Chapter Review · a rapid pre-test review (speedrun)

Geometric Inequalities: Chapter Review

A rapid review before the test: the chapter's vocabulary and notation, every formula with the conditions to use it, the standard problem types step by step, and the traps that cost points.

Vocabulary and notation

Lengths and angle measures
A side is a segment. In AB<ACAB<AC, the names ABAB and ACAC denote lengths. Write an angle measure as m∠ABCm\angle ABC, and compare like quantities.
Opposite side
The side that does not touch an angle's vertex. In triangle ABCABC, side BCBC is opposite angle AA. Rotation does not change the pairing.
Nondegenerate and degenerate
A nondegenerate triangle has three noncollinear vertices and positive side lengths. A degenerate triangle is a flat boundary arrangement, excluded from ordinary triangle theorems.
Hypotenuse
The side opposite the right angle of a right triangle. The two sides meeting at that angle are its legs.
Perpendicular foot and distance to a line
The perpendicular foot is where a perpendicular from the point meets the line. The distance is that perpendicular length; for a point already on the line, it is zero.
Third-side range
Every possible third-side length when two positive side lengths are fixed. Both geometric boundary values are excluded.
Broken path
A route made of consecutive straight segments. Its travel order matters: points on one line can still create backtracking.
Included angle and matched arms
The included angle is between the two chosen sides, at their common endpoint. In a hinge comparison, matched arms are two pairs of equal corresponding side lengths.

Formulas and theorems

  • Side-angle ordering in one triangle

    Each strict comparison below implies the other:

    AC>ABm∠B>m∠C\begin{gathered} AC>AB\\ m\angle B>m\angle C \end{gathered}
    Match each angle to its opposite sideTriangle ABC has A at the lower left, B above it, and C at the lower right. Solid side AC is across from the double-marked angle B. Dashed side AB is across from the single-marked angle C. The distinct line and arc patterns identify the pairs; no numerical comparison is given.ABCACAB
    Text description

    Triangle ABC has A at the lower left, B above it, and C at the lower right. Solid side AC is across from the double-marked angle B. Dashed side AB is across from the single-marked angle C. The distinct line and arc patterns identify the pairs; no numerical comparison is given.

    Use when Use one nondegenerate triangle ABC. AC faces angle B; AB faces angle C.

  • Equal sides and equal opposite angles

    Each equality below implies the other:

    AC=ABm∠B=m∠C\begin{gathered} AC=AB\\ m\angle B=m\angle C \end{gathered}

    Use when The two sides and their opposite angles belong to the same nondegenerate triangle ABC. These are the isosceles equality cases.

  • Exterior angle theorem and inequality

    e=m∠A+m∠Be>m∠Ae>m∠Be+m∠C=180∘\begin{gathered} e=m\angle A+m\angle B\\ e>m\angle A\\ e>m\angle B\\ e+m\angle C=180^\circ \end{gathered}
    Separate remote and adjacent interior anglesIn triangle ABC, B, C, and D lie in that order on a straight line. A is above the line. The exterior angle e is between CA and CD. The remote interior angles are at A and B; the adjacent interior angle is at C.ABCDe
    Text description

    In triangle ABC, B, C, and D lie in that order on a straight line. A is above the line. The exterior angle e is between CA and CD. The remote interior angles are at A and B; the adjacent interior angle is at C.

    Use when In nondegenerate triangle ABC, extend BC through C to D. Here e is the measure of exterior angle ACD. A and B are remote; C is adjacent.

  • Longest side opposite a right or obtuse angle

    In a right triangle, the hypotenuse is uniquely longest. In an obtuse triangle, the side opposite the obtuse angle is uniquely longest.

    Use when Use a nondegenerate triangle and an established right or obtuse angle. Identify its opposite side, regardless of orientation.

  • Perpendicular is uniquely shortest

    PH<PQ\begin{gathered} PH<PQ \end{gathered}
    Locate the unique closest endpointPoint P lies above a line through Q, H, and R in that order. PH is perpendicular to the line, with a right-angle square at H. PQ and PR are slanted segments to different endpoints. The right angles identify PH as the perpendicular distance; no numerical lengths are supplied.PQHR
    Text description

    Point P lies above a line through Q, H, and R in that order. PH is perpendicular to the line, with a right-angle square at H. PQ and PR are slanted segments to different endpoints. The right angles identify PH as the perpendicular distance; no numerical lengths are supplied.

    Use when P is outside a line, H is its perpendicular foot, and Q is any other point on that line. If P lies on the line, its distance to the line is zero.

  • Triangle inequality

    a+b>ca+c>bb+c>a\begin{gathered} a+b>c\\ a+c>b\\ b+c>a \end{gathered}

    Use when The positive lengths a, b, and c are sides of a nondegenerate triangle. All three comparisons are strict.

  • Positive side-triple and half-perimeter tests

    The proposed lengths form a triangle exactly when either equivalent test holds:

    a+b>cc<p2\begin{gathered} a+b>c\\ c<\frac p2 \end{gathered}

    Use when The proposed lengths a, b, c are positive and ordered from smallest to largest, allowing ties. Their perimeter is p. Equality is flat; a smaller shorter-side sum cannot close.

  • Complete third-side range

    ∣a−b∣<x<a+b\begin{gathered} |a-b|<x<a+b \end{gathered}
    Exclude both third-side boundariesA number line has an open circle at the difference of the fixed lengths and another at their sum. Only the segment strictly between the circles is highlighted. If the fixed lengths are equal, the difference endpoint is zero and remains excluded.|a - b|a + bpossible x
    Text description

    A number line has an open circle at the difference of the fixed lengths and another at their sum. Only the segment strictly between the circles is highlighted. If the fixed lengths are equal, the difference endpoint is zero and remains excluded.

    Use when The fixed lengths a and b are positive. Every x strictly inside the interval is possible. Equality is a flat or collapsed boundary; when a and b match, the excluded lower endpoint is zero.

  • Direct segment versus a broken path

    For any finite broken path, the direct distance is at most the total route length. For three legs:

    AB≤AP+PQ+QB\begin{gathered} AB\le AP+PQ+QB \end{gathered}
    Point order decides whether a route tiesTwo rows compare a direct segment with a route through P. The first row has A, P, B in that order and travels forward. The second has A, B, P in that order; traveling from A to P and back to B retraces BP. The second route is drawn on offset parallel tracks only to make the repeated travel visible.forward travelAPBovershoot and returnABP
    Text description

    Two rows compare a direct segment with a route through P. The first row has A, P, B in that order and travels forward. The second has A, B, P in that order; traveling from A to P and back to B retraces BP. The second route is drawn on offset parallel tracks only to make the repeated travel visible.

    Use when A and B are distinct, and the legs have positive lengths. Equality holds exactly when intermediate points lie on segment AB in travel order, without backtracking. An off-line turn or reversal makes the route longer.

  • Polygon inequality

    s<p−s\begin{gathered} s<p-s \end{gathered}

    Use when A nondegenerate polygon has vertices not all collinear, perimeter p, and chosen side length s. That side is strictly shorter than all the others together. This necessary bound alone does not construct the polygon.

  • Hinge theorem: SAS inequality

    The included-angle comparison implies the third-side comparison:

    m∠BAC>m∠EDFBC>EF\begin{gathered} m\angle BAC>m\angle EDF\\ BC>EF \end{gathered}
    Larger included angle in triangle ABC on the rightTriangle DEF on the left has a solid outline and the smaller included angle beta at D. Triangle ABC on the right has a dashed outline and the larger included angle alpha at A. Single ticks mark equal lengths DE and AB; double ticks mark equal lengths DF and AC. The opposite side BC is longer than EF. The diagram illustrates the stated hinge comparison.DEFABCβα
    Text description

    Triangle DEF on the left has a solid outline and the smaller included angle beta at D. Triangle ABC on the right has a dashed outline and the larger included angle alpha at A. Single ticks mark equal lengths DE and AB; double ticks mark equal lengths DF and AC. The opposite side BC is longer than EF. The diagram illustrates the stated hinge comparison.

    Use when Use two nondegenerate triangles ABC and DEF. AB equals DE in length, and AC equals DF in length. The compared angles are between those matched arms.

  • Converse hinge theorem

    The third-side comparison implies the included-angle comparison:

    BC>EFm∠BAC>m∠EDF\begin{gathered} BC>EF\\ m\angle BAC>m\angle EDF \end{gathered}

    Use when Use two nondegenerate triangles ABC and DEF with AB equal to DE and AC equal to DF in length. Reversing the strict side order reverses the included-angle order.

  • Hinge equality case

    Equal included angles give equal third sides by SAS. Equal third sides give equal included angles by SSS.

    Use when Both directions require two nondegenerate triangles with AB equal to DE and AC equal to DF in length. Every included angle is positive and less than 180 degrees.

Problem types, step by step

Rank sides or angles within a triangle

  1. Find any missing angles using the triangle sum, an exterior angle, or equal isosceles base angles. Keep every angle positive.
  2. Match each angle with its opposite side. Rank the given quantities, then transfer that order to the opposite quantities, preserving ties.

e.g. Given angles 37∘37^\circ, 64∘64^\circ, and 79∘79^\circ at A,B,CA,B,C, order the sides: shortest to longest, BC,AC,ABBC,AC,AB.

Combine comparisons and algebra

  1. Choose a triangle containing each pair being compared. Translate angle order into side order, or side order into angle order.
  2. Connect established inequalities through a common length when available. Solve expression comparisons and intersect them with all stated restrictions.
  3. For a largest-angle or largest-side task, compare with both competitors. Keep separate intervals and tie cases when the order changes.

e.g. In an existing triangle, angle BB exceeds CC; ACAC has length 7u+67u+6 and ABAB has length 4u+394u+39. Find the necessary comparison restriction: u>11u>11.

Compare distances to a line

  1. Locate the perpendicular foot and establish the right angle. Every segment to a different endpoint is a hypotenuse longer than the perpendicular.
  2. For endpoints on one side of the foot, use the stated point order and an exterior angle to compare slanted segments. Do not infer the perpendicular foot from a short list of sampled distances.

Classify proposed side lengths

  1. Reject zero or negative proposed lengths. For positive numerical lengths, identify a longest one.
  2. Compare the sum of the other two with the longest: greater gives a triangle, equality gives a flat boundary, and less gives failure to close.

e.g. Classify the positive lengths 13,18,3013,18,30: they form a nondegenerate triangle.

Find a third-side range or count allowed lengths

  1. Use the absolute difference of the fixed sides as the strict lower bound and their sum as the strict upper bound.
  2. For integers or specified multiples, retain only values strictly inside the interval. Count inclusively from the first allowed value to the last.

e.g. For fixed sides 1414 and 2323, find the integer third sides: 1010 through 3636, totaling 2727 possibilities.

Solve variable sides or a perimeter condition

  1. Use the perimeter to express a missing side when needed. Require every side expression to be positive.
  2. If the longest side can change, solve all three triangle inequalities. If its identity is established, use the longest-side or half-perimeter test.
  3. Intersect every restriction. For an exhaustive list of unordered integer triples, order the side lengths first and check every remaining integer choice.

Fit one length into several requirements

  1. Write a separate strict third-side interval for each triangle. Translate any perimeter cap or other requirement into its own restriction.
  2. Keep the overlap of all intervals, then apply integer or multiple restrictions. Check why each final boundary is excluded; different endpoints can fail different requirements.

Compare a route, shortcut, or polygon side

  1. Identify the direct endpoint segment and add all route legs. Use the non-strict path bound, becoming strict if a bend or reversal is established.
  2. Use point order for collinear routes; an overshoot is traveled twice. Compute savings as route length minus direct length.
  3. For a polygon, compare the chosen side with the sum of all others. A passed necessary bound does not determine an exact length or establish every shape condition.

e.g. A direct route ABAB is 2626 m; PP lies beyond BB with BPBP of 77 m. Find the route length from AA through PP to BB: 4040 m.

Compare two triangles by their hinges

  1. Identify both equal corresponding side pairs, counting a genuinely shared side if present. Locate their included angles.
  2. Recover missing openings by the triangle angle sum or by addition or subtraction of established angle parts. Use the hinge theorem for angle-to-side order, or its converse for side-to-angle order.
  3. Use SAS or SSS for equality. Compare only the opposite third sides or included angles justified by that result.

e.g. Triangles ABCABC and DEFDEF have corresponding arms 11,1811,18 and included angles 54∘,92∘54^\circ,92^\circ at A,DA,D. Compare their third sides: BC<EFBC<EF.

Find a complete hinge parameter range

  1. Use segment addition to recover any split arm. Solve the equations required to make both arm pairs equal, and check their common lengths are positive.
  2. Translate the requested strict order. Keep each included angle strictly between zero and 180 degrees; if third sides are specified, enforce their triangle ranges too.
  3. Intersect all restrictions and check boundary failures. Distinguish a necessary restriction for stated existing triangles from a full feasibility range for proposed data.

e.g. With matching arms 19,2819,28 and third sides s+9,3s−7s+9,3s-7, require the first opening smaller. Find all feasible ss: 8<s<188<s<18.

Exam traps

  • Trap Choosing an adjacent side or trusting the sketch

    Fix Use the side across from the angle vertex. Orientation and apparent size supply no measurement or incidence.

  • Trap Ordering sides in unrelated triangles from angles alone

    Fix Within-triangle ordering needs one triangle. A hinge comparison needs two equal corresponding side pairs, surrounding the compared angles.

  • Trap Forcing a unique largest part when two tie

    Fix Equal sides face equal angles. Preserve equalities while transferring the order; compare the proposed largest part with both competitors.

  • Trap Comparing an exterior angle with the adjacent angle

    Fix The strict exterior-angle inequality concerns the two remote interior angles. The adjacent interior angle is supplementary and can be larger.

  • Trap Calling the shortest sampled segment the distance to a line

    Fix The distance uses the perpendicular foot, compared with every point on the line. Being nearest among several named endpoints does not establish perpendicularity.

  • Trap Checking an arbitrary side sum or assuming a variable side is longest

    Fix For positive numerical lengths, test the two shorter against the longest. With variable lengths, justify the order or retain all three inequalities and positivity.

  • Trap Dropping the difference bound or including a boundary

    Fix A third side must exceed the absolute difference and stay below the sum. Both bounds are strict; remove endpoints before counting integers.

  • Trap Assuming collinearity guarantees an equal-length route

    Fix Equality requires forward travel on the direct segment in the stated order. Backtracking adds distance even on one line.

  • Trap Ordering two quantities because both are below the same reference

    Fix That information leaves their mutual order undecided. A transitive comparison needs a chain with a consistent direction.

  • Trap Using a known angle outside the matched arms

    Fix The included angle is at the common endpoint of the matched sides. One equal pair or a different labeled angle does not justify a hinge argument.

  • Trap Reading exact lengths or every angle from a hinge comparison

    Fix The hinge theorems give the third-side or included-angle order. They do not calculate exact lengths or make every corresponding angle larger.

  • Trap Applying the strict triangle inequality to flat auxiliary points

    Fix A construction can give collinear auxiliary points even when the original triangles are nondegenerate. Use the stated point order and segment addition with positive extra pieces, rather than the strict triangle inequality.

  • Trap Ruling out only equality in an indirect proof

    Fix Also exclude the reverse strict order, using the earlier forward theorem under its hypotheses. Only then is the desired order the remaining possibility.

Chapter Test Questions from across the chapter