Student: Billy Dorminy
Username: solafidefarms
ID#: 2112
USA Mathematical Talent Search
Year
17
Round
4
Problem
1
Let angle BPA be x degrees, let the radius of C1 be r, and let the point of tangency of C2 and
C3 be T .
We know that OB=OP since they are both radii of C1 , thus OPB is isosceles. We can then
find that P B = 2r cos x. We know that angle BOP is 180 − 2x degrees (since the other 2 angles of
OBP are x degrees), thus POA is 2x degrees. OP and OA are both radii of C1 , thus P A = 2r sin x.
Since PB and PQ are both radii of C2 , P Q = 2r cos x.
Now TP is a radii of C2 also, and the distance from A to the foot of the altitude from O to AP is
half of the distance from A to P since AOP is isosceles, so the distance from the foot of the altitude
from O to AP to T is 2r cos x − r sin x. Therefore AT = 2r(sin x − cos x). Since AR and AT are
radii,
of AQR is equal
p they are equal, so AR = 2r(sin x−cos x). Since RAQ is right, the
√ hypotenuse
2
2
2
to (2r(sin x − cos x)) + (2r(sin x + cos x)) , which simplifies to 2r 2 since sin x + cos2 x = 1.
Page 1 of Problem 1
Student: Billy Dorminy
Username: solafidefarms
ID#: 2112
USA Mathematical Talent Search
Year
17
Round
4
Problem
1
It is well-known that the area of a triangle is equal to abc/4R, where R is the radius of the
circumcircle.. The circle C4 , which we are trying to find, goes through A, R, and Q, thus it must
necessarily be the circumcircle of ARQ.
√ These
√ But RAQ is right, thus the area is also AR · AQ/2.
formulas give the same area, thus 2 2r3 (sin2 − cos2 x) = R(4r2 (sin2 x − cos2 x) → R = 2r. The
ratio of the areas of the circles C1 and C4 will be the square of the ratios of their radii by a
well-known theorem, thus the ratio of the areas is 2 .
Page 2 of Problem 1
Student: Billy Dorminy
Username: solafidefarms
ID#: 2112
USA Mathematical Talent Search
Year
17
Round
4
Problem
2
We know that since 1, a, b is an arithmetic sequence, a − 1 = b − a, thus b = 2a − 1. Similarly,
we know that 1c = dc since 1, c, d is a geometric sequence, thus d = c2 . We have now
3a − 1 = c + c2
We observe that between the nth hexagonal number, with sides of length n, and the n + 1th
hexagonal number, with sides of length n + 1, there is an extra layer of dots. In fact, this layer of
dots is composed of (n − 1)6 dots. Thus the formula for the number of dots in the nth hexagonal
number is equal to 1 + 6(1 + 2 + 3 + · · · + n) = 1 + 3(n)(n − 1).
When we substitute this in for a, we have 9(n)(n − 1) + 2 = c(c + 1). This factors to (3n −
2)(3n − 1) = c(c + 1), which indeed is true with c = 3n − 2. Thus any centered hexagonal number
is a possible value for a.
Now when we examine our equation 3a − 1 = c(c + 1) in modulus 3, we find that c ≡ 1 (mod 3),
thus c = 3x + 1. Our equation is now 3a − 1 = (3x + 1)(3x + 2) → 9x2 + 9x + 3 = 3a → a =
3x2 + 3x + 1 → a = 1 + 3(x)(x + 1). We can choose an n such that n = x + 1, thus our equation is
a = 1 + 3(n − 1)(n). Since a hexagonal number can be represented by 1 + 3(n)(n − 1), and every
number in the form 1 + 3(n)(n − 1) is hexagonal, all values of a are hexagonal.
W5
Page 1 of Problem 2
Student: Billy Dorminy
Username: solafidefarms
ID#: 2112
USA Mathematical Talent Search
Year
17
Round
4
Problem
3
Let us assume for some level the pot is at 100x and you have n strikes. If you wait to call until 2
more flips have gone by, you can only expect 100(x+2)
+ 34 (expected value if you have one fewer strike)
4
(since there is only a 1/4 chance of not getting a strike in that time). But if you wait only 1 more
turn, we can expect 100x+100
+ 21 (expected value if one fewer strike).
2
If you have 2 strikes and the pot is at 0, it is best to wait: you can only get 0 dollars if you
call at that time, while if you wait there is a 1/2 chance of getting 100 and 1/2 chance of getting
0, for an expected value of 50 dollars if you wait. If you flip successfully though and the pot is 100
dollars, if you get it immediately you get 100, while if you wait till next time you get 200/2+0=100,
so it is equally successful to pull on the 2nd or 3rd. There is a 1/4th chance of getting to the 2nd
round and getting 200 dollars, so the expected value of this round is 50. Thus if you have 2 strikes,
you can expect 50 dollars.
If you have 1 strike and the pot is 0, it is again best to wait. If you are at the 2nd flip with a
pot of 100, you have a 100 return if you end immediately, while if you flip and succeed you get 200
or if you lose you get 50, for expected value of 125. If you flip successfully, the pot is 200; if you
pull now you get 200, as opposed to 200 if you wait, so you should end the game now. Thus you
have a 1/4 chance of ending the game when you have 1 strike, and you will get an average of 50
dollars again, but the other 3/4 chance will get 50, for a total expected value of 175/2.
If you have 0 strikes, wait at least till the pot is 100. If you pull immediately you get 100; if you
wait you get 100+175/4, so you should wait. If you are at the 2rd round, you get 200 by ending
immediately; if you wait another round you get 150+175/8, so you should end now. Thus you get
an average of 50 from the first strike with a probability of 1/4 and an average of 175/2 from the
925
second and third strikes with a probability of 3/4, for a total expected value of
.
8
Page 1 of Problem 3
Student: Billy Dorminy
Username: solafidefarms
ID#: 2112
USA Mathematical Talent Search
Year
17
Round
4
Problem
4
Let x = a + bi, where a and b are real. We are given that x3 − 6x and x4 − 8x2 are both
rational. When these expressions are expanded, we have −b3 i − 3ab2 + 3a2 bi − 6bi + a3 − 6a and
b4 − 4ab3 i − 6a2 b2 + 8b2 + 4a3 bi − 16abi + a4 − 8a2 rational, thus the coefficient of i in each of these
expressions is 0. This gives us the system of equations
−b3 + 3a2 b − 6b = 0
and
−ab3 + a3 bi − 4abi = 0
.
These equations may be factored: the first is equal to b(−b2 + 3a2 − 6) = 0, but since b is
non-zero (since x is irrational and b is the irrational part of x) the equation is
b2 − 3a2 + 6 = 0
The other equation may be factored to ab(−b2 + a2 − 4) = 0, thus either a = 0 or b2 − a2 + 4 = 0.
If a = 0, by the first equation b2 + 6 = 0, which is impossible, since x2 ≥ 0. Thus we have the
system of equations
b2 − 3a2 + 6 = 0
b2 − a2 + 4 = 0
When we subtract the first equation from the second, we learn that 2a2 − 2 = 0, thus a2 = 1,
so a ∈ {−1, 1}.
No matter what a is, a2 = 1, thus we can solve for b. We have the equation b2 − 3a2 + 6 = 0 →
b2 + 3 = 0. But b2 ≥ 0, thus there are no solutions for x.
Page 1 of Problem 4
Student: Billy Dorminy
Username: solafidefarms
ID#: 2112
USA Mathematical Talent Search
Year
17
Round
4
Problem
5
It is well-known that the intersection of a plane and a cone such that the plane passes through
the cone completely results in an ellipse, whose area is abπ. (a and b are the lengths of the longest
and shortest distances between the endpoints of lines whose endpoints are on the ellipse and which
pass through the center of the ellipse.)
We firstly consider the vertical cross-section of the cone such that X lies on the cross-section.
Let G and F be the points of tangency of the sides of the cone and the sphere on the cross-section.
Let K be the point of intersection of P and C on the cross-section, and let J be the point of
tangency of B and KX. Also let the midpoint of KX be I, and let the intersection of AB and
KX be H. Finally, let a line parallel to the base of C and passing through B have its’ intersection
with the sides of C labeled D and E. (See above diagram)
(1) It is well-known that a line from the center of a circle to its point of tangency to another
line is perpendicular to the tangent line.
Also, since the cone is isosceles, the center of the inscribed sphere is on the median of the cone.
Thus the angles at B toward the base of the cone are 90 degrees each. Since ED is parallel to the
base of the cone and have the same vertex A, EDA is similar to the whole cone. Since the height
of the cone is the same as it’s width, DB = BE = BA. Thus angles AEB and ADB are both
equal to 45 degrees.
By (1), since G is the point of tangency of B and AE, angle BGE is 90 degrees, thus angle
GBE is 45 degrees. Similarly the measure of angle F BD is also 45 degrees. Thus since EBD is a
straight line F BG is 90 degrees.
Since BG is a radius of B and meets
√ line EGA at a right angle, BGA is a 45-45-90 triangle,
√
just like the cone, and thus AB = BG 2. We know that the total height of the cone is 12 + 12 2,
thus the radius of B is equal to 12.
Page 1 of Problem 5
Student: Billy Dorminy
Username: solafidefarms
ID#: 2112
USA Mathematical Talent Search
Year
17
Round
4
Problem
5
The Power of a Point theorem states that for a convex polygon which can have an inscribed
circle, the distance from a vertex to the point of tangency of one of the lines having an endpoint
at that vertex and the circle is the same as the distance from that vertex to the point of tangency
of the other line ending at that vertex and the circle. Thus AF = AG and, since XK is tangent to
B, F X = XJ and JK = KG.
Since F X = XJ, F B and BJ are both radii, and XB
√ is shared, XF B ≡ XJB. Similarly
JKB ≡ GKB. XF = 6, thus AX = XJ = 6 and XB = 6 5.
Let angle XBF be x degrees, where x = arctan 1/2 and 0 < x < 90. Now since XF B ≡ XJB
angle XBJ is also x degrees. Since F BG is 90 degrees, KJB ≡ KGB, and XBF + XBJ +
KBJ + KBG = F BG, angles KBJ and KBG are equal to 45 − x degrees. BJ is 12, thus
tan x+tan y
KJ = 12 tan 45 − x. By the tangent angle-addition formula, tan(x + y) = 1−tan
x tan y , thus KJ =
tan 45 − tan x
. We know tan x = 1/2 and tan 45 = 1, thus KJ = 4. Also, since their respective
12
1 + tan 45 tan x
triangles are congruent, KG = 4. Thus KX - the longest diameter of the ellipse - has length 10.
Since I is the midpoint of KX, KI = XI = 5. It is well-known that the shortest radius of an
ellipse is perpindicular to the longest radius of an ellipse, thus we wish to find the radius of the
horizontal cross-section of the cone which passes through the widest section of the ellipse. Thus
we wish to find the distance from I to DE, which can be done by finding the cosine of angle IBH
and the length of IB.
We know that XI = IK, thus BI is a median of XKB. The formula for the length of a median
p
√
2(a2 + b2 ) − c2
is
, thus IB = 145. Now we know that JB = 12, since JB is a radius of B, and
2
.
we have IB, thus we can find cos IBJ = √12
145
The angle which we desire to find the cosine of is 45 − IBJ − JBK − KBG; the cosine of this
is equal to, by the cosine addition formulas, cos(45) cos(IBJ + JBK + KBG) + sin(45) sin(IBJ +
cos(IBJ + 2KBG) + sin(IBJ + 2KBG)
√
JBK + KBG). This is equal to
, which is equal to
2
cos IBJ cos 2KBG − sin IBJ sin 2KBG + sin IBJ cos 2KBG + sin 2KBG cos IBJ
√
2
12
1
13 cos 2KBG + 11 sin 2KBG)
√ √
, and sin IBJ = √
, thus our equation is equal to
.
cos IBJ is equal to √
145
145
2 145
13 − 26 sin2 KBG + 22 sin KBG cos KBG)
√ √
By the double-angle formulae, this is equal to
, and
2 145
1
3
17
sin KBG = √ , cos KBG = √ , thus cos IBH = √ √
. Since our hypotenuse is equal to
10
10
2 145
√
17
145, the distance from I to DBE is √ .
2
√
By the Pythagorean Theorem, we find that the shortest distance from AB to I is 22 . Thus the
horizontal cross-section looks something like the following:
Page 2 of Problem 5
Student: Billy Dorminy
Username: solafidefarms
ID#: 2112
USA Mathematical Talent Search
Year
17
Round
4
Problem
5
LIM is the shortest radius of the ellipse, and is a chord of circle O. Also, OL and OM , the
lines from the center of the circle to the endpoints of the shortest diameter, are radii of O.
Now since the part of the cone above this new cross-section is similar to the whole cone (since
it is a horizontal cross-section), the radius of the new cone will be the same as the height. We can
√
find the height to be AB less the shortest distance
from I to DBE. We know that AB = 12 2
√
and the shortest distance from I to DBE is 172 2 , thus the height and thus the radius of circle O
are
√
7 2
2 .
Similarly, the distance OI is equal to the shortest distance from AB to I, namely √12 , thus we
√
6 by the Pythagorean Theorem. Thus the shortest radius of
can find the distance
IM
=
IN
=
2
√
the ellipse is 2 6, and the area of the ellipse - the part of the plane inside the cone - is
√
10 6π
Page 3 of Problem 5