AMC 10 Step by Step

2009 AMC 10B Problem 10

P1-10~2 minPrint

A flagpole is originally 55 meters tall. A hurricane snaps the flagpole at a point xx meters above the ground so that the upper part, still attached to the stump, touches the ground 11 meter away from the base. What is xx ?

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Key insight

The stump, the ground, and the broken top form a right triangle with legs x and 1 and long side 5 - x. When you expand, the x-squared terms cancel.

Solution

Draw the picture: the stump stands straight up, and the broken top leans from the top of the stump down to the ground. That is a right triangle.

1. The stump is xx meters tall. This is one leg.
2. The top touches the ground 11 meter from the base. This is the other leg.
3. The broken top is the rest of the pole, 5−x5 - x meters. It is the slanted side (the hypotenuse).
4. By the Pythagorean theorem:
x2+12=(5−x)2.x^2 + 1^2 = (5 - x)^2.
5. Expand the right side: (5−x)2=25−10x+x2(5 - x)^2 = 25 - 10x + x^2. So
x2+1=25−10x+x2.x^2 + 1 = 25 - 10x + x^2.
6. Both sides have x2x^2, so take it away from both: 1=25−10x1 = 25 - 10x. Then 10x=2410x = 24 and x=2.4x = 2.4.
7. Check: the legs are 2.42.4 and 11, the slanted side is 5−2.4=2.65 - 2.4 = 2.6. And 2.42+1=5.76+1=6.76=2.622.4^2 + 1 = 5.76 + 1 = 6.76 = 2.6^2. Correct.

The answer is (E) 2.4\boxed{\textbf{(E)}\ 2.4}.

Why this works

In a "snapped pole" problem, one leg plus the slanted side adds up to the known total height. Because xx is squared on both sides of the equation, the squares cancel and you only have a simple equation left, with no quadratic formula needed. The triangle 2.4,1,2.62.4, 1, 2.6 is a shrunken copy of the 12,5,1312, 5, 13 right triangle, which is a nice confirmation.

Alternative approach

Test the choices. For each xx, check whether x2+1x^2 + 1 equals (5−x)2(5 - x)^2. For x=2.4x = 2.4: 5.76+1=6.765.76 + 1 = 6.76 and 2.62=6.762.6^2 = 6.76. For x=2.3x = 2.3: 5.29+1=6.295.29 + 1 = 6.29 but 2.72=7.292.7^2 = 7.29. Only 2.42.4 works.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

The stump is vertical with height xx; the upper part has length 5−x5 - x and leans from the top of the stump to a point on the ground 11 meter from the base. These form a right triangle with legs xx and 11 and hypotenuse 5−x5 - x:
x2+1=(5−x)2=25−10x+x2.x^2 + 1 = (5 - x)^2 = 25 - 10x + x^2.
The x2x^2 terms cancel, leaving 1=25−10x1 = 25 - 10x, so 10x=2410x = 24 and x=2.4x = 2.4.

Check: legs 2.42.4 and 11, hypotenuse 2.62.6; indeed 2.42+1=5.76+1=6.76=2.622.4^2 + 1 = 5.76 + 1 = 6.76 = 2.6^2.

The answer is (E) 2.4\boxed{\textbf{(E)}\ 2.4}.

Review in the AoPS books

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Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site.