Geometry: Power of a Point & Mass Points

Study Sheet

Geometry: Power of a Point & Mass Points

Circle power and weighted balancing

Power of a Point

Tip
The constant product
ABCDP2634
Chords through PP: 2⋅6=3⋅42 \cdot 6 = 3 \cdot 4.

From a point PP, along any line through PP meeting a circle at A,BA,B, the product PA⋅PBPA\cdot PB is constant; for a tangent of length tt, t2=PA⋅PBt^2 = PA\cdot PB.

In plain terms. No matter which line you draw through a point, the two distances to the circle multiply to the same number.

Example. Chords ABAB and CDCD cross at PP with PA=2,PB=6,PC=3PA=2, PB=6, PC=3: then PD=2⋅63=4PD = \dfrac{2\cdot 6}{3} = 4.

Why it works. For two chords ABAB and CDCD through PP, the inscribed angles ∠PAC\angle PAC and ∠PDB\angle PDB stand on the same arc CBCB, so they are equal; so are ∠PCA\angle PCA and ∠PBD\angle PBD. The triangles PACPAC and PDBPDB are therefore similar, and PAPD=PCPB\tfrac{PA}{PD} = \tfrac{PC}{PB} cross-multiplies to PA⋅PB=PC⋅PDPA \cdot PB = PC \cdot PD. The same similar-triangle argument works when PP is outside the circle, and a tangent is the limiting secant whose two intersection points have merged. The common value is ∣d2−r2∣|d^2 - r^2|, where dd is the distance from PP to the centre (take the line through the centre to see it).

Example
Worked: tangent and secant
PABTO61812
PT2=PA⋅PBPT^2 = PA \cdot PB: 122=6⋅2412^2 = 6 \cdot 24, with PBPB the whole secant.

From a point PP outside a circle, the tangent segment PTPT has length 1212. A secant from PP meets the circle first at AA and then at BB, with AB=18AB = 18. Find PAPA.

Spotting it. A tangent and a secant from the same point: PT2=PA⋅PBPT^2 = PA \cdot PB, where PBPB is the WHOLE secant, not the chord.

Step 1. Let PA=xPA = x, so PB=x+18PB = x + 18 and x(x+18)=144x(x + 18) = 144.

Step 2. x2+18x−144=0x^2 + 18x - 144 = 0 factors as (x+24)(x−6)=0(x + 24)(x - 6) = 0. A length is positive, so PA=6PA = 6 (and PB=24PB = 24).

Check: 6⋅24=144=1226 \cdot 24 = 144 = 12^2 ✓

Example
Worked: two chords, one unknown split

Chords ABAB and CDCD of a circle meet at PP, with AP=4AP = 4 and PB=9PB = 9. If CD=15CD = 15 and CP<PDCP < PD, find CPCP.

Spotting it. Power of a point gives the PRODUCT CP⋅PDCP \cdot PD and the problem gives the SUM CP+PDCP + PD, so CPCP and PDPD are the roots of a quadratic.

Step 1. CP⋅PD=4⋅9=36CP \cdot PD = 4 \cdot 9 = 36 and CP+PD=15CP + PD = 15.

Step 2. They are the roots of t2−15t+36=(t−3)(t−12)t^2 - 15t + 36 = (t - 3)(t - 12), so CP=3CP = 3 and PD=12PD = 12.

Example
Worked: a point inside, from the centre
OPAB538
Every chord through PP splits into pieces with product 72−52=247^2 - 5^2 = 24; here 3⋅83 \cdot 8.

A point PP lies 55 units from the centre of a circle of radius 77. A chord through PP is divided by PP into two pieces, one of length 33. How long is the chord?

Spotting it. No second chord is given, but the power of PP can be computed from the centre: use the chord through the centre, the diameter through PP.

Step 1. On that diameter the pieces are 7−5=27 - 5 = 2 and 7+5=127 + 5 = 12, so every chord through PP has pieces multiplying to 2⋅12=24=r2−d22 \cdot 12 = 24 = r^2 - d^2.

Step 2. The other piece is 243=8\tfrac{24}{3} = 8, so the chord is 3+8=113 + 8 = 11.

Side note
Where it goes wrong

Multiplying by the chord instead of the whole secant. In the tangent–secant example, writing PT2=PA⋅ABPT^2 = PA \cdot AB gives 144=18x144 = 18x, so x=8x = 8; but then PB=26PB = 26 and 8⋅26=208≠1448 \cdot 26 = 208 \ne 144. Both factors are measured FROM PP: to the near point and to the far point.

Try it
Try it: two secants

Two secants are drawn from an external point PP. One meets the circle at AA and BB with PA=5PA = 5 and AB=7AB = 7; the other meets it at CC and DD with PC=4PC = 4 (CC between PP and DD). Find CDCD.

Show the answer

PA⋅PB=5⋅12=60=PC⋅PD=4⋅PDPA \cdot PB = 5 \cdot 12 = 60 = PC \cdot PD = 4 \cdot PD, so PD=15PD = 15 and CD=PD−PC=11CD = PD - PC = 11.

Mass Points and Coordinates

Concept
Mass points

Hang weights at the vertices so each side balances at its division point (weight ×\times length equal on both sides); the weights reveal all segment ratios.

In plain terms. Treat the triangle like a mobile: put weights so it balances, and the balance conditions give the ratios you want.

Example. If a cevian splits a side 2:32:3, put weights 33 and 22 at its endpoints so 3⋅2=2⋅33\cdot 2 = 2\cdot 3 balances at the division point.

Why it works. A lever with masses mBm_B and mCm_C balances at the point DD with mB⋅BD=mC⋅DCm_B \cdot BD = m_C \cdot DC, which is exactly the statement BD:DC=mC:mBBD : DC = m_C : m_B: the HEAVIER end is nearer the fulcrum. Choose masses at the vertices so that two cevian feet are balance points. The centre of mass of the three masses then lies on both cevians, so it is their intersection PP; and on the cevian ADAD it balances mAm_A at AA against the total mB+mCm_B + m_C sitting at DD, which gives AP:PD=(mB+mC):mAAP : PD = (m_B + m_C) : m_A.

Example
Worked: two ratios from masses
A (3)B (2)C (3)D (5)E (6)P32
Masses in brackets: each labelled point balances its segment, and P balances both cevians.

In triangle ABCABC, DD lies on BCBC with BD:DC=3:2BD : DC = 3 : 2, and EE is the midpoint of ACAC. Segments ADAD and BEBE meet at PP. Find AP:PDAP : PD and BP:PEBP : PE.

Spotting it. Two cevians, ratios on the sides, and a ratio ALONG a cevian asked for: mass points.

Step 1. BD:DC=3:2BD : DC = 3 : 2 puts the heavier mass at CC: mB=2m_B = 2, mC=3m_C = 3, so DD carries 55.

Step 2. EE is a midpoint, so AA must match CC: mA=3m_A = 3, and EE carries 3+3=63 + 3 = 6.

Step 3. On ADAD: AP:PD=mD:mA=5:3AP : PD = m_D : m_A = 5 : 3. On BEBE: BP:PE=mE:mB=6:2=3:1BP : PE = m_E : m_B = 6 : 2 = 3 : 1.

Example
Worked: an area from the ratios

In the same triangle, suppose [ABC]=40[ABC] = 40. Find [BPD][BPD].

Spotting it. Areas of triangles with a common apex are in the ratio of their bases; chain two such steps using the ratios just found.

Step 1. Triangles ABDABD and ABCABC share apex AA and have bases BDBD and BCBC, so [ABD]=35⋅40=24[ABD] = \tfrac35 \cdot 40 = 24.

Step 2. Triangles BPDBPD and BADBAD share apex BB and have bases PDPD and ADAD on the same line. PD:AD=3:8PD : AD = 3 : 8, so [BPD]=38⋅24=9[BPD] = \tfrac38 \cdot 24 = 9.

Example
Worked: coordinates do the rest
ABCDEF326
The shaded triangle is the square minus three right triangles: 36−9−3−1236 - 9 - 3 - 12.

Square ABCDABCD has side 66. EE is the midpoint of BCBC and FF is on CDCD with CF=2CF = 2. Find the area of triangle AEFAEF.

Spotting it. A square with points given by distances along its sides: place it on axes and every point has coordinates at once.

Step 1. A=(0,0)A = (0, 0), B=(6,0)B = (6, 0), C=(6,6)C = (6, 6), D=(0,6)D = (0, 6), so E=(6,3)E = (6, 3) and F=(4,6)F = (4, 6).

Step 2. With one vertex at the origin, the area is 12∣xEyF−xFyE∣=12∣6⋅6−4⋅3∣=12⋅24=12\tfrac12|x_Ey_F - x_Fy_E| = \tfrac12|6 \cdot 6 - 4 \cdot 3| = \tfrac12 \cdot 24 = 12.

Check by subtraction: 36−[ABE]−[ECF]−[FDA]=36−9−3−12=1236 - [ABE] - [ECF] - [FDA] = 36 - 9 - 3 - 12 = 12 ✓

Side note
Where it goes wrong

Putting the masses the wrong way round. For BD:DC=3:2BD : DC = 3 : 2 it is tempting to write mB=3m_B = 3, mC=2m_C = 2, copying the ratio. But DD is nearer CC, and a lever balances nearer the HEAVIER end, so CC needs the larger mass. With the masses swapped, the first worked example would give AP:PD=5:2AP : PD = 5 : 2, which is wrong. Check a mass assignment with mB⋅BD=mC⋅DCm_B \cdot BD = m_C \cdot DC: 2⋅3=3⋅22 \cdot 3 = 3 \cdot 2.

Try it
Try it: a cevian meets a median

In triangle ABCABC, FF is on ABAB with AF:FB=1:2AF : FB = 1 : 2, and DD is the midpoint of BCBC. Segments CFCF and ADAD meet at PP. Find AP:PDAP : PD.

Show the answer

AF:FB=1:2AF : FB = 1 : 2 gives mA=2m_A = 2, mB=1m_B = 1. DD is a midpoint, so mC=mB=1m_C = m_B = 1 and DD carries 22. Then AP:PD=mD:mA=2:2=1:1AP : PD = m_D : m_A = 2 : 2 = 1 : 1: PP is the midpoint of ADAD.

Concept
When in doubt, use coordinates

Place the figure on axes; use the distance formula, perpendicular slopes, and the shoelace area formula.

In plain terms. Give the points coordinates and the geometry turns into algebra you can just compute.

Example. A right angle at the origin with legs on the axes makes the third vertex, lengths, and area immediate.

Reminder — Distance and midpoint:  d=(x2−x1)2+(y2−y1)2,M=(x1+x22, y1+y22)d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}, \qquad M=\left(\frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2}\right)

Similar Triangles, Areas and the Shoelace Formula

Tip
Ratios: lengths, areas, volumes

Similar figures with ratio kk have areas in ratio k2k^2 and volumes in ratio k3k^3.

Two triangles with sides in ratio 2:32:3 have areas in ratio 4:94:9 — halving that to 2:32:3 is one of the most common lost points on the AMC 10. The same exponent rule explains why doubling a pizza's diameter quadruples the food, and why a scale model at 1:101:10 weighs a thousandth of the original.

Tip
The shoelace formula

For vertices (x1,y1),…,(xn,yn)(x_1,y_1), \dots, (x_n,y_n) in order, the area is 12∣∑(xiyi+1−xi+1yi)∣\tfrac12\left|\sum (x_iy_{i+1} - x_{i+1}y_i)\right|.

It handles any polygon given by coordinates, convex or not, with no decomposition: the triangle (0,0),(4,0),(1,3)(0,0), (4,0), (1,3) has area 12∣0⋅0−4⋅0+4⋅3−1⋅0+1⋅0−0⋅3∣=6\tfrac12|0\cdot0 - 4\cdot0 + 4\cdot3 - 1\cdot0 + 1\cdot0 - 0\cdot3| = 6. Two rules make it reliable: go around the polygon in one consistent direction, and repeat the first vertex at the end so no term is missed.

Example
Worked: an area ratio inside a triangle
ABCDE
Shared angle at A: [ADE]/[ABC] = (1/3)(1/4).

In triangle ABCABC, DD is on ABAB with AD:DB=1:2AD:DB = 1:2 and EE is on ACAC with AE:EC=1:3AE:EC = 1:3. What fraction of [ABC][ABC] is [ADE][ADE]?

Step 1 — triangles ADEADE and ABCABC share the angle at AA.

Step 2 — for a shared angle, [ADE][ABC]=AD⋅AEAB⋅AC\dfrac{[ADE]}{[ABC]} = \dfrac{AD\cdot AE}{AB\cdot AC}.

Step 3 — =13⋅14=112= \dfrac13\cdot\dfrac14 = \dfrac1{12}. The shared-angle area ratio is worth memorising; it replaces a full similarity argument in one line.

Try it
Try it: shoelace

Find the area of the triangle with vertices (0,0)(0,0), (5,1)(5,1), (2,6)(2,6).

Show the answer

12∣0⋅1−5⋅0+5⋅6−2⋅1+2⋅0−0⋅6∣=12∣30−2∣=14\tfrac12|0\cdot1 - 5\cdot0 + 5\cdot6 - 2\cdot1 + 2\cdot0 - 0\cdot6| = \tfrac12|30 - 2| = 14.

Example
Worked: the trapezoid and its diagonals
ABCDP124
The diagonals cut the trapezoid into four triangles with areas 3, 27, 9 and 9.

Trapezoid ABCDABCD has AB∥CDAB \parallel CD, AB=12AB = 12 and CD=4CD = 4. Its diagonals meet at PP, and [PCD]=3[PCD] = 3. Find the area of the trapezoid.

Spotting it. Parallel sides make similar triangles (PAB∼PCDPAB \sim PCD, by alternate angles), and similar triangles give a LENGTH ratio. The other two triangles are then reached by the shared-height rule.

Step 1. The ratio of similarity is ABCD=3\tfrac{AB}{CD} = 3, so [PAB]=32⋅3=27[PAB] = 3^2 \cdot 3 = 27 and AP:PC=3:1AP : PC = 3 : 1.

Step 2. Triangles PADPAD and PCDPCD share apex DD and have bases APAP and PCPC on one line, so [PAD]=3⋅[PCD]=9[PAD] = 3 \cdot [PCD] = 9. By the same argument with apex BB, [PBC]=13[PAB]=9[PBC] = \tfrac13[PAB] = 9.

Step 3. Total: 3+27+9+9=483 + 27 + 9 + 9 = 48.

Check with the trapezoid formula: [PCD]=3[PCD] = 3 with base 44 puts PP at height 32\tfrac32 above CDCD, a quarter of the full height, so the height is 66 and the area is 12+42⋅6=48\tfrac{12 + 4}{2} \cdot 6 = 48 ✓

Side note
Where it goes wrong

Scaling areas by the length ratio. In the trapezoid, "ABAB is 33 times CDCD, so [PAB]=3⋅3=9[PAB] = 3 \cdot 3 = 9" is wrong: similar figures scale areas by the SQUARE of the length ratio, giving 2727. And the two side triangles are not similar to anything; their areas come from shared heights, not from a similarity ratio at all.