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Equations and Inequalities: Star problems

Ten optional challenges to stretch your reasoning. Work on paper, use hints when you need them, and check the answer or full solution when you are ready. You can skip these problems and continue the course.

1 of 3 stars: Stretch
2 of 3 stars: Challenge
3 of 3 stars: Deep challenge

Stars indicate difficulty within this set.

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Problem 1 of 10

Problem 1 Two sets of landmarks

Difficulty: 1 of 3 stars, Stretch

Find every real number $x$ for which
$|x-2|+|x-7|+|x-11|+1=|x-1|+|x-8|+|x-12|.$
Explain why the solution set can contain intervals even though the two sides use different landmarks.
1. Hint 1
  Group each distance on the left with a distance to a nearby landmark on the right.
2. Hint 2
  Study $D(x)=\bigl(|x-2|-|x-1|\bigr)+\bigl(|x-7|-|x-8|\bigr)+\bigl(|x-11|-|x-12|\bigr)$ . Each bracket is constant outside the interval between its two landmarks.
Answer

$x\in(-\infty,1]\cup[8,11]$ .

Full solution

We need $D(x)=-1$ .
A difference of distances changes only while $x$ lies between its two landmarks.
For example, $|x-2|-|x-1|$ is $1$ for $x\le1$ , $3-2x$ for $1\le x\le2$ , and $-1$ for $x\ge2$ .

Combining the three brackets gives
$D(x)=\begin{cases}-1&x\le1,\\1-2x&1\le x\le2,\\-3&2\le x\le7,\\2x-17&7\le x\le8,\\-1&8\le x\le11,\\2x-23&11\le x\le12,\\1&x\ge12.\end{cases}$
Adjacent formulas agree at every shared endpoint.

The constant pieces equal to $-1$ give $(-\infty,1]$ and $[8,11]$ .
The other three nonconstant pieces reach $-1$ only at $1$ , $8$ , or $11$ , already included.
The remaining constant pieces contribute nothing.
Thus the list is complete.
Intervals arise because, between landmarks, the changes in the three paired differences can cancel exactly.

Answer

$x\in(-\infty,1]\cup[8,11]$ .

Key idea

Pair nearby distances before opening absolute values; constant pieces often explain entire intervals of solutions.
Problem 1: Mark complete
Problem 2 A setting that must work everywhere

Difficulty: 1 of 3 stars, Stretch

A real setting $a$ must make
$(a-1)x+a+2>0$
true for every permitted input $x$ .

(a) Find all settings that work when the permitted inputs form the closed interval $[-2,3]$ .

(b) Find all settings that work when the permitted inputs instead form the open interval $(-2,3)$ . Prove that checking infinitely many inputs requires only a small number of tests, and explain the changed endpoint decisions.

Builds on Linear Inequalities
1. Hint 1
  An affine expression at an interior point of an interval is a weighted average of its endpoint values.
2. Hint 2
  The endpoint values are $4-a$ and $4a-1$ . For an open interval, a zero endpoint value need not produce an allowed zero.
Answer

(a) $1/4<a<4$ . (b) $1/4\le a\le4$ .

Full solution

Write
$L(x)=(a-1)x+a+2.$
At the two endpoints, $L(-2)=4-a$ and $L(3)=4a-1$ .
For $-2\le x\le3$ ,
$L(x)=\frac{3-x}{5}L(-2)+\frac{x+2}{5}L(3).$
The coefficients are nonnegative and sum to $1$ .

On the closed interval both endpoint values must be strictly positive, and those two conditions suffice by the displayed weighted average.
They give $a<4$ and $a>1/4$ .

On the open interval, the endpoint values must be nonnegative: if either were negative, the affine formula would remain negative at sufficiently nearby interior inputs.
Nonnegativity gives
$1/4\le a\le4.$
At these settings the two endpoint values cannot both be zero.
For each interior input both averaging coefficients are positive, so its value is strictly positive.
In particular, $a=1/4$ produces its only endpoint zero at $3$ , and $a=4$ produces its only endpoint zero at $-2$ ; both zeros are excluded in part (b).

Answer

(a) $1/4<a<4$ . (b) $1/4\le a\le4$ .

Key idea

A strict inequality on an open interval can allow a zero at an excluded boundary.
Problem 2: Mark complete
Problem 3 The speed of the last section

Difficulty: 1 of 3 stars, Stretch

A route has three consecutive sections of lengths $2d$ , $3d$ , and $d$ , where $d>0$ . A vehicle travels the first section at speed $u>0$ , the second at speed $3u$ , and the third at a constant speed $w>0$ . There are no stops.

(a) Find the exact set of possible average speeds for the whole route as $w$ varies. For any attainable target speed $A$ , give the unique $w$ that attains it.

(b) Find $w$ if the target average is $9u/5$ . Can a finite final speed make the overall average equal to $2u$ ? Explain the obstruction in terms of travel time.
1. Hint 1
  The average speed is total distance divided by total time; the three section speeds cannot simply be averaged.
2. Hint 2
  The first two sections already take $3d/u$ units of time. Write the total time as $3d/u+d/w$ , then solve the average-speed formula for $w$ .
Answer

(a) Exactly $0<A<2u$ , with $w=Au/(6u-3A)$ . (b) $w=3u$ . The value $2u$ cannot be attained at any finite positive $w$ .

Full solution

The total distance is $6d$ .
The times spent on the three sections are $2d/u$ , $d/u$ , and $d/w$ , respectively.
Therefore the overall average is
$A=\frac{6d}{3d/u+d/w}=\frac{6uw}{3w+u}.$
All denominators are positive under the stated assumptions.

Since the final section takes strictly positive time, total time is strictly greater than $3d/u$ .
Thus
$0<A<6d/(3d/u)=2u.$
Solving the displayed formula gives $3Aw+Au=6uw$ , and hence
$w=Au/(6u-3A).$
For every target $0<A<2u$ , this is a unique positive finite speed.
Substitution recovers that target, proving that the full interval, with no gaps, is attainable.

For $A=9u/5$ , the formula gives $w=3u$ .
To average $2u$ over the full route would require total time exactly $3d/u$ , but the first two sections have already used all of that time.
The last positive-length section would have to take zero time.
Increasing its speed can bring the average arbitrarily close to $2u$ , but no finite speed reaches it.

Answer

(a) Exactly $0<A<2u$ , with $w=Au/(6u-3A)$ . (b) $w=3u$ . The value $2u$ cannot be attained at any finite positive $w$ .

Key idea

Solve a physical formula together with its positivity restrictions; an algebraic boundary may require an impossible zero time.
Problem 3: Mark complete
Problem 4 Two rounded readings

Difficulty: 2 of 3 stars, Challenge

A positive real number $x$ is measured in two ways. One instrument rounds $3x$ to the nearest integer, and another rounds $5x$ to the nearest integer. A value exactly halfway between consecutive integers is rounded upward. The second reported integer is exactly $7$ more than the first.

Find the complete set of possible values of $x$ . Your proof must reduce the possible integer reports to a finite list and must handle all interval endpoints correctly.

Builds on Linear Inequalities
1. Hint 1
  If the first report is $n$ , write one half-open interval for each unrounded reading.
2. Hint 2
  The two intervals for $x$ are $[(n-1/2)/3,(n+1/2)/3)$ and $[(n+13/2)/5,(n+15/2)/5)$ . They must overlap.
Answer

$x\in[31/10,19/6)\cup[33/10,37/10)\cup[23/6,39/10)$ .

Full solution

Let the first integer report be $n$ ; the second is $n+7$ .
The tie convention gives
$\frac{n-1/2}{3}\le x<\frac{n+1/2}{3},\qquad \frac{n+13/2}{5}\le x<\frac{n+15/2}{5}.$
For these intervals to overlap, each lower endpoint must be strictly smaller than the other upper endpoint.

Those two comparisons give $n<25/2$ and $n>17/2$ .
Hence the only integer reports to test are $n=9,10,11,12$ ; this is a proof of a finite bound, not a guessed search range.

Intersecting the two intervals for those four values gives, respectively,
$[31/10,19/6),\quad[33/10,7/2),\quad[7/2,37/10),\quad[23/6,39/10).$
Every value in each intersection gives exactly the stated pair of reports, so the conditions are sufficient as well as necessary.
The middle two intervals join because $7/2$ belongs to the interval for $n=11$ .
Their union gives the answer.
All other right endpoints remain excluded: at the relevant half-integer threshold an instrument rounds up and changes the required difference.

Answer

$x\in[31/10,19/6)\cup[33/10,37/10)\cup[23/6,39/10)$ .

Key idea

Rounding information is an interval constraint; derive finite bounds on the hidden integers before enumerating them.
Problem 4: Mark complete
Problem 5 Designing a disconnected solution set

Difficulty: 2 of 3 stars, Challenge

Find all real triples $(h,r,s)$ with $r>s>0$ for which the inequality
$\bigl||x-h|-r\bigr|\le s$
has solution set exactly $[-7,-3]\cup[1,5]$ .

Then give a necessary and sufficient condition on four real endpoints $A<B<C<D$ for $[A,B]\cup[C,D]$ to be representable by the same form with $r>s>0$ . Include formulas for the parameters when it is possible.
1. Hint 1
  First solve the inequality for the distance $|x-h|$ .
2. Hint 2
  The four endpoints, in order, are $h-r-s$ , $h-r+s$ , $h+r-s$ , and $h+r+s$ . Compare the lengths and centers of the two intervals.
Answer

For the given set, $(h,r,s)=(-1,4,2)$ . In general it is possible exactly when $B-A=D-C$ ; then $s=(B-A)/2$ , $h=(A+D)/2$ , and $r=(C+D-A-B)/4$ , uniquely.

Full solution

The inequality says
$r-s\le|x-h|\le r+s.$
Because $r>s>0$ , the inner radius is positive, and the solution set consists of two disjoint intervals:
$[h-r-s,h-r+s]\ \cup\ [h+r-s,h+r+s].$
Each has length $2s$ .
Thus the given intervals, both of length $4$ , force $s=2$ .
Their centers are $-5$ and $3$ , which equal $h-r$ and $h+r$ ; hence $h=-1$ and $r=4$ .
Direct substitution verifies the required set.

For general endpoints, equal interval lengths are necessary.
Suppose $B-A=D-C>0$ .
Their centers are $(A+B)/2$ and $(C+D)/2$ .
Set $h$ to the average of these centers, $r$ to half their difference, and $s$ to half the common length.
Equal lengths simplify $h$ to $(A+D)/2$ and give the stated formulas.

The gap between the intervals is $C-B=2(r-s)>0$ , so $r>s>0$ holds.
The four reconstructed endpoints are exactly $A,B,C,D$ , proving sufficiency.
Lengths and centers force all three parameters, proving uniqueness.

Answer

For the given set, $(h,r,s)=(-1,4,2)$ . In general it is possible exactly when $B-A=D-C$ ; then $s=(B-A)/2$ , $h=(A+D)/2$ , and $r=(C+D-A-B)/4$ , uniquely.

Key idea

An inverse construction is easiest when parameters are interpreted as centers, widths, and gaps.
Problem 5: Mark complete
Problem 6 Recovering two hidden locations

Difficulty: 2 of 3 stars, Challenge

Two unknown real locations satisfy $a\le b$ . An instrument reports the total distance from an input $x$ to the two locations: $D(x)=|x-a|+|x-b|$ . Its readings are
$D(0)=8,\qquad D(5)=4,\qquad D(10)=12.$
Find $a$ and $b$ and prove they are uniquely determined. Do not assume in advance that either location lies between $0$ and $10$ .
1. Hint 1
  Compare $D(0)+D(10)$ with the distance $10$ traversed once for each hidden location.
2. Hint 2
  For any real $t$ , $|t|+|10-t|\ge10$ , with equality exactly when $0\le t\le10$ . After locating both landmarks in that interval, use their sum and the reading at $5$ .
Answer

$a=2$ and $b=6$ .

Full solution

For any real $t$ , the route from $0$ to $t$ to $10$ has length
$|t|+|10-t|\ge10.$
Inspecting the three cases $t<0$ , $0\le t\le10$ , and $t>10$ shows equality occurs exactly in the middle case.

Here $D(0)+D(10)=20$ is the sum of two quantities each at least $10$ , one for $a$ and one for $b$ .
Both must equal $10$ , so
$0\le a\le b\le10.$
The reading at $0$ now gives $a+b=8$ .

Both locations cannot be at most $5$ , because then
$D(5)=10-(a+b)=2.$
Both cannot be at least $5$ , because their sum would be at least $10$ .
Consequently $a<5<b$ , and
$D(5)=b-a=4.$
Combining the sum and difference gives $a=2,b=6$ .
Their three readings are $2+6=8$ , $3+1=4$ , and $8+4=12$ , so they work.
The equality argument forced the permitted location interval, and each subsequent step was necessary; uniqueness follows.

Answer

$a=2$ and $b=6$ .

Key idea

Equality in a distance bound can locate unknown points before any equations are solved.
Problem 6: Mark complete
Problem 7 Equal changes in a mixture

Difficulty: 2 of 3 stars, Challenge

Two liquids have distinct concentrations $c$ and $d$ , measured in the same units. Mixing one unit of the first liquid with $t>0$ units of the second gives concentration
$C(t)=\frac{c+td}{1+t}.$
Volumes add and no substance is lost.

Let $r>1$ . Find all pairs $(t,r)$ for which $C(t)$ , $C(rt)$ , and $C(r^2t)$ are three distinct consecutive terms of an arithmetic progression. Prove that your condition works for any distinct concentrations $c,d$ , including when $d<c$ .
1. Hint 1
  Compare the two consecutive differences rather than solving separately for $c$ and $d$ .
2. Hint 2
  Write $C(t)=d+(c-d)/(1+t)$ . Both differences contain the same nonzero factor $(d-c)t(r-1)$ .
Answer

Exactly $r>1$ and $t=1/r$ . The three mixing ratios are then $1/r,1,r$ , and the middle concentration is $(c+d)/2$ .

Full solution

Rewriting $C(t)=d+(c-d)/(1+t)$ makes subtraction short.
The two increments are
$C(rt)-C(t)=\frac{(d-c)t(r-1)}{(1+t)(1+rt)},$ $C(r^2t)-C(rt)=\frac{(d-c)rt(r-1)}{(1+rt)(1+r^2t)}.$
Every denominator is positive.
The common factor $(d-c)t(r-1)$ is nonzero, since $c\ne d$ , $t>0$ , and $r>1$ .

Equality of the increments is therefore equivalent to
$\frac1{1+t}=\frac r{1+r^2t},$
or
$1+r^2t=r+rt.$
Factoring gives
$(r-1)(rt-1)=0.$
Since $r>1$ , necessarily $rt=1$ .

Conversely, when $t=1/r$ , this calculation reverses and the increments are equal.
They are nonzero, so all three concentrations are distinct.
Their common sign is the sign of $d-c$ : the progression increases when $d>c$ and decreases when $d<c$ .
The middle mixing ratio is $rt=1$ , giving middle concentration $(c+d)/2$ .
No cancellation discarded a permitted case.

Answer

Exactly $r>1$ and $t=1/r$ . The three mixing ratios are then $1/r,1,r$ , and the middle concentration is $(c+d)/2$ .

Key idea

For a fractional formula, comparing increments can eliminate unknown physical constants and expose the controlling ratio.
Problem 7: Mark complete
Problem 8 Agreement after different numbers of steps

Difficulty: 3 of 3 stars, Deep challenge

An unknown rule has the form $T(x)=ax+b$ , where $a,b$ are real. Write $T^{\circ j}(x)$ for the result of applying this same rule $j$ times. Fix integers $m>n\ge1$ .

For two distinct real starting values $u,v$ , it is known that
$T^{\circ m}(u)=T^{\circ n}(u),\qquad T^{\circ m}(v)=T^{\circ n}(v).$
Classify all possible rules $T$ , in terms of $m,n$ , and prove that each one satisfies the equality at every real starting value. Include constant rules.

Show by a counterexample that agreement at only one starting value would not give the same conclusion. Avoid expanding a long formula for the iterates.
1. Hint 1
  Track the difference between two starting values: one application of $T$ multiplies their difference by $a$ .
2. Hint 2
  The two given equalities imply $a^m(u-v)=a^n(u-v)$ . First classify the real possibilities for $a$ , then use the original rule to find the permitted $b$ .
Answer

For every $m>n\ge1$ , constant rules $T(x)=b$ and the identity $T(x)=x$ work. If and only if $m-n$ is even, all reflections $T(x)=-x+b$ also work. These are all possibilities. A one-input counterexample is $T(x)=2x$ at input $0$ .

Full solution

For any two inputs,
$T(u)-T(v)=a(u-v).$
Repeating this observation gives $T^{\circ j}(u)-T^{\circ j}(v)=a^j(u-v)$ for every positive integer $j$ .
Subtract the two given equalities to obtain
$(a^m-a^n)(u-v)=0.$
Since $u\ne v$ , we must have
$a^n(a^{m-n}-1)=0.$

One possibility is $a=0$ .
Then every application after the first gives $b$ , so all constant rules work because $n\ge1$ .
If $a\ne0$ , then $a^{m-n}=1$ .
A positive real number has a positive integer power equal to $1$ only when it is $1$ .
A negative real number can do so only when the exponent is even and the number is $-1$ .
Thus $a=1$ always remains, and $a=-1$ remains only when $m-n$ is even.

For $a=1$ , the rule adds $b$ at each step.
Equality of the two iterates gives $(m-n)b=0$ , so $b=0$ and the rule is the identity.
For $a=-1$ , applying the rule twice returns every input to itself.
If $m-n$ is even, the two iterate counts have the same parity, so their outputs agree for every input and every $b$ .
This verifies every classified rule and proves the global conclusion.

Finally, $T(x)=2x$ gives equality at $x=0$ for any $m,n$ , but at $x=1$ the outputs are the distinct numbers $2^m$ and $2^n$ .
One starting value cannot justify the global conclusion.

Answer

For every $m>n\ge1$ , constant rules $T(x)=b$ and the identity $T(x)=x$ work. If and only if $m-n$ is even, all reflections $T(x)=-x+b$ also work. These are all possibilities. A one-input counterexample is $T(x)=2x$ at input $0$ .

Key idea

For affine rules, differences remove the constant term; two distinct test inputs can force a global identity.
Problem 8: Mark complete
Problem 9 Agreement between imperfect instruments

Difficulty: 3 of 3 stars, Deep challenge

(a) Prove the following statement for a finite collection of nonempty closed, bounded intervals on the real line: if every pair of intervals overlaps, then all the intervals have at least one common point.

(b) Three instruments intended to measure $2x$ , $3x$ , and $5x$ report $7$ , $2$ , and $20$ , respectively. Every instrument has the same allowed absolute error $e\ge0$ . Find the least $e$ for which there is a real $x$ satisfying
$|2x-7|\le e,\qquad |3x-2|\le e,\qquad |5x-20|\le e.$
Find every such $x$ at the least error. Use part (a) to explain why checking compatibility two instruments at a time is sufficient here.

Builds on Linear Inequalities
1. Hint 1
  For part (a), look at the largest lower endpoint and the smallest upper endpoint.
2. Hint 2
  Two readings $px\approx u$ and $qx\approx v$ , with $p,q>0$ , allow a common $x$ exactly when $|qu-pv|\le(p+q)e$ . Derive this by comparing interval endpoints.
Answer

The least error is $e=25/4$ , with the unique true value $x=11/4$ .

Full solution

Write the intervals as $[l_i,u_i]$ .
Let $L$ be the largest lower endpoint and $U$ the smallest upper endpoint; both occur because the collection is finite.
The interval providing $L$ overlaps the interval providing $U$ , so $L\le U$ .
Every point of $[L,U]$ belongs to every interval.
This proves (a).

For part (b), the three possible-input intervals are $[(7-e)/2,(7+e)/2]$ , $[(2-e)/3,(2+e)/3]$ , and $[(20-e)/5,(20+e)/5]$ .
More generally, the intervals from $px\approx u$ and $qx\approx v$ overlap exactly when both cross-endpoint inequalities hold.
Multiplying by the positive $pq$ combines them into
$|qu-pv|\le(p+q)e.$

The three pairwise conditions are $e\ge17/5$ , $e\ge5/7$ , and $e\ge25/4$ .
By part (a), their conjunction is also sufficient, so the least error is $25/4$ .
At this value the second interval ends at $(2+25/4)/3=11/4$ , exactly where the third begins,
$(20-25/4)/5=11/4.$
Thus $x$ is forced.
The remaining error is $|2(11/4)-7|=3/2\le25/4$ , confirming feasibility.

Answer

The least error is $e=25/4$ , with the unique true value $x=11/4$ .

Key idea

One-dimensional interval geometry can turn a many-condition feasibility problem into pairwise compatibility tests.
Problem 9: Mark complete
Problem 10 A finite test for infinitely many inequalities

Difficulty: 3 of 3 stars, Deep challenge

Let $t_1<t_2<\cdots<t_n$ be real numbers with $n\ge2$ , and let $w_1,\ldots,w_n$ be positive real numbers. Define
$G(x)=\sum_{i=1}^{n}w_i|x-t_i|.$
Here the summation means to add one term for each $i=1,\ldots,n$ .

(a) Prove that a line $L(x)=ax+b$ satisfies $|L(x)|\le G(x)$ for every real $x$ if and only if it satisfies the inequality at the $n$ inputs $t_1,\ldots,t_n$ . Your proof must address the two unbounded outside intervals.

(b) Apply your result to find all real pairs $(a,b)$ for which
$|ax+b|\le|x-2|+2|x+1|$
holds for every real $x$ . Give the allowed region by exact linear inequalities.
1. Hint 1
  Between successive landmarks, $G$ is affine. Control both $G-L$ and $G+L$ there.
2. Hint 2
  Put $W=w_1+\cdots+w_n$ . Show that $G(t_1)+G(t_n)=W(t_n-t_1)$ ; the two endpoint tests then force $|a|\le W$ .
Answer

(b) Exactly the pairs satisfying $-3\le b-a\le3$ and $-6\le2a+b\le6$ . Part (a) is the stated equivalence.

Full solution

Necessity is immediate.
For sufficiency, assume all landmark tests hold and write
$W=\sum_iw_i.$
Between consecutive landmarks, each absolute-value term is affine, so both $G-L$ and $G+L$ are affine.
They are nonnegative at both endpoints and hence throughout that interval.

The outside intervals require an extra argument.
Since every $t_i$ lies between the extreme landmarks,
$G(t_1)+G(t_n)=\sum_iw_i\bigl((t_i-t_1)+(t_n-t_i)\bigr)=W(t_n-t_1).$
Therefore
$|a|(t_n-t_1)=|L(t_n)-L(t_1)|\le G(t_n)+G(t_1),$
giving $|a|\le W$ because $t_n>t_1$ .

For $x\ge t_n$ , the slope of $G$ is $W$ .
The slopes of $G-L$ and $G+L$ are $W-a$ and $W+a$ , both nonnegative, so neither can fall below its nonnegative value at $t_n$ .
For $x\le t_1$ , their slopes are $-W-a$ and $-W+a$ , both nonpositive; moving left cannot lower them.
Thus the inequality holds everywhere.

In (b), the landmarks are $-1$ and $2$ , and the values of the right side are $3$ and $6$ .
The theorem gives exactly $|b-a|\le3$ and $|2a+b|\le6$ , which are the four stated linear inequalities.
No separate test at infinity is needed: its slope restriction follows from these endpoint tests.

Answer

(b) Exactly the pairs satisfying $-3\le b-a\le3$ and $-6\le2a+b\le6$ . Part (a) is the stated equivalence.

Key idea

Finite breakpoint tests need a separate tail argument; endpoint values can supply the missing slope bound.
Problem 10: Mark complete