100GeometryProblems: Solutions - Art of Problem Solving

100 Geometry Problems: Solutions

Alvin Zou

April 26th, 2015

1. Let ra , rb , rc be the radii of the circles centered at A, B, C respectively. We then have the following system of

equations

?

?

?BC = 9 = rb + rc

AX = ra = 6 + rb

?

?

AY = ra = 5 + rc .

Solving yields rb = 4, rc = 5, and AX = ra = 10 .

2. Denote ¡ÏBAC = ¦Á and ¡ÏBAD = ¦Õ, then ¦Õ is our unknown. By AC = AD we get ¡ÏDAC = ¡ÏCAD = ¦Á ? ¦Õ

and thus ¡ÏACD = 180? ? 2(¦Á ? ¦Õ). By that we can now get ¡ÏABC = 180? ? (180? ? 2(¦Á ? ¦Õ)) ? ¦Á = ¦Á ? 2¦Õ.

And now using the condition we get

30? = ¡ÏCAB ? ¡ÏABC = ¦Á ? (¦Á ? 2¦Õ) = 2¦Õ ? ¦Õ = 15? .

3. We can draw the perpendicular from the center of the semicircle to CE (call this point X). Notice that the

radius of the semicircle is 1. Since ABCD is a square, BC and EA are tangents as well. We have

CX = BC = 2 and EX = EA. We can Pythagorean Theorem ¡÷EDC with legs DE = 2 ? EA and DC = 2

5

.

with a hypotenuse CE = 2 + EA. Solving, we get EA = 21 and CE =

2

4. The diagram:

A

M

6

B

6

D

3

C

It is given that ¡ÏAM D ¡«

= ¡ÏCM D. Since ¡ÏAM D and ¡ÏCDM are alternate interior angles and AB k DC,

¡ÏAM D ¡«

= ¡ÏCDM ?¡ú ¡ÏCM D ¡«

= ¡ÏCDM . Use the Base Angle Theorem to show DC ¡«

= M C. We know that

ABCD is a rectangle, so it follows that M C = 6. We notice that ¡÷BM C is a 30 ? 60 ? 90 triangle, and

¡ÏBM C = 30? . If we let x be the measure of ¡ÏAM D, then

2x + 30 = 180

2x = 150

x = 75

1

5. From F draw a perpendicular to AD and let the foot be G. Then, we quickly notice that triangles ABE,

EF G, GF D are congruent because of 2 equal sides and the right angle. (SsA) Let the side length of the

square be a then by the Pythagorean Theorem we get

1

302 ? ( a)2 =

3

a2

900 =

10 2

a

9

a2 = 810

6. First, we draw the diagram:

A

E

D

B

C

Since ¡ÏDEC = ¡ÏDBC = 90, quadrilateral DEBC is cyclic, from which the result follows due to same

inscribed arcs. Q.E.D.

7. Let ¡÷ABC be an equilateral triangle with side length a, and point D be on line AC such that

CD = AC = a, and point A and D are distinct. ¡ÏBCD = 120, and BC = CD, so ¡÷BCD

¡Ì is isosceles. Also,

¡ÏCBD = ¡ÏCBD = 30, so ¡ÏABD = 90. Thus, ¡÷ABD is a 30-60-90 triangle, so b = a 3, and

¡Ì

¡Ì

a 3

b

3.

a = a =

8. Drawing the diagram, we get:

C

L

M

N

A

B

Since quadrilateral ABLM is cyclic, ¡ÏM LB = 180 ? ¡ÏA = 90? = ¡ÏM LC. Thus ¡ÏCM L = 90 ? ¡ÏC. We also

have ¡ÏC = ¡ÏAN M since ALCN is cyclic. Then since ¡ÏM AN = 180 ? ¡ÏA = 90? , we get

¡ÏAM N = 90 ? ¡ÏAN M = 90 ? ¡ÏC = ¡ÏCM L.

Since the vertical angles are congruent, L, M, N are collinear. Q.E.D.

2

9. First, a diagram:

A

E

I

B

X

D

Y

C

Let D and E be the feet of the altitudes from I to BC and AB respectively. Since IX = IA and ID = IE we

have that ¡÷IDX ¡«

= ¡÷IEA by HL congruence (both are clearly right triangles). It is also easy to show that

¡÷IDX ¡«

= ¡÷IDY. So, AE = s ? a = XD = 12 XY, and thus XY = b + c ? a = 1400 + 1800 ? 2014 = 1186 .

10. Notice that, since ADBE is cyclic, we want to show that it is an isosceles trapezoid. Thus, it suffices to prove

that arc DB = arc AE. However, we have arc AB = arc AC and arc AD = arc CE (since AD = CE). Thus,

we have arc DB = arc AB - arc AD = arc AC - arc CE = arc AE. Q.E.D.

11. Let P be the shape¡¯s perimeter and A its area. Note that if we dilate the shape by a factor of k, its perimeter

becomes kP and its are becomes k 2 A. Thus, if we want our dilated shape to equiable, or kP = k 2 A, we

should dilate the shape by a factor of k = P

A . Q.E.D.

12. By simple angle chasing, we notice that triangle AEB is similar to EF C. Let the side length of the square be

a, EC = x and thus BE = a ? x. Because of similarity, it is

x

3

a

= ? x = a.

4

3

4

That yields BE = 14 a. Using the Pythagorean Theorem in triangle AEB the result is

1

17 2

162

a2 + ( a) = 16 ?

a = 16 ? a2 =

4

16

17

13. ¡ÏABN = ¡ÏAM N since they are subtended from the same chord. Since ¡ÏM XN = ¡ÏM Y N = 90? , it follows

that M , X, Y , and N are concyclic. Then, since M and X form angles subtended from the same chord,

¡ÏY XN = ¡ÏY M N = ¡ÏAM N = ¡ÏABN . Because BN is an extension of XN , and because

¡ÏY XN = ¡ÏABN , it follows that AB k XY . Q.E.D.

14. Solution 1: Let M be the intersection point of the diagonals of the square. Then reflect A, B about M and

we¡¯ll get C, D. Thus, the reflection of E about M which we call E ¡ä will give us DE ¡ä = 5 and E ¡ä C = 12, E ¡ä

must be F ! Thus

EM = M F.

Now notice that

(5, 12, 13) is a pythagorean triple which yields ¡ÏBEA = 90? . Obviously, ¡ÏAM B = 90? and

¡Ì

AM = M B = 2¡¤13

2 . Then AEBM is a cyclic quadrilateral and by Ptolemy we get

¡Ì

¡Ì

¡Ì

2 ¡¤ 13

2 ¡¤ 13

2 ¡¤ 17

¡¤5+

¡¤ 12 = 13 ¡¤ M E ? M E =

.

2

2

2

¡Ì

Therefore EF = 2 ¡¤ M E = 2 ¡¤ 17, so

¡Ì

EF 2 = ( 2 ¡¤ 17)2

= 2 ¡¤ 289

= 578 .

3

Solution 2:

E

A

B

D

C

F

Extend F C, EB, EA, F D as shown until the sides intersect. The two intersection points as well as point E

¡«

and F form a square,

¡Ì 2 as the newly formed triangles are congruent to triangles AEB and CF D by ASA =. So

2

then EF = (17 2) = 578 .

15. Let O be ¡÷ABC¡¯s circumcenter. Note that ¡ÏBF E = 12 ¡ÏBOE = 14 ¡ÏBOC = 21 ¡ÏA. Similarly, ¡ÏAEF = 12 ¡ÏB

and ¡ÏDF B = 21 ¡ÏC. Since ¡ÏDF E + ¡ÏAEF = ¡ÏDF B + ¡ÏBF E + ¡ÏAEF = 21 (¡ÏA + ¡ÏB + ¡ÏC) = 90? , it

follows that DF ¡Í AE. Q.E.D.

16. Call P the projection of D onto AB, and let Dp be the projection of D onto plane ABC. Since AB = 3, we

1

?

have DP = 2 ¡¤ 12

3 = 8. By definition, ¡ÏDDp P = 90 so DDp P is a 30-60-90 triangle. Then DDp = 2 DP = 4,

so

[ABC] ¡¤ DDp

= (15 ¡¤ 4)/3 = 20 .

VABCD =

3

4

17. Here¡¯s a colorful diagram:

P1

A3

O

P4

P2

P3

A1

Since the quadrilateral is orthodiagonal, we can write the statement we want to show as D2 = P1 P22 + P3 P42 .

Let A1 and A3 be the antipodes of P1 and P3 , respectively, and let O be the center of the circle. The claim is

that reflecting P1 P2 P3 P4 over the diameter of the circle that is parallel to P1 P3 results in the new

quadrilateral A3 P4 A1 P2 . We can prove this by noting that P1 OP3 ¡«

= A3 OA1 , so P1 goes to A3 and P3 goes to

A1 (and obviously P2 and P4 switch places). Now, after reflecting, we get that A1 P2 = P3 P4 , so now we want

to show that D2 = P1 P22 + A1 P22 which is true by the Pythagorean Theorem on ¡÷P1 P2 A1 . Q.E.D.

18. Notice that the midpoint of AB, the centers of the two circles, and the center of the sphere form a rectangle.

Thus, if we know the sides of the rectangle we can compute the distance from the center of the sphere to the

midpoint of AB. Then, since this line is perpendicular to AB, we can then compute the ¡Ì

radius of the sphere.

542 ? 212 and

First,

we

find

the

distances

from

the

centers

of

the

circles

to

midpoint

of

AB.

They

are

¡Ì

2 ? 212 . Thus, the distance from the center of the sphere to the midpoint of AB is

66

¡Ì

54¡Ì2 ? 212 + 662 ? 212 . Finally, we apply pythagorean theorem once more to get the radius of the sphere to

be 542 ? 212 + 662 ? 212 + 212 . Squaring, we get that R2 = 6831 .

19. Extend AB and CD to meet at X. We see that ¡ÏAXD is a right angle, so XM and XN are medians of

triangle XBC and XAD, respectively. We can also see that X, M, N are collinear. Hence, XN = AN and

XM = BM , so we easily find that M N = XN ? XM = 1004 ? 500 = 504 .

20. First, we know that EF ||AB by symmetry. Now we wish to show that DF ||EF , as this would imply that

D, E, F collinear. Since line l is tangent to the circle, ¡ÏDBC = ¡ÏA, and therefore ¡ÏBCD = 90 ? ¡ÏA. Notice

that quadrilateral BDCF is cyclic (opposite angles 90). Thus, we have ¡ÏBCD = ¡ÏDF B = 90 ? ¡ÏA.

However, we also have ¡ÏEF B = ¡ÏF BA = 90 ? ¡ÏA by parallel lines, so lines DF and EF make the same

angle with BF . This implies that lines DF and EF are parallel, which implies D, E, F collinear. Q.E.D.

21. Let the center of the circle with radius 1 be A. Let the circle shaded grey have center B and let the circle

shaded black that is adjacent to that grey circle be C. Let r be the radius that we want to find. Clearly,

BC = 2r, AB = 4 ? r and AC = r + 1. We can also see that, from symmetry, ¡ÏBAC = 60? . From Law of

Cosines, we have

4r2 = (r + 1)2 + (4 ? r)2 ? 2(r + 1)(4 ? r) cos 60? )

This reduces down to r2 + 9r ? 13, and plugging into the quadratic formula:

¡Ì

?9 + 133

r=

2

We arrive at ?9 + 133 + 2 = 126 .

5

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