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Geometric Inequalities: Chapter Test

20 multiple-choice questions and 10 core practice problems, drawn from across the chapter and mixed together.

Core practice

10 problems from across the chapter. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 Two pieces of a triangular frame

D lies strictly between B and C. The nondegenerate triangles ABD and ADC satisfy the following equalities.
$AB=BD.$ $AD=DC.$
Angle BAD measures 28 degrees. Find the three angles of triangle ABC, then list its sides from shortest to longest with reasons.
1. Hint 1
  Use the equal sides in triangle ABD to find its remaining angles.
2. Hint 2
  Use the straight angle at D, then the equal sides in triangle ADC.
3. Hint 3
  Angle BAC consists of two adjacent parts. Match each full angle of ABC with its opposite side.
Answer

Angles A, B, C of triangle ABC are 42, 124, 14 degrees. The side order is
$AB<BC.$ $BC<AC.$

Full solution

In triangle ABD, equal sides AB and BD give equal opposite angles ADB and BAD.
Hence
$m\angle ADB=28^\circ.$

The angle sum in ABD gives
$m\angle ABD=124^\circ.$

Since D is between B and C, angles ADB and ADC are supplementary.
Thus
$m\angle ADC=152^\circ.$

In triangle ADC, equal sides AD and DC make angles DAC and ACD equal.
Each is half of the remaining 28 degrees, so each measures 14 degrees.

Angle BAC consists of BAD and DAC, giving
$m\angle BAC=42^\circ.$

The full angles at B and C are 124 and 14 degrees.
Their opposite sides therefore have order
$AB<BC.$ $BC<AC.$

Answer

Angles A, B, C of triangle ABC are 42, 124, 14 degrees. The side order is
$AB<BC.$ $BC<AC.$

Key idea

Track the two isosceles triangles before comparing sides of the whole triangle.
Problem 2 An integer parameter

Two sides of a nondegenerate triangle are 15 and 26 units. The third side has length 2n minus 1 units, where n is an integer.

The angle opposite the third side must be strictly larger than the angle opposite the 15-unit side. Find every possible n and count the choices. Explain what fails at the integer immediately below and the integer immediately above your list.

Text description of this figure

A triangle has given sides 26 and 15 units, with its third side labeled 2n minus 1. The parameter has not been solved.
1. Hint 1
  Find the strict feasible interval for the third side.
2. Hint 2
  Translate the stated angle order into a side comparison within this same triangle.
3. Hint 3
  Intersect both requirements before counting; the two excluded neighbors may fail for different reasons.
Answer

The integers are 9 through 20, inclusive: 12 choices. At n of 8 the two compared angles tie; at n of 21 the triangle is flat.

Full solution

The triangle requirement gives
$11<2n-1<41.$

Solving this interval gives
$6<n<21.$

Within one triangle, the larger angle is opposite the longer side.
The additional requirement gives
$2n-1>15.$

Thus n must exceed 8.
Intersecting both requirements gives
$8<n<21.$

The integers are 9 through 20, so the count is
$20-9+1=12.$

At n of 8, the third side is 15.
The triangle is valid, but the two equal sides have equal opposite angles, violating the strict angle order.

At n of 21, the third side is 41, and
$15+26=41.$
The arrangement is flat.
Every listed integer passes both strict conditions, so the list is exhaustive.

Answer

The integers are 9 through 20, inclusive: 12 choices. At n of 8 the two compared angles tie; at n of 21 the triangle is flat.

Key idea

A feasible triangle can still fail an additional strict angle comparison.
Problem 3 Two triangular frames

In triangle ABC, the side lengths in units are
$AB=11.$ $AC=19.$ $BC=17.$
In triangle DEF, the side lengths in units are
$DE=11.$ $DF=19.$ $EF=23.$
Compare angles A and D. Then order the three angles of DEF from smallest to largest.

Milo compares BC with DF and concludes that angle E is larger than angle A. Does that comparison establish his conclusion? Explain.
1. Hint 1
  For the first comparison, identify two matching corresponding side pairs and their included angles.
2. Hint 2
  For the order within DEF, match each angle with its opposite side.
3. Hint 3
  Check whether the theorem relating opposite sides and angles permits a comparison using sides from different triangles.
Answer

Angle A is smaller than angle D. Within DEF, the order is
$m\angle F<m\angle E.$ $m\angle E<m\angle D.$
Milo’s comparison does not establish the claimed cross-triangle angle order.

Full solution

Both triangles have arms 11 and 19.
Their third sides, 17 and 23, lie strictly between 8 and 30, so both triples are feasible.

The corresponding arm pairs match.
The longer third side EF faces the larger included angle, giving
$m\angle A<m\angle D.$

Within DEF, side lengths have order
$DE<DF.$ $DF<EF.$
Their opposite angles therefore have order
$m\angle F<m\angle E.$ $m\angle E<m\angle D.$

Milo compares opposite sides from two different triangles.
The rule that orders opposite sides and angles applies within one triangle; it does not justify that comparison across these triangles.
The matching-arm theorem compares A with D, and does not by itself compare A with E.
The claimed argument therefore fails to establish its conclusion.

Answer

Angle A is smaller than angle D. Within DEF, the order is
$m\angle F<m\angle E.$ $m\angle E<m\angle D.$
Milo’s comparison does not establish the claimed cross-triangle angle order.

Key idea

Distinguish a within-triangle opposite-side comparison from a matched-arm comparison between triangles.
Problem 4 Across a line

P and Q lie on opposite sides of line l. Their perpendicular feet H and K on l are distinct. The perpendicular lengths, in meters, are
$PH=7.$ $QK=10.$
Could PQ have length 17 meters? Prove your conclusion for every such placement, rather than relying on a measured drawing. Then explain what changes if H and K are allowed to coincide.

Text description of this figure

P and Q are on opposite sides of line l. The distinct feet H and K are marked with right angles. PH is 7 meters and QK is 10 meters. Segment PQ is shown without a length or a derived comparison.
1. Hint 1
  The segment from P to Q crosses the line at some point R.
2. Hint 2
  Compare each part PR and RQ with the perpendicular from the same point. Include the possibility that R is one of the feet.
3. Hint 3
  Can both comparisons be equalities when the feet are distinct?
Answer

With distinct feet,
$PQ>17.$
If the feet coincide, P and Q lie on the same perpendicular on opposite sides of l, and PQ is 17 meters.

Full solution

Let R be the point where segment PQ crosses l.
It lies between P and Q, so
$PQ=PR+RQ.$

If R differs from H, then PR is the hypotenuse of right triangle PHR and is strictly longer than PH.
If R equals H, PR equals PH.
Thus
$PR\ge PH.$

Likewise, if R differs from K, RQ is the hypotenuse of right triangle QKR and is strictly longer than QK; if R equals K, they are equal.
Thus
$RQ\ge QK.$

Because H and K are distinct, R cannot equal both.
At least one of these comparisons is strict.
Adding gives
$PQ>PH+QK.$

The given perpendicular lengths sum to 17 meters, so a length of 17 is impossible with distinct feet.

If H and K coincide, the two perpendiculars lie on the same line.
Since P and Q are on opposite sides of l, the common foot lies between them.
Segment addition then gives
$PQ=17.$

Answer

With distinct feet,
$PQ>17.$
If the feet coincide, P and Q lie on the same perpendicular on opposite sides of l, and PQ is 17 meters.

Key idea

Separate equality cases before adding two shortest-distance comparisons.
Problem 5 Three proposed lengths

Proposed side lengths are $3t-5$ , $t+7$ , and $2t+6$ units, where t is real.

A claim says that every t greater than five-thirds gives a nondegenerate triangle because all three lengths are positive. Assess the claim. Find the complete correct range, and give one value that disproves the claim.
1. Hint 1
  Positivity is necessary, but the three lengths must also pass all strict side-sum comparisons.
2. Hint 2
  Simplify each comparison independently, then test a value allowed by the claim but outside the final range.
Answer

The claim is false. The complete range is
$t>2.$
At t of 2 the positive lengths 1, 9, 10 meet flat.

Full solution

The first two lengths sum to 4t plus 2.
They must exceed the third length:
$4t+2>2t+6.$
This simplifies to
$t>2.$

The first and third lengths sum to 5t plus 1.
They must exceed the second:
$5t+1>t+7.$
This simplifies to
$t>\frac32.$

The second and third lengths sum to 3t plus 13.
They must exceed the first:
$3t+13>3t-5.$
This reduces to 13 greater than negative 5, which always holds.

All lengths are positive when t exceeds five-thirds.
Intersecting positivity with the three comparisons leaves exactly
$t>2.$
Every value in this interval passes all tests and forms a triangle.

For a counterexample to the claim, take t equal to 2.
It exceeds five-thirds, and all lengths are positive, but
$1+9=10.$
This is flat rather than nondegenerate.

Answer

The claim is false. The complete range is
$t>2.$
At t of 2 the positive lengths 1, 9, 10 meet flat.

Key idea

Positive lengths still need every strict triangle comparison.
Problem 6 Three gaps

Triangles ABC, ABD, and ABE share AB, and AC, AD, AE are equal positive lengths. Ray AC lies strictly inside angle BAD.

The given angle measures are
$m\angle BAC=44^\circ.$
Also,
$m\angle CAD=37^\circ.$
E lies on the opposite side of line AB from C and D, with
$m\angle BAE=62^\circ.$
List BC, BD, BE from shortest to longest. Explain why the positions of the triangles do not prevent the comparison.
1. Hint 1
  The opening BAD contains the two adjacent given parts.
2. Hint 2
  Each comparison uses the same arm AB and one of the equal arms AC, AD, AE. Compare the included openings.
Answer

$BC<BE.$ $BE<BD.$

Full solution

Because AC lies inside BAD, angle addition gives
$m\angle BAD=44^\circ+37^\circ.$
Thus angle BAD is 81 degrees.

Each pair of triangles has the common side AB and a second equal arm from AC, AD, AE.
The included openings are 44 degrees for ABC, 62 for ABE, and 81 for ABD.

The hinge theorem orders the opposite gaps in the same way:
$BC<BE.$ $BE<BD.$

The theorem compares the matched lengths and openings.
It does not require the triangles to lie on the same side of their shared line.

Answer

$BC<BE.$ $BE<BD.$

Key idea

A shared arm and equal remaining arms let several openings order several gaps.
Problem 7 A route with a limit

A and B are 30 meters apart. P is off line AB. The straight legs AP and PB have lengths $r+4$ and $2r-3$ meters, respectively, where r is real.

The route through P must be at most 52 meters long. Find the complete possible range of r. Explain what happens at each endpoint and why every included value is feasible.

Text description of this figure

A nondegenerate triangular route has direct side AB of 30 meters, with legs AP labeled r plus 4 and PB labeled 2r minus 3. No derived range is shown.
1. Hint 1
  The three lengths must form a nondegenerate triangle, as well as meeting the route-length limit.
2. Hint 2
  Combine all strict triangle comparisons with the non-strict cap on the sum of the two route legs.
3. Hint 3
  A complete range needs the reverse check: every remaining parameter must give positive lengths passing every test.
Answer

$\frac{29}{3}<r\le17.$
The lower boundary is flat; the upper boundary is a valid 52-meter route.

Full solution

The two route legs sum to 3r plus 1.
Their sum must exceed the direct distance, giving
$3r+1>30.$
Hence
$r>\frac{29}{3}.$

The comparison of AP plus AB with PB gives
$r+34>2r-3.$
Hence
$r<37.$

The remaining triangle comparison gives
$2r+27>r+4.$
Hence
$r>-23.$

Positivity requires r greater than negative 4 and r greater than three-halves.
Both are already ensured by r greater than 29/3.

The route limit gives
$3r+1\le52.$
Hence
$r\le17.$

The overlap is
$\frac{29}{3}<r\le17.$
Every value makes both legs positive and satisfies all three strict side comparisons, so it gives a nondegenerate triangle with the permitted route length.

At the lower boundary the route legs sum to 30, so the arrangement is flat and P cannot be off the line.

At the upper boundary the legs are 21 and 31.
Their difference is 10, their sum is 52, and 30 lies strictly between those bounds.
The route is valid and reaches the allowed cap.

Answer

$\frac{29}{3}<r\le17.$
The lower boundary is flat; the upper boundary is a valid 52-meter route.

Key idea

A strict geometric boundary and an inclusive practical cap can coexist.
Problem 8 An interior point

A lies off line BC, and D is strictly between B and C. The lengths satisfy
$AB<AD.$
Prove that AC is the unique longest side of triangle ADC. Justify any angle comparison you use.

Text description of this figure

D is strictly between B and C on a line, while A is off that line. AD joins the interior point to A. The only side comparison given is AB less than AD.
1. Hint 1
  First compare the opposite angles within triangle ABD.
2. Hint 2
  Use the sum of two interior angles to bound angle ADB below 90 degrees.
3. Hint 3
  Angles ADB and ADC form a straight angle. Identify the largest angle of triangle ADC.
Answer

AC is the unique longest side of triangle ADC.

Full solution

In triangle ABD, AB is opposite angle ADB and AD is opposite angle ABD.
The given side order gives
$m\angle ADB<m\angle ABD.$

Their sum is less than 180 degrees because angle BAD is positive.
Twice the smaller angle is less than that sum, so
$2m\angle ADB<180^\circ.$
Hence angle ADB is less than 90 degrees.

Since D lies between B and C, rays DB and DC are opposite.
Angles ADB and ADC are supplementary, so
$m\angle ADC>90^\circ.$

The other two angles of nondegenerate triangle ADC are positive and together total less than 90 degrees.
Each is therefore smaller than angle ADC.

AC is opposite this unique largest angle, so it is longer than both AD and DC.

Answer

AC is the unique longest side of triangle ADC.

Key idea

A side comparison in one triangle can control an adjacent angle in another.
Problem 9 Two related frames

Triangles ABC and DEF have equal corresponding arms: AB and DE are each 15 units, while AC and DF are each 22 units.

Their third sides BC and EF have lengths g and g plus 6 units. The sum of the two triangle perimeters is 110 units.

Find both third sides, verify that both triangles are feasible, and compare the included angles at A and D.
1. Hint 1
  Account for the two fixed arms in each perimeter before using the relation between the third sides.
2. Hint 2
  After finding the gaps, check each against the difference and sum of 15 and 22.
3. Hint 3
  The equal arm pairs let the longer third side identify the larger included angle.
Answer

BC is 15 units and EF is 21 units. Both triangles are feasible, and
$m\angle A<m\angle D.$

Full solution

Each triangle has fixed arms totaling 37 units.
The fixed arms contribute 74 to the combined perimeter, while the two third sides contribute 2g plus 6.

Combining terms gives
$2g+80=110.$
Subtracting 80 and dividing by 2 gives
$g=15.$
Thus EF is 21 units.

The allowable third-side interval for arms 15 and 22 is
$7<x<37.$
Both 15 and 21 lie strictly inside it, so both triples are positive and form nondegenerate triangles.

Both corresponding arm pairs match, and EF is longer than BC.
The converse hinge theorem therefore gives
$m\angle A<m\angle D.$

Answer

BC is 15 units and EF is 21 units. Both triangles are feasible, and
$m\angle A<m\angle D.$

Key idea

A perimeter relation can supply the third sides before their openings are compared.
Problem 10 A moving point

V, T, W lie on a line in that order, with T strictly between V and W.

S may lie anywhere in the plane subject to the condition
$TS=TW.$
Determine every possible comparison between VS and VW. State exactly which allowed position of S gives equality, and justify your classification for all positions, including collinear and coincident points.

One allowed example; S is not restricted to this position.

Text description of this figure

One allowed example has V, T, W on a line in that order and S off the line. TS and TW have matching single ticks. The question permits other positions of S. No comparison between VS and VW is marked.
1. Hint 1
  Compare the direct segment VS with the two-leg route through T, and keep equality separate from strictness.
2. Hint 2
  For collinear S, first decide which side of T it lies on. The given equal lengths restrict the same-side case.
3. Hint 3
  On the opposite side from W, S can lie between V and T, coincide with V, or lie beyond V. Use segment addition in each arrangement.
Answer

Always
$VS\le VW.$
Equality occurs exactly when S coincides with W. Every other allowed position gives
$VS<VW.$

Full solution

Both VT and TW are positive.
Since TS equals TW, S cannot coincide with T.

If S is off line VW, triangle VTS is nondegenerate, so
$VS<VT+TS.$

Replacing the equal length TS by TW gives
$VT+TS=VT+TW.$

The stated order V, T, W gives
$VT+TW=VW.$
Hence VS is strictly shorter than VW.

If S is collinear and on the same side of T as W, the given equality of distances from T forces S to be W.
In this case,
$VS=VW.$

It remains to consider collinear S on the opposite side of T from W.
There are three positions relative to V.

If S lies strictly between V and T, the order V, S, T, W gives
$VW=VS+ST+TW.$
Both added lengths are positive, so VS is strictly shorter than VW.

If S coincides with V, then
$VS=0.$
Since VW is positive, the comparison is again strict.

If S lies beyond V, the order S, V, T, W gives
$TS=TV+VS.$

Substituting the given equality of TS and TW gives
$TW=TV+VS.$

Segment addition along V, T, W gives
$VW=VT+TW.$

Substituting for TW gives
$VW=VT+TV+VS.$
The two added lengths VT and TV are positive, so VS is strictly shorter than VW.

Every point S is either off the line or collinear.
The same-side case and the three opposite-side positions exhaust the collinear possibilities, since S cannot be T.
Thus VS never exceeds VW, equality occurs exactly at W, and every other allowed position gives strict inequality.
Both equality and strict inequality are possible.

Independent examples: marks or named lengths record the equality within each panel.

Answer

Always
$VS\le VW.$
Equality occurs exactly when S coincides with W. Every other allowed position gives
$VS<VW.$

Key idea

A universal comparison needs every point position and an exact equality condition.

Multiple choice

Core practice