
= ∠NXJ since they are vertical angles. Then ∆HOX ~ ∆NJX by Angle-Angle. Let HX = a. Then we have
OH/a = 10/(5/2), which gives OH = 4a. Then note that ∆HON is isosceles with OH = HN: from the
Inscribed Angle Theorem we know that ∠HNO = ∠HJO, and we already know that ∠HJO = ∠HJN =
∠HON, so ∠HON = ∠HNO. Thus also HN = 4a. Let M be the midpoint of ON. Then MN = [2+(5/2)]/2 =
9/4 and MX = XN – MN = 1/4. Congruent triangles tell us that HMN = HMX = 90, so the Pythagorean
Theorem on ∆HMN gives HM2 = 16a2 – 81/16. Pythagorean Theorem on ∆HMX gives HM2 + MX2 =
16a2 – 81/16 + 1/16 = a2. Solving this equation gives a2 = 1/3. Then OH2 = (4a)2 = 16/3.
Question 9
Find the sum of the answers to the two questions below.
a) How many noncongruent triangles with integer side lengths have perimeter 9?
Answer: 3
Solution: Note that by the Triangle Inequality, the length of a side of a triangle must be less than
half its perimeter. Thus our search is restricted to side lengths of 4 or less. Only (3, 3, 3), (2, 3, 4),
and (1, 4, 4) work.
b) The lattice pictured below has evenly spaced rows of evenly spaced points, with alternate rows
shifted one half spacing. In how many ways can three points be chosen from this lattice such that
they are the vertices of a (non-degenerate!) triangle?
Answer: 110
Solution: There are 10C3 = 120 ways to choose three points from the lattice. But if the three
points are collinear, they cannot form a triangle. We can choose three collinear points in the
following ways: 1 for the top row, 1 for the bottom row, 4C3 = 4 for the middle row, and 4 for the
diagonals with three points in them. This makes a total of 10 exclusions. Then the answer is 120 –
10 = 110.
Question 10
Find the sum of the answers to the two questions below.
a) Quadrilateral JOHN has the following interior angle measures: J = 15, O = 240, H = 15. If JN =
NH = 8 6, find the area of JOHN.
Answer: 192 − 64 3
Solution: In this solution, we denote the area of a polygon P as [P]. Since JOHN is concave,
[JOHN] = [JHN] – [JHO]. ∠N = 360 – 15 – 240 – 15 = 90, so ∆JHN is a right triangle and so
!
[JHN] = (1/2) 8 6 = 192. We now find [JHO]. In ∆JHO, we have interior angle ∠O = 360 –
240 = 120. By Pythagorean Theorem on ∆JHN, JH = 16 3. Let P be the foot of the altitude from
O. Then since ∆JHO is isosceles (∆NJO is congruent to ∆NHO by SAS, so JO = HO), P bisects
JH and OP bisects ∠JOH, so JP = 8 3 and ∠JOP = 60. So OP = 8 and thus [JHO] = (1/
2) 16 3 8 = 64 3. Therefore [JOHN] = [JHN] – [JHO] = 192 − 64 3.
b) A right triangle has legs of length 3 and 4. A rectangle has two vertices on the triangle’s
hypotenuse and the remaining two vertices on the triangle’s legs. Find the maximal area of the
rectangle.
Answer: 3
Solution: The answer can be found using similar triangles or analytic geometry. We provide a
quicker solution. Reflect each of the three vertices of the triangle about the side of the rectangle it
is closest to. If not all three of the reflections coincide, the three smaller right triangles either
more than or less than completely cover the rectangle when rearranged. In the former/latter case,
it can be shown that the side parallel to the hypotenuse can be moved closer/farther (respectively)
to the hypotenuse to increase the area of the rectangle. Then the maximum area is attained when
all three of the reflections coincide, which is when the reflection of the vertex with the right angle
lies on the hypotenuse, which is when the side parallel to the hypotenuse spans the midpoints of
the legs. Then the area of the rectangle is half the area of the triangle, or 3.
Solution idea credit: Henry Maltby, Canyon Crest Academy (San Diego, CA)
Question 11
Find the sum of the answers to the two questions below.
a) Triangle MAX has MA = 11, AX = 13, and XM = 20.
The distances from P to MA and P to AX are both 3.
Find two times the distance from P to XM.
Answer: 7
General solution: Use areas.
Detailed solution: In this solution, [P] denotes the area of polygon P. Using Heron’s Formula,
[ABC] = 22 11 9 (2) = 66. But we also have [ABC] = [ABP] + [BCP] + [CAP] = (11)(3)/2
+ (13)(3)/2 + (20)(h)/2 = 66, where h is the desired distance. Solving for h, we get h = 7/2, so the
answer is 7.
b) Trapezoid ABCD has AB || CD and AB = 3, BC = DA = 13, and CD = 15. AC and BD intersect
at E. Let F be on AD such that EF ⊥ AD. Find 13(EF).
Answer: 30
General solution: Use areas.
Detailed solution: In this solution, [P] denotes the area of polygon P. It is easy to find that the
height of the trapezoid is 12. Let M and N be on AB and CD respectively such that MN is an
altitude of the trapezoid passing through E. Note that EM and EN are altitudes of ∆AEB and
∆DEC respectively. Let EM = x. Since ∆AEB ~ ∆DEC and CD/AB = 15/3 = 5, EN = 5x. EM +
EN = 12, so x + 5x = 12, so x = 2. Then EN = 10. [DEC] = (15)(10)/2 = 75 and [DAC] =
(15)(12)/2 = 90, so [DAE] = [DAC] – [DEC] = 15. Then we have [DAE] = (AD)(EF)/2 =
(13/2)(EF) = 15, which gives EF = 30/13, so 13(EF) = 30.
Question 12
Find the sum of the areas of the two convex quadrilaterals described below.
a) ABCD, where AB = 6, BC = 6, CD = 5, DA = 5, and BD = 8.
Answer: 15 + 8 5
Solution: It can be shown with congruent triangles that the diagonals are perpendicular and XB =
XD. (This quadrilateral is a kite.) Let the intersection of the diagonals be X. Then we have XB =
XD = 4. By the Pythagorean Theorem, XC = 3 and XA = 2 5. We can then add up the areas of
the four right triangles to get the total area of 15 + 8 5.
b) EFGH, where EF = FG = GH = HE = 4 and EG = 8
Answer: 16 5
Solution: EFG and EHG are congruent, so we can find the area of one of them and double it to
get the total area. The altitude of isosceles EFG from F splits the base into two equal segments, so
the length of this altitude is 6! − 4! = 2 5. The area of EFG is then 8 5, so the total area is
16 5.
Geometry Team Answer Sheet:
1. 31
2. 10 3 + 162
3. 4
4. 1234
5. 56
6. 26 – 3 2
7. 176
8. 16/3
9. 113
10. 195 − 64 3
11. 37
12. 15 + 24 5