Triangle Inequality - Math 1

1. Triangle ABC has sides of lengths $a$, $b$, and $c$. If $a + b = 15$ and $c = 8$, what is the range of possible values for side $a$?
   Answer: $3.5 < a < 11.5$ Given $a + b = 15$ and $c = 8$, we have $b = 15 - a$. For a valid triangle, three conditions must hold: (1) $a + b > c$ gives $a + (15-a) > 8$, which simplifies to $15 > 8$ (always true). (2) $a + c > b$ gives $a + 8 > 15 - a$, which simplifies to $2a > 7$, so $a > 3.5$. (3) $b + c > a$ gives $(15-a) + 8 > a$, which simplifies to $23 > 2a$, so $a < 11.5$. We also need $a > 0$ and $b = 15 - a > 0$, giving $0 < a < 15$. Combining all constraints: $3.5 < a < 11.5$.
2. A student incorrectly states: "If three lengths $a \leq b \leq c$ satisfy $a + b > c$, then they form a triangle." Which of the following represents the most likely reasoning error behind this incorrect statement?
   Answer: The student only checked one of the three triangle inequality conditions The triangle inequality theorem states that for any triangle with sides $a$, $b$, and $c$, all three conditions must hold: $a + b > c$, $a + c > b$, and $b + c > a$. When $a \leq b \leq c$, the condition $a + b > c$ is indeed the most restrictive (since $c$ is the largest side), but the other two conditions $a + c > b$ and $b + c > a$ are not automatically satisfied just because $a + b > c$ holds. The student's error is checking only one of the three required triangle inequality conditions. While this one condition is the most likely to be violated when $c$ is the longest side, it's not sufficient by itself to guarantee that the three lengths form a triangle.
3. A triangle has integer side lengths. Two of the sides have lengths $7$ and $12$. How many possible integer values are there for the length of the third side?
   Answer: $13$ possible values Let the third side have length $c$. By triangle inequality: (1) $7 + 12 > c$, so $c < 19$. (2) $7 + c > 12$, so $c > 5$. (3) $12 + c > 7$, so $c > -5$ (automatically satisfied for positive $c$). Therefore, $5 < c < 19$. Since $c$ must be a positive integer, we have $c \in \{6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18\}$. Counting these values gives us $18 - 6 + 1 = 13$ possible integer values for the third side.
4. A triangle has sides with lengths in the ratio $3:4:x$. For what values of $x$ can this triangle exist?
   Answer: $1 < x < 7$ Let the sides have lengths $3k$, $4k$, and $xk$ for some positive constant $k$. Since $k > 0$, we can divide all triangle inequalities by $k$ to get inequalities in terms of the ratios $3$, $4$, and $x$. The triangle inequalities become: (1) $3 + 4 > x$, so $x < 7$. (2) $3 + x > 4$, so $x > 1$. (3) $4 + x > 3$, so $x > -1$. Since $x$ represents a ratio of lengths, $x > 0$. Combining these constraints: $x > 0$, $x > 1$, and $x < 7$, we get $1 < x < 7$. Choice B incorrectly allows $x \leq 1$, choice C misses the upper bound, and choice D misses both the lower bound $x > 1$ and upper bound.
5. A triangle has sides with lengths in the ratio $3:4:k$ where $k$ is a positive real number. What is the range of possible values for $k$?
   Answer: $1 < k < 7$ Let the sides have lengths $3t$, $4t$, and $kt$ where $t > 0$. By the triangle inequality: (1) $3t + 4t > kt$ gives $7t > kt$, so $k < 7$. (2) $3t + kt > 4t$ gives $kt > t$, so $k > 1$. (3) $4t + kt > 3t$ gives $kt > -t$, so $k > -1$ (automatically satisfied since $k > 0$). Therefore, $1 < k < 7$.
6. Three towns A, B, and C are located such that the distance from A to B is $15$ km, the distance from B to C is $8$ km, and the distance from A to C is $d$ km. If the three towns do NOT form a triangle (i.e., they are collinear), what are the possible values of $d$?
   Answer: $d = 7$ or $d = 23$ For three points to be collinear (not form a triangle), the triangle inequality must become an equality. This happens when one distance equals the sum of the other two. We have three cases: (1) $d = 15 + 8 = 23$ (C is beyond B from A), (2) $15 = d + 8$, so $d = 7$ (C is between A and B), (3) $8 = 15 + d$, so $d = -7$ (impossible since distance is positive). Therefore, $d = 7$ or $d = 23$. Choice C gives the range where they DO form a triangle, choice D gives where they don't form a triangle but includes impossible values, and choice B gives incorrect specific values.
7. A triangle has two sides of lengths 7 and 12. If the third side has an integer length, how many possible triangles can be formed?
   Answer: 17 triangles Let the third side have length $s$. By the triangle inequality: (1) $7 + 12 > s$ gives $s < 19$. (2) $7 + s > 12$ gives $s > 5$. (3) $12 + s > 7$ gives $s > -5$ (always satisfied for positive $s$). Combining: $5 < s < 19$. Since $s$ must be a positive integer, $s \in \{6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18\}$. Counting these values: there are 17 possible integer lengths.
8. In a triangle, the longest side is twice the length of the shortest side. If the shortest side has length $s$ and the middle side has length $m$, what constraints must $m$ satisfy?
   Answer: $s < m < 2s$ When you see a triangle problem involving side lengths, you need to apply the triangle inequality theorem, which states that the sum of any two sides must be greater than the third side. Given that the shortest side has length $s$, the middle side has length $m$, and the longest side has length $2s$, you must check all three triangle inequality conditions: First, $s + m > 2s$, which simplifies to $m > s$. Second, $s + 2s > m$, giving us $m < 3s$. Third, $m + 2s > s$, which is automatically satisfied since all lengths are positive. From the first two conditions, we get $s < m < 3s$. However, since we're told that $m$ is the middle side length, it must be between the shortest and longest sides: $s < m < 2s$. Looking at the answer choices: Choice A gives $\frac{s}{2} < m < \frac{3s}{2}$, but this allows $m$ to be less than $s$, violating our requirement that $m$ is the middle length. Choice B includes $m = s$ and $m = 2s$, but these would make sides equal, contradicting the distinct shortest, middle, and longest designations. Choice C gives $s < m < 3s$, which satisfies the triangle inequality but doesn't ensure $m$ is actually the middle side length—it could be longer than $2s$. Choice D correctly captures both the triangle inequality constraints and ensures $m$ is truly the middle side. Study tip: Always check that your constraints match the given relationships between sides, not just the triangle inequality.
9. Three towns A, B, and C are connected by straight roads. The distance from A to B is 15 km, and from B to C is 8 km. If a new road is built directly from A to C, and the three roads form a triangle, what is the range of possible lengths for the new road?
   Answer: $7 < AC < 23$ When you see three points forming a triangle with two known side lengths, you're dealing with the triangle inequality theorem. This fundamental geometric principle states that the sum of any two sides of a triangle must be greater than the third side, and the difference of any two sides must be less than the third side. With $AB = 15$ km and $BC = 8$ km, you need to find the possible values for $AC$. The triangle inequality gives you two key constraints: First: $AB + BC > AC$, so $15 + 8 > AC$, which means $AC < 23$ Second: $AC + BC > AB$, so $AC + 8 > 15$, which means $AC > 7$ Therefore, $7 < AC < 23$. Looking at the wrong answers: Choice A ($0 < AC < 23$) ignores the lower bound completely—you can't have a triangle where one side is extremely small compared to the others. Choice B ($7 ≤ AC ≤ 23$) incorrectly includes the boundary values. If $AC = 7$, then you'd have $7 + 8 = 15$, which creates a straight line, not a triangle. Similarly, if $AC = 23$, you get $15 + 8 = 23$, again forming a straight line. Choice C ($8 < AC < 15$) severely restricts the range and misses many valid triangle configurations. The correct answer is D: $7 < AC < 23$. Study tip: For triangle inequality problems, always check that the sum of any two sides exceeds the third, and remember that equality conditions create degenerate triangles (straight lines), so use strict inequalities for true triangles.
10. Three sticks have lengths $a$, $b$, and $c$ where $a \leq b \leq c$. If $a + b = c + 2$, which of the following best describes when these sticks can form a triangle?
    Answer: Always when all lengths are positive, since $a + b > c$ and the other inequalities follow from $a \leq b \leq c$ Given $a + b = c + 2$ and $a \leq b \leq c$, we need to check all three triangle inequalities. First, $a + b > c$ becomes $(c + 2) > c$, which simplifies to $2 > 0$, always true. Second, $a + c > b$: since $c \geq b$, we have $a + c \geq a + b > b$ (the last inequality holds when $a > 0$). Third, $b + c > a$: since $b \geq a$ and $c > 0$, we have $b + c \geq a + c > a$. Therefore, when all lengths are positive, all triangle inequalities are satisfied. Choice A is incomplete reasoning, B incorrectly concludes never, and C adds an unnecessary condition on $a > 2$.
11. The sides of triangle PQR have lengths $a$, $a+d$, and $a+2d$ where $a > 0$ and $d > 0$. Which condition must be satisfied for these to form a valid triangle?
    Answer: $d < a$ The sides are $a$, $a+d$, and $a+2d$ with $a, d > 0$. Since $a+2d$ is the longest side, the critical triangle inequality is $a + (a+d) > a+2d$, which simplifies to $2a + d > a + 2d$, giving $a > d$, or equivalently $d < a$. The other inequalities: $a + (a+2d) > a+d$ gives $2a + 2d > a + d$, so $a + d > 0$ (always true), and $(a+d) + (a+2d) > a$ gives $2a + 3d > a$, so $a + 3d > 0$ (always true). Therefore, the condition is $d < a$.

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