Properties of figures, and mathematics in human activity (Mathematics A)

By the angle bisector property, BD:DC=AB:AC=6:4=3:2. Hence BD=BC×3/5=5×3/5=3. Getting the ratio the wrong way round as BD:DC=4:6 gives BD=2, which is in fact the length of DC. The segment BD, nearer to B, corresponds to the side AB that meets B.

The centroid divides a median in the ratio 2:1 measured from the vertex, so AG:GM=2:1 and AG=9×2/3=6. The value 3 is the length of GM, and 9/2 comes from confusing the centroid with the midpoint and halving.

The point where the three perpendicular bisectors meet is the circumcentre. A point on a perpendicular bisector is equidistant from the two ends of the segment, so the circumcentre is equidistant from the three vertices and is the centre of the circumscribed circle. The incentre comes from the internal angle bisectors, the centroid from the medians, and the orthocentre from the perpendiculars dropped from the vertices to the opposite sides.

The point where the internal angle bisectors meet is the incentre. A point on an angle bisector is equidistant from the two sides of the angle, so the incentre is equidistant from the three sides and is the centre of the inscribed circle. The point equidistant from the three vertices is the circumcentre, which is easy to confuse, so learn incentre, inscribed circle and equidistant from the sides as one package.

The median length theorem gives AB²+AC²=2(AM²+BM²). With BM=4 we get 49+25=2(AM²+16), so 74=2AM²+32, AM²=21 and AM=√21. Forgetting the factor 2 on the right, and so failing to divide, turns 74−16=58 into √58. Use BM=4, half of BC.

BD:DC=AB:AC=8:6=4:3, so DC=BC×3/7=7×3/7=3. The value 4 is the length of BD, so check which segment is asked for. The value 7/2 comes from confusing D with the midpoint.

In a triangle, the larger the side, the larger the angle opposite it. The longest side is AB=7 and the angle opposite it is ∠C, so ∠C is the largest; in fact the law of cosines gives ∠C=120°. To avoid picking ∠A or ∠B, the angles at the ends of the longest side, always check which angle faces the side.

Ceva's theorem gives (AF/FB)×(BD/DC)×(CE/EA)=1. From (AF/FB)×(2/3)×(1/2)=1 we get AF/FB=3, so AF:FB=3:1. Reversing the direction of the ratios, which must follow one trip round from vertex to dividing point to vertex, gives 1:3, so write them starting from A and going once round.

∠IBC+∠ICB=(∠B+∠C)/2=(180°−70°)/2=55°, so ∠BIC=180°−55°=125°. By formula, ∠BIC=90°+∠A/2=90°+35°=125°. The value 140° is 2∠A, confusing this with the circumcentre formula ∠BOC=2∠A.

Since O is the centre of the circumscribed circle, ∠BOC is the central angle on the arc BC and ∠A is the inscribed angle on the same arc. Hence ∠BOC=2∠A=100°. The value 115° confuses this with the incentre formula 90°+∠A/2, and 25° halves instead of doubling.

An inscribed angle is half the central angle, so 120°÷2=60°. The value 240° doubles instead. Check on a diagram which way the relation inscribed angle × 2 = central angle goes.

The central angle on the diameter AB is 180°, so the inscribed angle ∠ACB is half of it, 90°, by Thales' theorem. It is the same wherever C lies. The value 180° simply repeats the central angle.

In a cyclic quadrilateral, opposite interior angles add to 180°, so ∠C=180°−95°=85°. The value 95° comes from the misunderstanding that opposite angles are equal, whereas they in fact add to 180°.

By the tangent-chord angle theorem, the angle ∠TAB between the tangent and the chord AB equals the inscribed angle ∠ACB on that chord, so it is 65°. The value 130° doubles it by confusing inscribed and central angles, and 115° subtracts from 180°.

The power of a point gives PA×PB=PC×PD, so 3×4=2×PD and PD=6. The value 5 comes from working with sums and differences such as 3+4−2. The power of a point always equates products of distances.

The tangent form of the power of a point gives PT²=PA×PB=4×9=36, so PT=6. Answering 36 forgets to take the square root. The value 13/2 is (4+9)/2 and comes from taking the average instead of the product.

PB=PA+AB=3+5=8. The power of a point gives PA×PB=PC×PD, so 3×8=4×PD and PD=6. Taking the 5 of AB as PB and writing 3×5=4×PD gives 15/4. For a secant from an external point, use the distance to the farther intersection.

The two tangents from an external point have equal length, so writing the tangent lengths from A, B and C as x, y and z gives x+y=9, y+z=10 and z+x=7. Adding all of them gives x+y+z=13, the semi-perimeter s, so AP=x=s−BC=13−10=3. The value 6 is BP, which is s−CA, and 4 is the tangent length from C, which is s−AB, so mind which vertex the tangent comes from.

The length of a common external tangent is √(d²−(r₁−r₂)²)=√(100−(4−2)²)=√96=4√6. The value 8=√(100−36) uses the sum of the radii, (4+2)², which gives the common internal tangent instead. Use the difference for an external tangent and the sum for an internal one.

Inscribed angles are proportional to their arcs, so the three interior angles divide 180° in the ratio 3:4:5, giving 45°, 60° and 75°. The largest is the angle ∠B on the largest arc CA, namely 75°. The value 90° confuses this with the side ratio 3:4:5 of a right triangle.

540=2×270=2×2×135=2²×135, and 135=27×5=3³×5, so 540=2²×3³×5. Checking, 4×27×5=540. The values 2³×3²×5=360 and 2²×3²×5=180 do not match. When an exponent looks wrong, multiply back to check.

12=2²×3 and 18=2×3². The least common multiple takes the larger exponent of each prime, giving 2²×3²=36. The value 6 is the greatest common divisor. Simply using the product, as in 216=12×18÷1, is wrong; from G×L=a×b you can also get L=12×18÷6=36. The value 72 is a common multiple of the two numbers but not the least one.

For two numbers a and b we have G×L=a×b, so 6×180=30×b and b=1080÷30=36. Checking, gcd(30,36)=6 and lcm(30,36)=180, which fits. The values 60 and 90 are easy to guess from divisor and multiple relations, but G×L=ab pins the answer down to one.

273=63×4+21 and 63=21×3+0. The remainder has reached 0, so the last divisor, 21, is the greatest common divisor. From 273=3×7×13 and 63=3²×7 we also get gcd=3×7=21. The values 7 and 3 are common divisors but not the greatest.

319=143×2+33, 143=33×4+11, and 33=11×3+0, so the greatest common divisor is 11. The value 33 is an intermediate remainder; the algorithm continues until the remainder is 0. From 319=11×29 and 143=11×13 we can also confirm it.

110101₂=1×32+1×16+0×8+1×4+0×2+1×1=32+16+4+1=53. The value 43 comes from applying the place values in reverse order, which is the same as reading 101011₂. Note that the leftmost digit is the highest place, worth 2⁵=32.

123÷5=24 remainder 3, 24÷5=4 remainder 4, and 4÷5=0 remainder 4. Reading the remainders from the bottom up gives 443₅. Checking, 4×25+4×5+3=123. Reading the remainders from the top down gives 344₅, the typical mistake.

In decimal, 1011₂=11 and 110₂=6, and the sum is 17=16+1=10001₂. Working column by column in binary and tracking the carry 1+1=10 also gives 10001₂. Dropping one carry produces values such as 1111₂, which is 15, or 10011₂, which is 19.

From 3(x−2)+5(y+1)=0 we get 3(x−2)=−5(y+1). Since 3 and 5 are coprime, x−2 must be a multiple of 5, giving x=2+5k and y=−1−3k. The variable x picks up the other coefficient 5 and y picks up the other coefficient 3, with opposite signs. The pair x=2+3k, y=−1−5k works at k=0 but fails at k=1, where 3×5+5×(−6)=−15≠1, and x=2−5k, y=−1−3k also fails at k=1, where 3×(−3)+5×(−4)=−29≠1. Substituting to check is the sure way.

Writing the number as N gives N=7x+3=13y+5, hence 7x−13y=2. One solution is x=4, y=2, and then N=7×4+3=31. Checking, 31=7×4+3 and 31=13×2+5, which meets both conditions. Solutions recur every 7×13=91, so the next one is 122, which is not the smallest. The value 18=13+5 leaves remainder 4 on division by 7, so it is wrong.