Solutions

HMMT 2014
Saturday 22 February 2014
Geometry
1. Let O1 and O2 be concentric circles with radii 4 and 6, respectively. A chord AB is drawn in O1 with
length 2. Extend AB to intersect O2 in points C and D. Find CD.
√
Answer: 2 21 Let O be the common center of O1 and O2 , and let M be the midpoint of AB.
√
√
2
2
Then
√ OM ⊥ AB,
√ so by the Pythagorean Theorem, OM = 4 − 1 = 15. Thus CD = 2CM =
2 62 − 15 = 2 21.
2. Point P and line ℓ are such that the distance from P to ℓ is 12. Given that T is a point on ℓ such that
P T = 13, find the radius of the circle passing through P and tangent to ℓ at T .
Answer: 169/24 Let O be the center of the given circle, Q be the foot of the altitude from P to
ℓ, and M be the midpoint of P T . Then since OM ⊥ P T and ∠OT P = ∠T P Q, ∆OM P ∼ ∆T QP .
13/2
169
Thus the OP = T P · PPM
Q = 13 · 12 = 24
3. ABC is a triangle such that BC = 10, CA = 12. Let M be the midpoint of side AC. Given that BM
is parallel to the external bisector of ∠A, find area of triangle ABC. (Lines AB and AC form two
angles, one of which is ∠BAC. The external bisector of ∠A is the line that bisects the other angle.)
√
Answer: 8 14 Since BM is parallel to the external bisector of ∠A = ∠BAM , it is perpendicular
to the angle bisector of ∠BAM . Thus BA = BM = 12 BC = 6. By Heron’s formula, the area of
p
√
∆ABC is therefore (14)(8)(4)(2) = 8 14.
4. In quadrilateral ABCD, ∠DAC = 98◦ , ∠DBC = 82◦ , ∠BCD = 70◦ , and BC = AD. Find ∠ACD.
B
A
82◦
98◦
C
D
Answer: 28 Let B ′ be the reflection of B across CD. Note that AD = BC, and ∠DAC+∠CB ′ D =
180◦ , so ACB ′ D is a cyclic trapezoid. Thus, ACB ′ D is an isosceles trapezoid, so ∠ACB ′ = 98◦ . Note
that ∠DCB ′ = ∠BCD = 70◦ , so ∠ACD = ∠ACB ′ − ∠DCB ′ = 98◦ − 70◦ = 28◦ .
5. Let C be a circle in the xy plane with radius 1 and center (0, 0, 0), and let P be a point in space
with coordinates (3, 4, 8). Find the largest possible radius of a sphere that is contained entirely in the
slanted cone with base C and vertex P .
√
Answer: 3 − 5 Consider the plane passing through P that is perpendicular to the plane of the
circle. The intersection of the plane with the cone and sphere is a cross section consisting of a circle
inscribed in a triangle with a vertex P . By symmetry, this circle is a great circle of the sphere, and
hence has the same radius. The other two vertices of the triangle are the points of intersection between
the plane and the unit circle, so the other two vertices are ( 53 , 54 , 0), (− 35 , − 54 , 0).
Using the formula A = rs
√ and using the distance formula to find the side lengths, we find that
2∗8 √
5.
=
=
3
−
r = 2A
2s
2+10+4 5
6. In quadrilateral ABCD, we have AB = 5, BC = 6, CD = 5, DA = 4, and ∠ABC = 90◦ . Let AC and
BE
.
BD meet at E. Compute ED
√
√
Answer:
3 We find that AC = 61, and applying the law of cosines to triangle ACD tells us that
√
BE
∠ADC = 120. Then ED
is the ratio of the areas of triangles ABC and ADC, which is (5)(6)√3 = 3.
(4)(5)
2
7. Triangle ABC has sides AB = 14, BC = 13, and CA = 15. It is inscribed in circle Γ, which has center
O. Let M be the midpoint of AB, let B ′ be the point on Γ diametrically opposite B, and let X be the
intersection of AO and M B ′ . Find the length of AX.
Answer:
′
65/12 Since B ′ B is a diameter, ∠B ′ AB = 90◦ , so B ′ A k OM , so
A
2
=B
OM = 2, so AX = 3 R, where R =
it all together gives AX = 65
12 .
AX
XO
abc
4A
=
(13)(14)(15)
4(84)
=
65
8
OM
B′ A
=
BM
BA
= 21 . Thus
is the circumradius of ABC. Putting
8. Let ABC be a triangle with sides AB = 6, BC = 10, and CA = 8. Let M and N be the midpoints of
BA and BC, respectively. Choose the point Y on ray CM so that the circumcircle of triangle AM Y
is tangent to AN . Find the area of triangle N AY .
Answer:
600/73 Let G = AN ∩ CM be the centroid of ABC. Then GA = 32 GN =
√
1
1
2
2
3 CM = 3 8 + 3 =
√
73
3 .
By power of a point, (GM )(GY ) = GA2 , so GY =
2
GA
GY
=
10
3 and
(10/3)2
√
73
3
GM =
=
100
√ .
3 73
Thus
[GAY ] [N AY ]
·
[GAM ] [GAY ]
1
GY N A
= [ABC] ·
·
6
GM GA
600
100 3
· =
=4·
73 2
73
[N AY ] = [GAM ] ·
9. Two circles are said to be orthogonal if they intersect in two points, and their tangents at either point of
intersection are perpendicular. Two circles ω1 and ω√
2 with radii 10 and 13, respectively, are externally
tangent at point P . Another circle ω3 with radius 2 2 passes through P and is orthogonal to both ω1
and ω2 . A fourth circle ω4 , orthogonal to ω3 , is externally tangent to ω1 and ω2 . Compute the radius
of ω4 .
Answer:
92
61
Let ωi have center Oi and radius ri . Since ω3 is orthogonal to ω1 , ω2 , ω4 , it has equal
r32
to each of them. Thus O3 is the radical center of ω1 , ω2 , ω4 , which is equidistant to the three
power
sides of △O1 O2 O4 and therefore its incenter.
For distinct i, j ∈ {1, 2, 4}, ωi ∩ ωj lies on the circles with diameters O3 Oi and O3 Oj , and hence ω3
130r4
1 r2 r4
= 23+r
=⇒ r4 = 92
itself. It follows that ω3 is the incircle of △O1 O2 O4 , so 8 = r32 = r1r+r
61 .
2 +r4
4
Comment: The condition P ∈ ω3 is unnecessary.
10. Let ABC be a triangle with AB = 13, BC = 14, and CA = 15. Let Γ be the circumcircle of ABC, let
d Circle ω1 is internally tangent to
O be its circumcenter, and let M be the midpoint of minor arc BC.
Γ at A, and circle ω2 , centered at M , is externally tangent to ω1 at a point T . Ray AT meets segment
BC at point S, such that BS − CS = 4/15. Find the radius of ω2 .
Answer:
1235/108
2
15 .
Let N be the midpoint of BC. Notice that BS − CS =
4
15
means that
NS =
let lines M N and AS meet at P , and let D be the foot of the altitude from A to BC.
2/15
SN
3
Then BD = 5 and AD = 12, so DN = 2 and DS = 32
15 . Thus N P = AD SD = 12 32/15 = 4 . Now
q
√
¡ 65 ¢2
(13)(14)(15)
27
OB 2 − BN 2 =
− 72 = 33
= 65
OB = R = abc
4A =
4(84)
8 , so ON =
8
8 . Thus OP = 8
and P M = OM − OP = 19
4 . By Monge’s theorem, the exsimilicenter of ω1 and Γ (which is A), the
insimilicenter of ω1 and ω2 (which is T ), and the insimilicenter of ω2 and Γ (call this P ′ ) are collinear.
But notice that this means P ′ = OM ∩ AT = P . From this we get
MP
38
radius of ω2
=
=
.
R
OP
27
Thus the radius of ω2 is
65
8
·
38
27
=
1235
108 .

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