AMC 10 · 2008 · #15
Grade 8 number-theoryHow many right triangles have integer leg lengths a and b and a hypotenuse of length b+1, where b<100?
Pick an answer.
AMC 10 2008 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A right triangle has legs of whole-number lengths $a$ and $b$, and its hypotenuse is exactly one longer than the leg $b$, so the hypotenuse is $b+1$. With the extra rule $b<100$, count how many such triangles exist.
Givens: The two legs $a$ and $b$ are positive integers.; The hypotenuse has length $b+1$, one more than leg $b$.; The leg $b$ satisfies $b<100$.
Unknowns: The number of triangles $(a,b,b+1)$ that meet all the rules.
Understand
Restated: A right triangle has legs of whole-number lengths $a$ and $b$, and its hypotenuse is exactly one longer than the leg $b$, so the hypotenuse is $b+1$. With the extra rule $b<100$, count how many such triangles exist.
Givens: The two legs $a$ and $b$ are positive integers.; The hypotenuse has length $b+1$, one more than leg $b$.; The leg $b$ satisfies $b<100$.
Plan
Primary tool: #4 Introduce a Variable
Secondary: #2 Make a Systematic List, #3 Eliminate Possibilities
The three sides are already named $a$, $b$, and $b+1$, so tool #4 (Introduce a Variable) leads: writing the Pythagorean relation $a^2+b^2=(b+1)^2$ and cleaning it up collapses the whole problem into one tidy equation, $a^2=2b+1$. That equation ties $b$ directly to $a$, turning 'count the triangles' into 'count the allowed values of $a$'. Tool #3 (Eliminate Possibilities) then throws out every even $a$, because $2b+1$ is odd, so $a$ must be odd for $b$ to be a whole number. Finally tool #2 (Make a Systematic List) walks the odd values of $a$ that keep $b$ a positive integer under $100$ and counts them.
Execute — Answer: A
8.G.B.7 Step 1 Write the Pythagorean equation
- The legs are $a$ and $b$ and the hypotenuse is $b+1$.
- For a right triangle the two legs squared add up to the hypotenuse squared, so $a^2+b^2=(b+1)^2$.
- This single equation captures every triangle the problem allows.
💡 In any right triangle the squares on the two legs together equal the square on the hypotenuse.
6.EE.A.3 Step 2 Simplify to a clean form
- Expand the right side: $(b+1)^2=b^2+2b+1$.
- The equation becomes $a^2+b^2=b^2+2b+1$.
- The $b^2$ terms are on both sides, so they cancel, leaving $a^2=2b+1$.
- Solving for $b$ gives $b=\frac{a^2-1}{2}$, which says: once you pick a leg $a$, the other leg $b$ is completely determined.
💡 Because the two $b^2$ terms match, they wash out and the messy square shrinks to a simple line.
6.EE.B.5 Step 3 Keep only odd legs a
- The value $2b+1$ is always odd, and it equals $a^2$.
- A square is odd only when its root is odd, so $a$ must be odd.
- Checked the other way: $b=\frac{a^2-1}{2}$ is a whole number only when $a^2-1$ is even, which again forces $a$ to be odd.
- Every even $a$ is thrown out.
- Also $a=1$ gives $b=0$, which is no triangle, so the smallest usable leg is $a=3$.
💡 Half of an even number is whole, so $a^2-1$ must be even, which only happens when $a$ is odd.
7.EE.B.4 Step 4 List the odd legs under the limit
- Now apply $b<100$.
- Since $b=\frac{a^2-1}{2}$, the rule $b<100$ means $\frac{a^2-1}{2}<100$, so $a^2<201$, giving $a\le 14$.
- Combined with 'odd and at least $3$', the allowed legs are $a=3,5,7,9,11,13$.
- Each one gives a valid triangle: $a=3\to b=4$, $a=5\to b=12$, $a=7\to b=24$, $a=9\to b=40$, $a=11\to b=60$, $a=13\to b=84$.
- That is $6$ triangles, so the answer is $(A)$.
💡 The limit $b<100$ caps $a$ at $13$, and counting the odd legs from $3$ up gives the total directly.
8.G.B.7 The legs are $a$ and $b$ and the hypotenuse is $b+1$. For a right triangle the t 6.EE.A.3 Expand the right side: $(b+1)^2=b^2+2b+1$. The equation becomes $a^2+b^2=b^2+2b+ 6.EE.B.5 The value $2b+1$ is always odd, and it equals $a^2$. A square is odd only when i 7.EE.B.4 Now apply $b<100$. Since $b=\frac{a^2-1}{2}$, the rule $b<100$ means $\frac{a^2- Review
Reasonableness: Each listed pair really is a right triangle: $(3,4,5)$, $(5,12,13)$, $(7,24,25)$, $(9,40,41)$, $(11,60,61)$, $(13,84,85)$ — every one is a Pythagorean triple, and in each the hypotenuse is exactly one more than the longer leg $b$, matching the rule. The next odd leg $a=15$ would give $b=112$, which breaks $b<100$, so stopping at $13$ is correct. Six triangles matches choice $(A)$, the smallest option, which fits since the $b<100$ cap is fairly tight.
Alternative: Instead of counting $a$, count $b$. The relation $a^2=2b+1$ says $2b+1$ must be a perfect square. So look for odd perfect squares $9,25,49,81,121,169,225,\dots$ below $201$ (since $2b+1<201$): they are $9,25,49,81,121,169$ — six of them — each giving a valid $b=\frac{a^2-1}{2}<100$. Same count, $6$.
CCSS standards used (min grade 8)
8.G.B.7Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Writing $a^2+b^2=(b+1)^2$ from the right-triangle sides $a$, $b$, and $b+1$.)6.EE.A.3Apply the properties of operations to generate equivalent expressions (Expanding $(b+1)^2$ and cancelling $b^2$ to reduce the equation to $a^2=2b+1$.)6.EE.B.5Understand solving an equation or inequality as a process of finding values (Deciding which values of $a$ make $b=\frac{a^2-1}{2}$ a positive whole number (odd $a\ge 3$).)7.EE.B.4Use variables to represent quantities and construct simple equations and inequalities (Turning $b<100$ into $a^2<201$, bounding $a\le 13$, and listing and counting the valid legs.)
⭐ When the hypotenuse is just one more than a leg, the Pythagorean equation collapses to $a^2=2b+1$, so you only have to count the odd legs that stay in range.
⭐ When the hypotenuse is just one more than a leg, the Pythagorean equation collapses to $a^2=2b+1$, so you only have to count the odd legs that stay in range.
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