2017 AMC 8 Problem 16

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To help you master 2017 AMC 8 Problem 16, this page includes:

The original AMC problem to try, plus the correct answer

Clear, step-by-step solutions using different methods

Key ideas, useful formulas, and takeaways you can apply to similar problems

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2017 AMC8 Problem 16

Solution

Solution 1

Area Ratio Method

Since $△ A B D$ and $△ A C D$ share the same altitude from point $A$ , their areas are proportional to their base lengths on the line $BC$ .

If two triangles share the same height, the ratio of their areas is equal to the ratio of their bases.

Step 1: Set up variables for the unknown segments

Let $x$ be the length of segment $B D$ . Since point $D$ lies on the hypotenuse $BC$ and the total length is $5$ , the remaining segment $C D$ has a length of $5 - x$ .

B D = x

C D = 5 - x

Step 2: Use the equal perimeters condition

Write the perimeter of each smaller triangle by summing its three sides. Since the problem states these perimeters are equal, we can set the two expressions equal to each other.

Perimeter of △ A C D = 3 + (5 - x) + A D

Perimeter of △ A B D = 4 + x + A D

3 + (5 - x) + A D = 4 + x + A D

Step 3: Solve for the segment lengths

Notice that the shared side length cancels out from both sides of the equation. We can then isolate the variable to find the exact lengths of the two segments on the hypotenuse.

8 - x = 4 + x

4 = 2 x

x = 2

B D = 2

C D = 3

Step 4: Find the total area of the large triangle

Check the side lengths of the large triangle. Because they satisfy the Pythagorean relationship, the triangle is a right triangle with the right angle between the two shorter legs. We can use these legs as the base and height to find the total area.

3^{2} + 4^{2} = 9 + 16 = 25 = 5^{2}

Area (△ A BC) = \frac{1}{2} \cdot 4 \cdot 3 = 6

Step 5: Use the area ratio to find the target area

The two smaller triangles share the same top vertex, and their bases lie on the same line, meaning they have the exact same height. Therefore, the fraction of the total area that belongs to the smaller triangle is exactly the fraction of the total base it occupies.

\frac{Area ( △ A B D )}{Area ( △ A BC )} = \frac{B D}{BC} = \frac{2}{5}

Area (△ A B D) = \frac{2}{5} \cdot 6 = \frac{12}{5}

Correct answer: D. $\frac{12}{5}$
The exact answer is

\frac{12}{5}

, which corresponds to choice D. We can verify this by calculating the area of

△ A C D

, which occupies the remaining

\frac{3}{5}

of the base. Its area would be

\frac{3}{5} \cdot 6 = \frac{18}{5}

. The sum of the two areas is

\frac{12}{5} + \frac{18}{5} = \frac{30}{5} = 6

, which perfectly matches the total area of

△ A BC

.

Solution 2

Base and Altitude Method

We can use the bottom side $A B$ as the base of $△ A B D$ and find its corresponding height by drawing a perpendicular line from $D$ to $A B$ .

If two right triangles share an acute angle, they are similar, meaning the ratio of their corresponding side lengths is constant.

Step 1: Find the length of the segment on the hypotenuse

Using the same perimeter logic as in the first method, we determine the length of the segment connecting the bottom-right vertex to point $D$ .

B D = 2

Step 2: Draw the altitude to the bottom base

To find the area of the triangle using the bottom side as the base, we need the corresponding height. We draw a perpendicular line from point $D$ straight down to the segment $A B$ , intersecting at a new point $E$ .

Step 3: Use similar triangles to find the height

The new small triangle is a right triangle that shares an acute angle with the large right triangle, making them similar. The ratio of their hypotenuses gives us the scale factor. We apply this scale factor to the vertical leg of the large triangle to find the height of the small triangle.

△ EB D \sim △ A BC

\frac{D E}{A C} = \frac{B D}{BC} = \frac{2}{5}

D E = \frac{2}{5} \cdot 3 = \frac{6}{5}

Step 4: Calculate the target area

Now that we have the base and the corresponding height, we can directly calculate the area of the triangle.

Area (△ A B D) = \frac{1}{2} \cdot A B \cdot D E

Area (△ A B D) = \frac{1}{2} \cdot 4 \cdot \frac{6}{5} = \frac{12}{5}

Correct answer: D. $\frac{12}{5}$
The exact answer is

\frac{12}{5}

, corresponding to choice D. We can verify this by splitting

△ A B D

into two smaller right triangles at the altitude

D E

. The base

BE

is

\frac{2}{5} \cdot 4 = \frac{8}{5}

, giving an area of

\frac{1}{2} \cdot \frac{8}{5} \cdot \frac{6}{5} = \frac{24}{25}

. The remaining base is

A E = 4 - \frac{8}{5} = \frac{12}{5}

, giving an area of

\frac{1}{2} \cdot \frac{12}{5} \cdot \frac{6}{5} = \frac{36}{25}

. Their sum is

\frac{24}{25} + \frac{36}{25} = \frac{60}{25} = \frac{12}{5}

, confirming our result.

Adapted Problems to Practice

1 / 10

What Formulas Do You Need for This Problem?

Triangles with a Shared Altitude:

If two triangles share the same height, the ratio of their areas is equal to the ratio of their bases.

Applicable when a larger triangle is split by a straight line drawn from one vertex to the opposite side. Both $△ A B D$ and $△ A BC$ have bases along the line $BC$ and share the same top vertex $A$ , meaning the perpendicular distance from $A$ to line $BC$ is identical for both.

Converse of the Pythagorean Theorem:

If the side lengths $a, b, c$ of a triangle satisfy $a^{2} + b^{2} = c^{2}$ , then the triangle is a right triangle.

Applicable when all three side lengths of a triangle are known. The side lengths of $△ A BC$ are $3, 4,$ and $5$ . Since $3^{2} + 4^{2} = 5^{2}$ , the angle opposite the side of length $5$ is a right angle.

Similar Right Triangles:

If two right triangles share an acute angle, they are similar, meaning the ratio of their corresponding side lengths is constant.

Applicable when a perpendicular is dropped from a point on the hypotenuse to one of the legs. Dropping a perpendicular from $D$ to $A B$ creates a small right triangle that shares $∠ B$ with the large right triangle $△ A BC$ .

Key Takeaway: Remember This Strategy for Next Time

A triangle split into two smaller triangles by a line from a vertex to the opposite side.

Try: Use the fact that the smaller triangles share the same height, so the ratio of their areas equals the ratio of their bases.

Why: The area formula is $\frac{1}{2} bh$ . Since the height $h$ is constant for both triangles, the area depends entirely on the length of the base $b$ .

A problem involving the perimeters of adjacent triangles that share a side.

Try: Write out the full perimeter equations and set them equal. The shared side will cancel out, leaving a simple relationship between the remaining segments.

Why: The shared boundary contributes equally to both perimeters, so the difference in perimeters depends only on the unshared outer boundary.

A point on the hypotenuse of a right triangle.

Try: Drop perpendiculars to the legs to create smaller similar right triangles.

Why: The new small triangles share an acute angle with the original large right triangle, guaranteeing similarity and allowing you to find lengths using proportions.

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2017 AMC 8 Problem 16

In This Page:

2017 AMC8 Problem 16