Star problems Advanced ← Back to chapter

The Language of Algebra: 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.

0 of 10 completed

Progress saved in this browser.

Problem 1 of 10

Problem 1 Three nearby products

Difficulty: 1 of 3 stars, Stretch

Without calculating any of the four-digit products separately, evaluate
$\frac{2027\cdot2032-2029\cdot2030}{2028\cdot2031-2029\cdot2030}.$
Explain why the answer is unchanged if every number appearing in the expression is increased by the same real number.
1. Hint 1
  All three products have factors with the same sum. Give the smallest number a short name.
2. Hint 2
  Write the three products as $n(n+5)$ , $(n+1)(n+4)$ , and $(n+2)(n+3)$ . Compare their expanded forms.
Answer

$3$ , unchanged under every common real shift.

Full solution

Let $n$ be the smallest number.
The three products are $n(n+5)=n^2+5n$ , $(n+1)(n+4)=n^2+5n+4$ , and
$(n+2)(n+3)=n^2+5n+6.$
They share the entire variable part; only the constants differ.

The numerator is therefore $-6$ , and the denominator is $-2$ .
Their quotient is $3$ .
Increasing every displayed number by a real number $t$ merely replaces $n$ by $n+t$ .
The same cancellations still occur, so the quotient remains $3$ .
The denominator stays $-2$ even if some individual factors become zero or negative, so no common shift creates an undefined expression.

Answer

$3$ , unchanged under every common real shift.

Key idea

Products with the same sum of factors often share a common variable part after expansion.
Problem 1: Mark complete
Problem 2 Information hidden in a ratio

Difficulty: 1 of 3 stars, Stretch

Positive real numbers $a$ and $b$ satisfy $\frac ab+\frac ba=7$ . Find the exact values of
$\frac{(a+b)^2}{a^2+b^2}\qquad\text{and}\qquad\frac{(a-b)^2}{a^2+b^2}.$
Explain why there is no need to determine $a$ or $b$ separately.

Builds on Algebraic Fractions
1. Hint 1
  Combine the two fractions in the given information.
2. Hint 2
  The condition gives $a^2+b^2=7ab$ . Expand each requested numerator.
Answer

$\frac97$ and $\frac57$ , respectively.

Full solution

Combining the given fractions gives $(a^2+b^2)/(ab)=7$ , so
$a^2+b^2=7ab.$
The product $ab$ is positive and may safely be canceled.

Expand
$(a+b)^2=a^2+2ab+b^2=9ab.$
Dividing by $a^2+b^2=7ab$ gives $9/7$ .
Similarly, $(a-b)^2=a^2-2ab+b^2=5ab$ , which gives $5/7$ .

The condition determines the ratio between $a^2+b^2$ and $ab$ , and that ratio is all either answer uses.
Individual values of $a$ and $b$ would contain unnecessary information.
In fact, multiplying both numbers by any positive factor preserves both the hypothesis and the requested values.

Answer

$\frac97$ and $\frac57$ , respectively.

Key idea

Determine the combination of quantities a question needs before trying to determine the quantities themselves.
Problem 2: Mark complete
Problem 3 When roots preserve addition

Difficulty: 1 of 3 stars, Stretch

Let $n$ be a positive integer. For odd $n$ , $\sqrt[n]{u}$ means the unique real $n$ th root of $u$ ; for even $n$ , it means the nonnegative real $n$ th root, defined when $u\ge0$ .

For each $n$ , find all pairs of real numbers $(a,b)$ for which
$\sqrt[n]{a^n}+\sqrt[n]{b^n}=\sqrt[n]{(a+b)^n}.$
Explain why every displayed root is defined, and prove your classification.

Builds on Fractional Exponents and Radicals
1. Hint 1
  The answer depends on whether the power retains or loses the sign of its input.
2. Hint 2
  For even $n$ , compare the cases in which $a,b$ have the same sign or opposite signs. In the opposite-sign case, write the two nonzero magnitudes as positive numbers.
Answer

For odd $n$ , every real pair works. For even $n$ , exactly the pairs with $ab\ge0$ work, including pairs with a zero entry.

Full solution

If $n$ is odd, taking an $n$ th power and its unique real $n$ th root returns the original number, including its sign.
Thus the equation becomes $a+b=a+b$ , and every real pair works.
All odd roots are defined.

If $n$ is even, every quantity inside a root is an even power and is therefore nonnegative, so again all displayed roots exist.
The root of a powered number equals that number when it is nonnegative and its negative when it is negative.

If $a,b\ge0$ , the equation becomes $a+b=a+b$ .
If $a,b\le0$ , it becomes
$-a-b=-(a+b).$
These cases include zero entries.
If $a,b$ have opposite nonzero signs, let their positive magnitudes be $u,v$ .
The left side is $u+v$ , but the right side is $u-v$ when $u\ge v$ and $v-u$ when $v\ge u$ .
In either case it is strictly smaller, because both $u,v$ are positive.

Therefore, for even $n$ , the equation holds exactly when the two numbers have the same weak sign, equivalently $ab\ge0$ .

Answer

For odd $n$ , every real pair works. For even $n$ , exactly the pairs with $ab\ge0$ work, including pairs with a zero entry.

Key idea

An even power loses sign information; track that loss before moving a root across addition.
Problem 3: Mark complete
Problem 4 Three fractions that cooperate

Difficulty: 2 of 3 stars, Challenge

Positive real numbers $x,y,z$ satisfy $xyz=1$ . Prove that
$\frac1{1+x+xy}+\frac1{1+y+yz}+\frac1{1+z+zx}=1.$
Your proof should explain why three apparently different denominators can be coordinated.

Builds on Algebraic Fractions
1. Hint 1
  Try multiplying one of the denominators by one of the variables.
2. Hint 2
  If $D=1+x+xy$ , then $x(1+y+yz)=D$ . Find a similar relation for the third denominator.
Answer

The sum is always $1$ .

Full solution

Let $D=1+x+xy$ .
The condition $xyz=1$ gives
$x(1+y+yz)=x+xy+xyz=D.$

It also gives
$xy(1+z+zx)=xy+xyz+x^2yz=xy+1+x=D.$

Since all variables are positive, $D$ and all three original denominators are positive.
We may therefore rewrite the three fractions as $1/D$ , $x/D$ , and $xy/D$ , respectively.
Their sum is
$(1+x+xy)/D=1.$

The useful common denominator did not come from multiplying three large expressions together.
Instead, the product condition supplied exactly the scaling factors needed to turn each denominator into $D$ .
Keeping the given relation available during simplification prevents unnecessary expansion.

Answer

The sum is always $1$ .

Key idea

A constraint linking variables may reveal a much smaller common denominator than a mechanical calculation would produce.
Problem 4: Mark complete
Problem 5 Four number cards and a constant

Difficulty: 2 of 3 stars, Challenge

Choose four distinct cards from $1,2,3,4,5,6$ and place their numbers in the positions $a,b,c,d$ , with $a<b$ and $c<d$ . You want
$(x+a)(x+b)-(x+c)(x+d)$
to have the same value for every real $x$ .

What is the largest possible value of that constant? Find every placement that attains it, and prove that no larger value is possible.
1. Hint 1
  Which coefficient must disappear when you expand the expression?
2. Hint 2
  The pairs must have equal sums. List the possible distinct pairs for each shared sum, and compare their products.
Answer

The maximum is $6$ , attained only by $(a,b,c,d)=(3,4,1,6)$ .

Full solution

Expansion gives $(a+b-c-d)x+(ab-cd)$ .
This is constant exactly when $a+b=c+d$ ; otherwise changing $x$ changes the value.
We must therefore compare products of disjoint pairs having the same sum.

The shared sum must be between $5$ and $9$ : smaller or larger sums provide only one available pair.
At sum $5$ , the pairs are $(1,4),(2,3)$ and their products differ by $2$ .
At sum $6$ , the pairs $(1,5),(2,4)$ give difference $3$ .
At sum $7$ , the pairs $(1,6),(2,5),(3,4)$ have products $6,10,12$ , so the largest difference is $6$ .

At sum $8$ , $(2,6),(3,5)$ give difference $3$ ; at sum $9$ , $(3,6),(4,5)$ give difference $2$ .
These cases exhaust all equal-sum choices.
Thus the maximum positive difference is $12-6=6$ , and the ordering conditions give the unique placement $(3,4,1,6)$ .
Substitution confirms that its expression is always $6$ .

Answer

The maximum is $6$ , attained only by $(a,b,c,d)=(3,4,1,6)$ .

Key idea

First translate a condition that holds for every input into a restriction on coefficients; then optimize within that smaller set.
Problem 5: Mark complete
Problem 6 Zeros that disappear with the denominator

Difficulty: 2 of 3 stars, Challenge

Consider the expression
$R=\frac{\dfrac{x^2-4}{x-2}-\dfrac{x^2-9}{x-3}}{\dfrac1{x-2}-\dfrac1{x-3}}.$
(a) Find its exact real domain, checking the large denominator as well as the small ones.

(b) Simplify $R$ . Does the original expression ever equal zero? Explain why the simplified formula alone could lead to a wrong answer.

Builds on Algebraic Fractions, Expanding and Factoring Expressions
1. Hint 1
  You can check the factorizations of $x^2-4$ and $x^2-9$ by multiplication.
2. Hint 2
  On the original domain, the large numerator becomes $-1$ . Put the large denominator over $(x-2)(x-3)$ .
Answer

Domain: $x\ne2,3$ . On that domain, $R=(x-2)(x-3)$ , and $R$ is never zero.

Full solution

The small denominators require $x\ne2$ and $x\ne3$ .
For such $x$ , the large denominator is
$\frac1{x-2}-\frac1{x-3}=\frac{(x-3)-(x-2)}{(x-2)(x-3)}=-\frac1{(x-2)(x-3)}.$

This is never zero, so it adds no further exclusions.
Multiplication verifies $x^2-4=(x-2)(x+2)$ and
$x^2-9=(x-3)(x+3).$
Consequently the large numerator is $(x+2)-(x+3)=-1$ , and division gives $R=(x-2)(x-3)$ on the stated domain.

A product of two real numbers is zero only if one of its factors is zero.
Here that would require $x=2$ or $x=3$ , but both were excluded by the original expression.
Thus $R$ never vanishes.
Simplification preserves values where the original exists; it does not automatically add the missing inputs.

Answer

Domain: $x\ne2,3$ . On that domain, $R=(x-2)(x-3)$ , and $R$ is never zero.

Key idea

Carry domain restrictions through every cancellation, especially when studying the zeros of a simplified expression.
Problem 6: Mark complete
Problem 7 Build an identity from three tests

Difficulty: 2 of 3 stars, Challenge

Find real constants $A,B,C$ such that
$A(x-1)(x-2)+B(x-2)(x-3)+C(x-1)(x-3)=1$
for every real $x$ . Prove both that your constants work for every $x$ and that there is no other choice.
1. Hint 1
  Choose inputs that make two of the three terms disappear.
2. Hint 2
  Testing $x=1,2,3$ determines the three constants separately. Afterward, expand to verify the full identity.
Answer

$A=B=\frac12$ and $C=-1$ .

Full solution

Any identity valid for every real $x$ must work at $x=1$ .
At that input only the term with $B$ remains, giving $2B=1$ , so $B=1/2$ .
At $x=2$ only the term with $C$ remains, giving $-C=1$ , so $C=-1$ .
At $x=3$ only the term with $A$ remains, giving $2A=1$ , so $A=1/2$ .
Every possible solution is therefore forced to use these constants.

Testing three inputs alone is not our verification for all inputs.
Substitute the constants and expand the entire left side:
$\tfrac12(x^2-3x+2)+\tfrac12(x^2-5x+6)-(x^2-4x+3)=1.$

The $x^2$ terms and the $x$ terms both cancel, and the constants total $1$ .
Hence the candidate works for every real $x$ .
The forced values from the three special inputs establish uniqueness.

Answer

$A=B=\frac12$ and $C=-1$ .

Key idea

Choose substitutions that eliminate terms to discover a candidate, then use an identity to prove it works everywhere.
Problem 7: Mark complete
Problem 8 A zero sum fixes a cubic-looking ratio

Difficulty: 3 of 3 stars, Deep challenge

Nonzero real numbers $a,b,c$ satisfy $a+b+c=0$ . Find and prove the value of
$\frac{a^2}{bc}+\frac{b^2}{ca}+\frac{c^2}{ab}.$
Does having that value force $a+b+c=0$ ? If not, give a counterexample. You may derive any needed identity directly by expansion.

Builds on Algebraic Fractions
1. Hint 1
  Use the common denominator $abc$ and replace $c$ by $-(a+b)$ .
2. Hint 2
  Expand $(a+b)^3$ by multiplying $(a+b)^2$ by $(a+b)$ . Compare $a^3+b^3+c^3$ with $abc$ .
Answer

The value is $3$ . The converse fails; for example, $a=b=c=1$ gives $3$ but has sum $3$ .

Full solution

Because $abc\ne0$ , the requested expression is $(a^3+b^3+c^3)/(abc)$ .
The zero-sum condition says $c=-(a+b)$ .
Direct multiplication gives $(a+b)^3=a^3+3a^2b+3ab^2+b^3$ , so
$a^3+b^3+c^3=a^3+b^3-(a+b)^3=-3ab(a+b)=3abc.$

Dividing by $abc$ proves that the expression equals $3$ .
This computation is valid even though the variables cannot all be positive: no ordering or positivity was assumed, and all denominators are nonzero by hypothesis.

The converse is false.
Taking $a=b=c=1$ makes each of the three fractions equal $1$ , so their sum is again $3$ , while $a+b+c=3$ .
Proving that one condition guarantees an outcome does not prove that the condition is the only way to obtain that outcome.

Answer

The value is $3$ . The converse fails; for example, $a=b=c=1$ gives $3$ but has sum $3$ .

Key idea

Use a linear relation to eliminate a variable inside a higher-degree expression, and distinguish a statement from its converse.
Problem 8: Mark complete
Problem 9 Recover three integers from symmetric clues

Difficulty: 3 of 3 stars, Deep challenge

Find all ordered triples of nonnegative integers $(a,b,c)$ satisfying
$a+b+c=10,\qquad ab+bc+ca=31.$
Your solution must prove that the list is complete; guessing a triple is not enough.
1. Hint 1
  Expand $(a+b+c)^2$ to obtain a useful restriction involving squares.
2. Hint 2
  First put the numbers in increasing order. The smallest is $0,1,2$ , or $3$ . For each choice, determine the sum and product of the other two.
Answer

All six permutations of $(2,3,5)$ .

Full solution

The square of the first equation gives $100=a^2+b^2+c^2+62$ , so the sum of squares is $38$ .
The equations are unchanged by permuting the numbers.
We may therefore first suppose
$a\le b\le c.$
Since their sum is $10$ , the smallest number $a$ is one of $0,1,2,3$ .

If $a=0$ , then $b+c=10$ and $bc=31$ .
But $(b-c)^2=(b+c)^2-4bc=100-124<0$ , impossible for real numbers.
If $a=1$ , then $b+c=9$ and $bc=22$ , giving $(b-c)^2=81-88<0$ , also impossible.

If $a=2$ , then $b+c=8$ and $bc=15$ .
The positive integer factor pairs of $15$ are $(1,15)$ and $(3,5)$ , and only $(3,5)$ has sum $8$ .
If $a=3$ , the other two numbers have sum $7$ and product $10$ , forcing $(2,5)$ ; that contradicts $a\le b$ .

Thus the sole increasing triple is $(2,3,5)$ .
It satisfies both clues, and its three distinct entries have six different orderings.
These are all the ordered solutions.

Answer

All six permutations of $(2,3,5)$ .

Key idea

Symmetric clues let you impose an order temporarily; a bound on the smallest entry can turn an infinite search into a few complete cases.
Problem 9: Mark complete
Problem 10 When two nested radicals add to an integer

Difficulty: 3 of 3 stars, Deep challenge

Find every positive integer $q$ for which
$\sqrt{30+\sqrt q}+\sqrt{30-\sqrt q}$
is a real integer. Give the integer value for each $q$ , and justify that no others work. All square roots are nonnegative real square roots.

Builds on Fractional Exponents and Radicals
1. Hint 1
  Call the whole expression $k$ and square it. Also determine which $q$ make both radicals real.
2. Hint 2
  You obtain $k^2=60+2\sqrt{900-q}$ . Bound the possible integers $k$ , then test them without losing the sign condition on the remaining square root.
Answer

$q=896$ gives $8$ , and $q=500$ gives $10$ .

Full solution

The radicals are real exactly when $q\le900$ ; combined with the hypothesis,
$1\le q\le900.$
Let their sum be the integer $k$ .
Squaring and using nonnegative square roots gives
$k^2=60+2\sqrt{(30+\sqrt q)(30-\sqrt q)}=60+2\sqrt{900-q}.$

Thus
$60\le k^2<120.$
Since $k$ is nonnegative, the only possible integer values are $8,9,10$ .
For $k=8$ , the equation gives $\sqrt{900-q}=2$ , so $q=896$ .
For $k=9$ , it gives $\sqrt{900-q}=21/2$ , so $q=900-441/4=3159/4$ , which is not an integer.
For $k=10$ , it gives $\sqrt{900-q}=20$ , so $q=500$ .

Both remaining candidates lie in the original domain.
At $q=896$ the square of the original, nonnegative expression is $60+2\sqrt4=64$ , so its value is $8$ .
At $q=500$ its square is $60+2\sqrt{400}=100$ , so its value is $10$ .
This verifies the candidates in the original expression and completes the classification.

Answer

$q=896$ gives $8$ , and $q=500$ gives $10$ .

Key idea

Bound an integer-valued expression before searching; after squaring, retain the sign and domain information needed to verify candidates.
Problem 10: Mark complete