Comprehensive Study Notes: Edexcel GCSE Mathematics Higher Tier Paper 1 (May 2017)

Examination Structure and Assessment Requirements

• Paper Identification:

  • Qualification: Pearson Edexcel Level 1 / Level 2 GCSE (9–1)
  • Subject: Mathematics
  • Tier: Higher Tier
  • Paper Reference: 1MA1/1H (Paper 1: Non-Calculator)
  • Session Date: Thursday 25 May 2017 – Morning
  • Total Time Allowed: 1 hour 30 minutes (90 minutes)
  • Total Marks: 80 marks

• Required Mathematical Equipment:

  • Ruler graduated in centimetres and millimetres
  • Protractor
  • Pair of compasses
  • Pen (black ink or ball-point pen)
  • HB pencil
  • Eraser
  • Tracing paper (optional / permitted)

• Examination Constraints and Rubric:

  • Calculators are strictly prohibited.
  • Diagrams are not drawn to scale unless explicitly stated.
  • All stages of working out must be clearly shown to secure full marks.

Bivariate Data Analysis and Scatter Graphs

• Context and Data Characteristics:

  • Data collected represents fourteen British towns on a single day.
  • Explanatory / Independent Variable ($x$-axis): Number of hours of sunshine, plotted across a scale from $7$ to $17\,\text{hours}$.
  • Response / Dependent Variable ($y$-axis): Maximum temperature ($^{\circ}\text{C}$), plotted across a scale from $10\,^{\circ}\text{C}$ to $21\,^{\circ}\text{C}$.

• Identification of Outliers:

  • Definition: An outlier is an anomalous data point that deviates significantly from the overall linear trend displayed by the rest of the dataset.
  • The distinct outlier point on the graph is located at coordinates $(10, 19)$, representing a town with $10\,\text{hours}$ of sunshine and an unusually high maximum temperature of $19\,^{\circ}\text{C}$.

• Correlation Classification:

  • Excluding the identified outlier, the plotted data points follow an upward trend from bottom-left to top-right.
  • The relationship represents positive correlation (as the number of sunshine hours increases, maximum temperature also increases).

• Interpolation and Estimation:

  • Given a maximum temperature of $16.4\,^{\circ}\text{C}$ in another British town on the same day, hours of sunshine are estimated by constructing a straight line of best fit through the correlated cluster.
  • Reading horizontally from $y = 16.4\,^{\circ}\text{C}$ down to the sunshine axis yields an estimate of approximately $12.8\,\text{hours}$ (acceptable range: $12.3\,\text{hours}$ to $13.2\,\text{hours}$).

• Evaluation of Statistical Claims:

  • Statement under evaluation: "Temperatures are higher on days when there is more sunshine."
  • Validity analysis: The scatter graph supports this claim because the points exhibit a positive correlation, demonstrating that higher sunshine durations correspond to higher maximum temperatures.
  • Critical qualification: The data set records fourteen towns on a single day rather than multiple days, but the positive correlation observed on this day aligns with the premise that greater sunshine duration is associated with higher temperatures.

Prime Factor Decomposition

• Prime Factorization of Composite Number $56$:
  • Step 1: Divide by the smallest prime factor, $2$:     $56 \div 2 = 28$
  • Step 2: Divide $28$ by $2$:     $28 \div 2 = 14$
  • Step 3: Divide $14$ by $2$:     $14 \div 2 = 7$
  • Step 4: The quotient $7$ is prime.
  • Expressed as a product of prime factors:     $56 = 2 \times 2 \times 2 \times 7$
  • Expressed in index notation:     $56 = 2^3 \times 7$

Arithmetic Operations with Decimals

• Decimal Long Multiplication for $54.6 \times 4.3$:
  • Identify decimal places: $54.6$ has $1$ decimal place; $4.3$ has $1$ decimal place. The final product will have $1 + 1 = 2$ decimal places.
  • Perform integer multiplication of $546 \times 43$:
  • Multiply by units ($3$):       $546 \times 3 = 1638$
  • Multiply by tens ($40$):       $546 \times 40 = 21840$
  • Sum the partial products:       $21840 + 1638 = 23478$
  • Reinsert decimal point two places from the right:     $54.6 \times 4.3 = 234.78$

Geometric Proof and Area of Squares

• Geometric Setup:

  • Large square $ABCD$ has each side partitioned into two segments of lengths $x\,\text{cm}$ and $3\,\text{cm}$.
  • Total side length of square $ABCD$:     $\text{Side} = x + 3$
  • Given area of square $ABCD = 10\,\text{cm}^2$.

• Algebraic Derivation:

  • Calculate area as side length squared:     $(x + 3)^2 = 10$
  • Expand binomial expression:     $x^2 + 6x + 9 = 10$
  • Subtract $9$ from both sides of the equation:     $x^2 + 6x = 10 - 9$$x^2 + 6x = 1$

Applied Trigonometry and Frame Mass Calculation

• Frame Structure and Piece Dimensions:

  • Frame consists of $5$ straight metal segments:
  • Two horizontal lengths: $12\,\text{m}$ each
  • Two vertical widths: $5\,\text{m}$ each
  • One internal diagonal brace connecting opposite corners

• Application of Pythagoras' Theorem for Diagonal Length:

  • In right-angled triangle formed by the length, width, and diagonal ($d$):     $d^2 = 12^2 + 5^2$$d^2 = 144 + 25 = 169$$d = \sqrt{169} = 13\,\text{m}$

• Total Length Calculation:

  • Sum lengths of all 5 metal pieces:     $\text{Total length} = 12 + 12 + 5 + 5 + 13 = 47\,\text{m}$

• Total Mass Calculation:

  • Mass per unit length: $1.5\,\text{kg/m}$
  • Multiply total length by unit mass:     $\text{Total weight} = 47\,\text{m} \times 1.5\,\text{kg/m} = 47 \times \frac{3}{2} = \frac{141}{2} = 70.5\,\text{kg}$

Coordinate Geometry of Parallel Lines

• Parallel Line Criteria:

  • Two non-vertical lines are parallel if and only if their gradients ($m$) are identical.

• Gradient of Line $L_1$:

  • Given equation:     $y = 3x - 2$
  • Written in slope-intercept form $y = mx + c$, gradient $m_1 = 3$.

• Gradient of Line $L_2$:

  • Given implicit linear equation:     $3y - 9x + 5 = 0$
  • Rearrange into slope-intercept form:     $3y = 9x - 5$$y = 3x - \frac{5}{3}$
  • Gradient $m_2 = 3$.

• Conclusion:

  • Because $m_1 = m_2 = 3$, lines $L_1$ and $L_2$ have the same gradient and are parallel.

Combined Mean and Weighted Averages

• Class Data Parameters:

  • Number of boys ($n_b$): $10$
  • Number of girls ($n_g$): $20$
  • Total class size ($N$): $10 + 20 = 30$
  • Mean mark of entire class ($\bar{x}_{\text{class}}$) = $60$
  • Mean mark of girls ($\bar{x}_g$) = $54$

• Step-by-Step Aggregate Score Calculations:

  • Calculate total marks scored by the entire class:     $\Sigma x_{\text{class}} = N \times \bar{x}_{\text{class}} = 30 \times 60 = 1800$
  • Calculate total marks scored by girls:     $\Sigma x_g = n_g \times \bar{x}_g = 20 \times 54 = 1080$
  • Calculate total marks scored by boys:     $\Sigma x_b = \Sigma x_{\text{class}} - \Sigma x_g = 1800 - 1080 = 720$
  • Compute mean mark for boys ($\bar{x}_b$):     $\bar{x}_b = \frac{\Sigma x_b}{n_b} = \frac{720}{10} = 72$

Standard Index Notation and Operations

• Conversion to Ordinary Number:

  • Number: $7.97 \times 10^{-6}$
  • Shift decimal point $6$ places to the left:     $7.97 \times 10^{-6} = 0.00000797$

• Division in Standard Form:

  • Operation:     $(2.52 \times 10^5) \div (4 \times 10^{-3})$
  • Divide mantissas:     $\frac{2.52}{4} = 0.63$
  • Apply laws of indices to powers of ten:     $\frac{10^5}{10^{-3}} = 10^{5 - (-3)} = 10^8$
  • Combine preliminary expression:     $0.63 \times 10^8$
  • Adjust to standard scientific notation form ($A \times 10^n$ where $1 \le A < 10$):     $0.63 \times 10^8 = (6.3 \times 10^{-1}) \times 10^8 = 6.3 \times 10^7$

Reverse Percentage and Tax Calculations

• Value Added Tax (VAT) Inversion:

  • Given VAT rate: $20\%$
  • Total retail price paid by Jules: £$600$
  • Multiplier relation: The post-tax price corresponds to $100\% + 20\% = 120\% = 1.20$ of the pre-tax base cost.

• Determination of Original Base Price:

  • Let $P$ be the price with no VAT added:     $1.20 \times P = 600$$P = \frac{600}{1.20} = \frac{6000}{12} = 500$
  • Pre-tax price: £$500$

Polynomial Triple Bracket Expansion

• Expansion of $(x + 1)(x + 2)(x + 3)$ into $ax^3 + bx^2 + cx + d$:
  • Step 1: Expand the first two linear binomials:     $(x + 1)(x + 2) = x(x + 2) + 1(x + 2) = x^2 + 2x + x + 2 = x^2 + 3x + 2$
  • Step 2: Multiply the resulting quadratic trinomial by the third binomial $(x + 3)$:     $(x^2 + 3x + 2)(x + 3) = x(x^2 + 3x + 2) + 3(x^2 + 3x + 2)$
  • Step 3: Expand each term across parentheses:     $= (x^3 + 3x^2 + 2x) + (3x^2 + 9x + 6)$
  • Step 4: Collect like terms:     $x^3 + (3x^2 + 3x^2) + (2x + 9x) + 6 = x^3 + 6x^2 + 11x + 6$
  • Verification: Coefficients are $a = 1$, $b = 6$, $c = 11$, $d = 6$, all of which are positive integers.

Graphical Analysis of Quadratic Functions

• Graphical Attributes of $y = \text{f}(x)$:

  • Parabolic curve with minimum turning point at the vertex.
  • Reading coordinates of the turning point directly from the vertex: $(1, -3)$.

• Roots Estimation for $\text{f}(x) = 0$:

  • Roots represent the points of intersection where the parabola crosses the horizontal axis ($y = 0$).
  • The curve crosses the $x$-axis at approximately $x = -0.7$ and $x = 2.7$ (exact roots based on $\text{f}(x) = (x - 1)^2 - 3$ are $1 \pm \sqrt{3} \approx -0.73$ and $2.73$).

• Value Estimation for $\text{f}(1.5)$:

  • Locate $x = 1.5$ on the horizontal axis and read vertically downward to the curve.
  • The corresponding $y$-value is approximately $-2.75$ (acceptable reading range: $-2.7$ to $-2.8$).

Indices: Fractional and Negative Exponents

• Evaluation of $81^{-\frac{1}{2}}$:

  • Apply reciprocal rule for negative power: $a^{-n} = \frac{1}{a^n}$$81^{-\frac{1}{2}} = \frac{1}{81^{\frac{1}{2}}}$
  • Apply radical rule for fractional power: $a^{\frac{1}{2}} = \sqrt{a}$$81^{\frac{1}{2}} = \sqrt{81} = 9$
  • Combine operations:     $81^{-\frac{1}{2}} = \frac{1}{9}$

• Evaluation of $\left(\frac{64}{125}\right)^{\frac{2}{3}}$:

  • Apply root power rule: $\left(\frac{a}{b}\right)^{\frac{m}{n}} = \left(\sqrt[n]{\frac{a}{b}}\right)^m$
  • Compute cube root of base fraction:     $\sqrt[3]{\frac{64}{125}} = \frac{\sqrt[3]{64}}{\sqrt[3]{125}} = \frac{4}{5}$
  • Square the resulting quotient:     $\left(\frac{4}{5}\right)^2 = \frac{16}{25}$

Inverse Proportionality and Algebraic Modeling

• Relationship Formulation:

  • Given: $y$ is inversely proportional to the square of $x$:     $y \propto \frac{1}{x^2} \implies y = \frac{k}{x^2}$
  • Given table coordinates:
  • At $x = 1$, $y = 9$
  • At $x = 2$, $y = 2\frac{1}{4} = \frac{9}{4}$
  • At $x = 3$, $y = 1$
  • At $x = 4$, $y = \frac{9}{16}$

• Constant of Proportionality ($k$):

  • Substitute point $(1, 9)$ into model:     $9 = \frac{k}{1^2} \implies k = 9$
  • General equation governing relationship:     $y = \frac{9}{x^2}$

• Calculation of Positive $x$ for $y = 16$:

  • Substitute $y = 16$ into the derived formula:     $16 = \frac{9}{x^2}$$x^2 = \frac{9}{16}$
  • Take positive square root:     $x = \sqrt{\frac{9}{16}} = \frac{3}{4} = 0.75$

Compound Ratio and Categorical Probability

• System Composition and Initial Ratios:

  • Game shapes categorized by color (White, Black) and geometry (Circles, Squares).
  • Overall color ratio: $\text{White} : \text{Black} = 3 : 7$
  • Fraction of White shapes: $\frac{3}{10}$
  • Fraction of Black shapes: $\frac{7}{10}$

• Sub-Category Proportions:

  • White shapes:
  • Ratio of circles to squares: $4 : 5$
  • Total white parts: $4 + 5 = 9$
  • Fraction of white shapes that are circles: $\frac{4}{9}$
  • Total fraction of all shapes that are white circles:       $\frac{3}{10} \times \frac{4}{9} = \frac{12}{90} = \frac{2}{15}$
  • Black shapes:
  • Ratio of circles to squares: $2 : 5$
  • Total black parts: $2 + 5 = 7$
  • Fraction of black shapes that are circles: $\frac{2}{7}$
  • Total fraction of all shapes that are black circles:       $\frac{7}{10} \times \frac{2}{7} = \frac{2}{10} = \frac{1}{5} = \frac{3}{15}$

• Overall Proportion of Circles:

  • Sum the circle contributions across both color groups:     $\text{Fraction of circles} = \frac{2}{15} + \frac{3}{15} = \frac{5}{15} = \frac{1}{3}$

Estimation and Error Propagation in Solid Geometry

• Cone Formulas and Given Parameters:

  • Formula for volume of a cone:     $V = \frac{1}{3}\pi r^2 h$
  • Actual measured values: $V = 98\,\text{cm}^3$, radius $r = 5.13\,\text{cm}$

• Height Estimation Strategy:

  • Round measured and constant parameters to $1$ significant figure:
  • Volume $V \approx 100\,\text{cm}^3$
  • Radius $r \approx 5\,\text{cm}$
  • Constant $\pi \approx 3$
  • Substitute rounded values into cone formula:     $100 = \frac{1}{3} \times 3 \times 5^2 \times h$$100 = 25h$$h = \frac{100}{25} = 4\,\text{cm}$

• Directional Error Propagation Analysis:

  • Express height explicitly in terms of variables:     $h = \frac{3V}{\pi r^2}$
  • Evaluation of rounding effects:
  • Numerator $V$ was rounded UP from $98$ to $100$.
  • Denominator factor $\pi$ was rounded DOWN from $3.14159...$ to $3$.
  • Denominator factor $r$ was rounded DOWN from $5.13$ to $5$, causing $r^2$ to be rounded DOWN from $26.3169$ to $25$.
  • Combined effect: A larger numerator divided by a smaller denominator produces an overestimate.
  • Conclusion: The calculated estimate ($4\,\text{cm}$) will be greater (more) than John's calculator value.

Formal Algebraic Parity Proof

• Proposition:

  • For any integer $n > 1$, the algebraic expression $n^2 - 2 - (n - 2)^2$ is always an even number.

• Step-by-Step Algebraic Expansion and Reduction:

  • Expand the subtracted squared binomial:     $(n - 2)^2 = n^2 - 4n + 4$
  • Substitute into the full expression and distribute the negative sign:     $n^2 - 2 - (n - 2)^2 = n^2 - 2 - (n^2 - 4n + 4)$$= n^2 - 2 - n^2 + 4n - 4$
  • Group and cancel terms:     $= (n^2 - n^2) + 4n - (2 + 4)$$= 4n - 6$
  • Factorize out $2$ to demonstrate divisibility:     $4n - 6 = 2(2n - 3)$

• Parity Conclusion:

  • Because $n$ is an integer, $2n - 3$ is also an integer.
  • Any integer multiplied by $2$ yields an even integer.
  • Thus, $n^2 - 2 - (n - 2)^2$ is always even for all integers $n > 1$.

Probability Without Replacement

• Bag Contents and Sample Space:

  • Green counters ($G$): $7$
  • Blue counters ($B$): $2$
  • Total counters: $7 + 2 = 9$
  • Two counters drawn sequentially at random without replacement.

• Combined Mutually Exclusive Event Paths:

  • Event condition: Exactly one counter of each color is selected.
  • Path 1: Green first, then Blue ($GB$):     $P(G_1 \cap B_2) = \frac{7}{9} \times \frac{2}{8} = \frac{14}{72}$
  • Path 2: Blue first, then Green ($BG$):     $P(B_1 \cap G_2) = \frac{2}{9} \times \frac{7}{8} = \frac{14}{72}$

• Total Probability Calculation:

  • Sum probabilities of disjoint paths:     $P(\text{one of each}) = \frac{14}{72} + \frac{14}{72} = \frac{28}{72}$
  • Reduce fraction to simplest terms:     $\frac{28}{72} = \frac{7}{18}$

Coordinate Geometry and Rhombus Diagonals

• Geometric Properties of a Rhombus:

  • The diagonals of a rhombus intersect at right angles (they are perpendicular bisectors of one another).
  • Therefore, diagonal $AC$ is perpendicular to diagonal $DB$.

• Perpendicular Gradient Determination:

  • Equation of diagonal $DB$:     $y = \frac{1}{2}x + 6$
  • Gradient of $DB$: $m_{DB} = \frac{1}{2}$
  • Negative reciprocal rule for perpendicular lines ($m_1 \times m_2 = -1$):     $m_{AC} = -\frac{1}{\frac{1}{2}} = -2$

• Equation of Line Passing Through Point $A(5, 11)$:

  • Point-slope equation of a straight line:     $y - y_1 = m(x - x_1)$$y - 11 = -2(x - 5)$
  • Expand and rearrange into standard form:     $y - 11 = -2x + 10$$y = -2x + 21$$2x + y = 21$

Vector Geometry and Collinearity

• Vector Definitions in Parallelogram $OABC$:

  • Defined baseline vectors:     $\overrightarrow{OA} = \mathbf{a}$$\overrightarrow{OC} = \mathbf{c}$
  • By parallelogram properties:     $\overrightarrow{CB} = \mathbf{a}$$\overrightarrow{AB} = \mathbf{c}$

• Midpoint Vector Formulation:

  • Vector diagonal $\overrightarrow{AC}$:     $\overrightarrow{AC} = \overrightarrow{AO} + \overrightarrow{OC} = -\mathbf{a} + \mathbf{c} = \mathbf{c} - \mathbf{a}$
  • Point $X$ is the midpoint of segment $AC$:     $\overrightarrow{AX} = \frac{1}{2}\overrightarrow{AC} = \frac{1}{2}(\mathbf{c} - \mathbf{a})$
  • Position vector of $X$ relative to origin $O$:     $\overrightarrow{OX} = \overrightarrow{OA} + \overrightarrow{AX} = \mathbf{a} + \frac{1}{2}(\mathbf{c} - \mathbf{a}) = \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{c}$

• Straight Line Collinearity on Extended Line $OCD$:

  • Points $O$, $C$, and $D$ lie along a single continuous straight line.
  • Ratio $OC : CD = k : 1$
  • Vector $\overrightarrow{CD}$ in terms of $\mathbf{c}$:     $\overrightarrow{CD} = \frac{1}{k}\overrightarrow{OC} = \frac{1}{k}\mathbf{c}$
  • Position vector of point $D$:     $\overrightarrow{OD} = \overrightarrow{OC} + \overrightarrow{CD} = \mathbf{c} + \frac{1}{k}\mathbf{c} = \left(1 + \frac{1}{k}\right)\mathbf{c}$

• Vector $\overrightarrow{XD}$ and Solving for Scalar $k$:

  • Vector subtraction to express $\overrightarrow{XD}$:     $\overrightarrow{XD} = \overrightarrow{OD} - \overrightarrow{OX} = \left(1 + \frac{1}{k}\right)\mathbf{c} - \left(\frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{c}\right)$$\overrightarrow{XD} = \left(1 + \frac{1}{k} - \frac{1}{2}\right)\mathbf{c} - \frac{1}{2}\mathbf{a} = \left(\frac{1}{2} + \frac{1}{k}\right)\mathbf{c} - \frac{1}{2}\mathbf{a}$
  • Equate to given identity $\overrightarrow{XD} = 3\mathbf{c} - \frac{1}{2}\mathbf{a}$:     $\frac{1}{2} + \frac{1}{k} = 3$$\frac{1}{k} = 3 - \frac{1}{2} = \frac{5}{2}$$k = \frac{2}{5} = 0.4$

Non-Linear Simultaneous Systems

• Simultaneous Equation System:

  • Quadratic circle equation (1):     $x^2 + y^2 = 25$
  • Linear equation (2):     $y - 3x = 13$

• Algebraic Substitution Method:

  • Express $y$ explicitly from (2):     $y = 3x + 13$
  • Substitute into quadratic equation (1):     $x^2 + (3x + 13)^2 = 25$
  • Expand binomial:     $x^2 + (9x^2 + 78x + 169) = 25$$10x^2 + 78x + 169 - 25 = 0$$10x^2 + 78x + 144 = 0$
  • Simplify by dividing entire quadratic by $2$:     $5x^2 + 39x + 72 = 0$

• Quadratic Factorization:

  • Factor pairs for product $ac = 5 \times 72 = 360$ that sum to $b = 39$:     $15 \times 24 = 360$$15 + 24 = 39$
  • Split linear term and factor by grouping:     $5x^2 + 15x + 24x + 72 = 0$$5x(x + 3) + 24(x + 3) = 0$$(5x + 24)(x + 3) = 0$
  • Roots for $x$:     $x_1 = -3$$x_2 = -\frac{24}{5} = -4.8$

• Determining Corresponding $y$-Values:

  • For $x = -3$:     $y = 3(-3) + 13 = -9 + 13 = 4$
  • For $x = -4.8$:     $y = 3(-4.8) + 13 = -14.4 + 13 = -1.4$
  • Solutions:     $x = -3, y = 4$$x = -4.8, y = -1.4$

Congruence Proof in Quadrilaterals

• Given Conditions in Quadrilateral $ABCD$:

  • Side $AB = CD$
  • Interior angle $\angle ABC = \angle BCD$

• Geometric Triangle Congruence Proof for $AC = BD$:

  • Construct and compare triangles $\triangle ABC$ and $\triangle DCB$:
  • Side 1: $AB = CD$ (given)
  • Included Angle: $\angle ABC = \angle DCB$ (given)
  • Side 2: $BC = CB$ (shared common side)
  • Criterion: Triangles $\triangle ABC$ and $\triangle DCB$ are congruent by the Side-Angle-Side (SAS) postulate:     $\triangle ABC \cong \triangle DCB$
  • Conclusion: Corresponding parts of congruent triangles are equal (CPCTC):     $AC = BD$

Advanced Trigonometry and Cosine Rule in Hexagonal Assemblies

• Structural Setup and Invariant Geometry:

  • Hexagon $ABCDEF$ is formed by two congruent parallelograms $ABEF$ and $CBED$.
  • Parallelograms share the interior central edge $BE$.
  • Side lengths: $AB = BC = x\,\text{cm}$.
  • By congruence of parallelograms:     $AB = FE = BC = ED = x\,\text{cm}$$AF = BE = CD$
  • Given interior angle $\angle ABC = 30^{\circ}$.
  • Defined point distances: $P$ lies on $AF$ and $Q$ lies on $CD$ such that $BP = BQ = 10\,\text{cm}$.

• Distance Between Points $A$ and $C$:

  • Apply the Cosine Rule to triangle $\triangle ABC$:     $AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(\angle ABC)$
  • Substitute known lengths and exact trigonometric constant $\cos(30^{\circ}) = \frac{\sqrt{3}}{2}$:     $AC^2 = x^2 + x^2 - 2(x)(x)\left(\frac{\sqrt{3}}{2}\right)$$AC^2 = 2x^2 - \sqrt{3}x^2 = (2 - \sqrt{3})x^2$

• Parallelogram Parallelism and Length Equivalence $PQ = AC$:

  • In parallelogram $ABEF$, segment $AF$ is parallel to $BE$.
  • In parallelogram $CBED$, segment $CD$ is parallel to $BE$.
  • Since both lines $AF$ and $CD$ are parallel to central line $BE$, line $AF$ is parallel to line $CD$.
  • By symmetry across the angle bisector of $\angle ABC$ along $BE$, the distances from $B$ to $P$ and $B$ to $Q$ are equal ($BP = BQ = 10\,\text{cm}$).
  • The vector displacement along the parallel edges is equal:     $\overrightarrow{AP} = \overrightarrow{CQ}$
  • Connecting vector between $P$ and $Q$:     $\overrightarrow{PQ} = \overrightarrow{PA} + \overrightarrow{AC} + \overrightarrow{CQ} = -\overrightarrow{AP} + \overrightarrow{AC} + \overrightarrow{AP} = \overrightarrow{AC}$
  • Because $\overrightarrow{PQ} = \overrightarrow{AC}$, the lengths are identical:     $PQ = AC$$PQ^2 = AC^2 = (2 - \sqrt{3})x^2$

• Proof of the Cosine Identity in Triangle $\triangle PBQ$:

  • Apply the Cosine Rule to triangle $\triangle PBQ$ with angle $\angle PBQ$:     $PQ^2 = BP^2 + BQ^2 - 2(BP)(BQ)\cos(\angle PBQ)$
  • Substitute known side lengths $BP = 10$ and $BQ = 10$:     $PQ^2 = 10^2 + 10^2 - 2(10)(10)\cos(\angle PBQ)$$PQ^2 = 100 + 100 - 200\cos(\angle PBQ)$$PQ^2 = 200 - 200\cos(\angle PBQ)$
  • Rearrange to solve for $\cos(\angle PBQ)$:     $200\cos(\angle PBQ) = 200 - PQ^2$$\cos(\angle PBQ) = \frac{200 - PQ^2}{200} = 1 - \frac{PQ^2}{200}$
  • Substitute $PQ^2 = (2 - \sqrt{3})x^2$ into the expression:     $\cos(\angle PBQ) = 1 - \frac{(2 - \sqrt{3})x^2}{200}$