Download Geometry I Solutions to Section A of the mock exam

Geometry I
Solutions to Section A of the mock exam
A1. `, m are parallel means that either they coincide or they do not intersect.
If ` meets n in a point P then both ` and n are lines parallel to m passing through
P . So they are equal. In either case, they are parallel!
A2. The computer’s solution:
For (d), it was not necessary to know the precise orientation of the ellipse.
A3. Let h = g −1 ◦ f . Then h(P ) = P if and only if f (P ) = g(P ). So h has a unique
fixed point and must be (a) rotation.
A4. The centroid is the intersection of the medians. The circumcentre is the centre
of the circle passing through the vertices (or the intersection of the perpendicular
bisectors of the sides). The orthocentre is the intersection of the altitudes.
(a) 13 (z1 + z2 + z3 )
(b) z1 + z2 + z3
A5. We substitute y = x + 1 and see which map becomes the identity. The answer
is (c).
On the x-axis, the number 7 (for example) moves to 8 units left of the mirror and
becomes −9. This way, we see that the reflection is (x, y) 7→ (−x − 2, y).
B6. If the point is P and the line `, drop a perpendicular from P to `. The distance
is |P D| where D is the foot of the perpendicular on `.
1. If the feet of the two perpendiculars are A and B then right-angled triangles
4OP A and 4OP B have two angles in common. They are therefore similar and
(having a common side) congruent. Therefore |P A| = |P B|.
2. With the same notation, the two triangles have all their sides equal, since the
third is determined by Pythagoras’ theorem. They are therefore similar (by SAS or
SSS), and ∠P OA = ∠P OB. It follows that OP bisects the angle and P ∈ s.
B7. 1. Use the method of q4 on Sheet 3.
2. Construct the line through H that is parallel to AG and extend BA to meet it.
Then by similar triangles, we see that
|BH|
|BO|
=
.
p
p+q
Using Pythagoras, and (1) to replace p + q by pq , we get
p
p p
1 + p2 ,
|BH| = q 1 + p2 =
p−1
the last equality by (1) again.
B8. All three parts can be done using appropriate right-angled triangles:
√
1. r = 21 tan(π/6) = 1/(2 3).
√
2. r cos(π/6) = 12 so r = 1/ 3.
3. ∠BAC = π/2 by Thales, so 2r sin θ = r and θ = π/6.
B9. 1. We do get sin A = sin B = sin C. But the question should have stated that all
angles are no greater than π/2, in which case the result is now obvious. Otherwise,
one has to appeal to the cosine rule.
2. It is
a
b
c
=
=
,
sin A
sin B
sin C
obtained by using the first term only in the series expansion sin x = x − 61 x3 + · · ·
about x = 0 (with x equal to a, b, c in turn).
3. Each equality is proved by constructing an altitude of the triangle and applying
elementary trigonometry to the two resulting triangles.