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Advanced Geometry
Adithya B., Brian L., William W., Daniel X.
8/19
§1 Introduction
In recent years, the AIME has included advanced geometry problems of an “olympiad
flavor.” These problems, while not necessarily requiring the advanced geometric techniques
of olympiads, are qualitatively different from other geometry problems. There will be
a lot of geometric work at the beginning of the problem, with the (oftentimes much
simpler) computation coming at the end. In these problems, it is especially important to
make geometric observations before diving into calculation. We’ll discuss two methods of
making these observations that also see common usage in olympiad geometry, namely
angle chasing and radical axes.
§2 Angle Chasing
To angle chase, it is useful to know a few useful theorems.
Theorem 2.1 (Inscribed Angle Theorem)
Let A, B, and C be inscribed in a circle O. Then, ∠AOB = 2∠ACB. Also, angles
that subtend the same arc are equal. That is, we have ∠ACB = ∠ADB in the
diagram below.
C D
O
B
A
Proof. Let ∠ACO = α and ∠BCO = β. Then, since OA = OC, ∠OAC = α and
◦ ◦
∠AOC=180 −2α. Similarly, ∠BOC =180 −2β. Therefore, we get,
◦
∠AOB=360 −∠AOC−∠BOC=2(α+β)=2∠ACB.
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Adithya B., Brian L., William W., Daniel X. (8/19) Advanced Geometry
The second part of the theorem follows directly from the fact that ∠ACB is independent
of the position of C.
Remark2.2. It should also be noted that by continuity this result is true even when D = A
or D = B. In that case, if TA is a tangent to circle O, we have ∠AOB = 2∠TAB and
∠TAB=∠ACB. Thelatter result is quite important in problems involving tangents to
circles and inscribed angles.
Now, with that in mind, we will move onto the most important concept in angle
chasing: cyclic quadrilaterals.
Theorem 2.3 (cyclic quadrilateral)
Let ABCD be a convex quadrilateral. Then, if the points A,B,C,D all lie on one
circle, we say ABCD is cyclic. The following three statements are equivalent:
1. ABCD is cyclic
◦
2. ∠ABC+∠CDA=180
3. ∠ABD=∠ACD
B
A
C
D
Proof. The forward direction follows from the Inscribed Angle Theorem, but the converse
requires more care.
It cannot be stressed enough how important cyclic quadrilaterals are. They will
show up in many of the examples below, and they will help to deduce important angle
relationships to solve the problems.
Sometimes, cyclic quadrilaterals may be given in the problem when there are 4 points
on the same circle. In this case, we can use the angle properties of cyclic quadrilaterals
to make other deductions. For example, cyclic quadrilaterals may lead to an important
pair of similar triangles, which would help to finish the problem.
Other times, we can prove that four points in our diagram is cyclic from one method,
and then use the resulting angle properties to make other deductions about the diagram.
Example 2.4 (2016 PUMaC Geometry #7)
Let ABCDbeacyclicquadrilateral with circumcircle ω and let AC and BD intersect
at X. Let the line through A parallel to BD intersect line CD at E and ω at Y 6= A.
If AB = 10,AD = 24,XA = 17, and XB = 21, then the area of △DEY can be
written in simplest form as m. Find m+n.
n
2
Adithya B., Brian L., William W., Daniel X. (8/19) Advanced Geometry
B
A
Y
X
C
D
E
Solution. First, using the parallel condition, we can find ∠DEY = ∠CDB = ∠XAB.
Also, ∠AYD =∠ABD, so DEY ∼XAB.
Now, AY k BD, so AYBD is an isosceles trapezoid. So, DY = AB = 10, and we
have
10 2 100 400
[DEY]= 21 [AXB]= 441 ·84= 21
So, our answer is 400 +21 = 421 .
Example 2.5 (2019 AIME I #13)
Triangle ABC has side lengths AB = 4, BC = 5, and CA = 6. Points D and E
are on ray AB with AB < AD < AE. The point F 6= C is a point of intersection
of the circumcircles of △ACD and △EBC satisfying DF = 2 and EF = 7. Then
√
BEcanbeexpressed as a+b c, where a, b, c, and d are positive integers such that a
d
and d are relatively prime, and c is not divisible by the square of any prime. Find
a+b+c+d.
A
C
B
X F
D
E
Solution. Notice that ∠BEF = ∠BCF and∠ADF = ∠ACF. Thus,∠DFE =∠ADF−
∠BEF =∠ACF−∠BCF =∠ACB. Now,byLawofCosines,wecanfindcos∠ACB =
3
Adithya B., Brian L., William W., Daniel X. (8/19) Advanced Geometry
2 2 2 q √
5 +6 −4 3 2 2 3
2·5·6 = 4. Now, we can find DE = 2 +7 −2·2·7· 4 =4 2. Now, let x = BX,
y = DX. Note that we have ACX ∼ FDX, so 3 = AC = AX, so FX = x+4. Now,
7DF EFFX FX 3 7
note that we also have BCX ∼ FEX, so we have 5 = BC = BX, so FX = 5x as
well. Setting these two values equal gives x = 5. Now, note that, by Power of a
4 √
Point, AX · DX = CX ·FX = BX ·EX, so (x+4)y = x(y +4 2), so we can find
√ √ √ √
y = x 2 = 5 2. Now, we can find the value of BE to be x+y +4 2 = 5+21 2, so our
4 4
answer is 5 +21+2+4 = 32 .
Example 2.6 (2019 AIME II #11)
Triangle ABC has side lengths AB = 7,BC = 8, and CA = 9. Circle ω passes
1
through B and is tangent to line AC at A. Circle ω2 passes through C and is tangent
to line AB at A. Let K be the intersection of circles ω1 and ω2 not equal to A. Then
AK=m,wheremandnarerelatively prime positive integers. Find m+n.
n
A
ω2
7 9
K
ω1
B 8 C
Solution. The tangency condition yields ∠ABK = ∠KAC and ∠KAB = ∠ACK. There-
fore, △BKA ∼ △AKC. Since AB = 7, the ratio between the two triangles is 7. Let
AK 7 AC 9 9 9
AK=x. Then, KC = 9, so KC = 7x. Now, we can also note that
◦ ◦ ◦
∠AKC=180 −∠KAC−∠KCA=180 −∠KAC−∠BAK=180 −∠A.
Therefore, we can apply the Law of Cosines on AKC to solve for x. First, we can
determine cosA from the Law of Cosines on △ABC. Note that,
2 2 2
cosA= 7 +9 −8 = 11.
2·7·9 21
Therefore, cos∠AKC = −11. From the Law of Cosines on AKC, we find,
21
2 81 2 9 11
x +49x +2x· 7x· 21 =81.
2 81 2 66 2 2
x +49x +49x =4x =81.
Solving yields x = 9, so the answer is 9 + 2 = 011 .
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