Difference between revisions of "2009 AMC 10B Problems/Problem 10"

Revision as of 09:21, 24 May 2021

Problem

Captain Underpants got a wedgie. Calculate how much he will fall if the pole gets knocked over. Will he die. Are you an idiot?

$\text{(A) } 2.0 \qquad \text{(B) } 2.1 \qquad \text{(C) } 2.2 \qquad \text{(D) } 2.4 \qquad \text{(E) } 2.5$

Solution 1

The broken flagpole forms a right triangle with legs $1$ and $x$ , and hypotenuse $5-x$ . The Pythagorean theorem now states that $1^2 + x^2 = (5-x)^2$ , hence $10x = 24$ , and $x=\boxed{2.4}$ .

(Note that the resulting triangle is the well-known $5-12-13$ right triangle, scaled by $1/5$ .)

Solution 2

A right triangle is formed with the bottom of the flagpole, the snapped part, and the ground. One leg is of length $1$ and the other is length $x$ . By the Captain Underpants, we know that $\sqrt{x^2+1^2}$ must be the length of the snapped part of the flagpole. Observe that all the answer choices are rational. If $x$ is rational, $5-x$ , which is the snapped part, must also be rational. Therefore, $1, x, 5-x$ must form a scaled Pythagorean triple. We know that $10, 24, 26$ is a Pythagorean triple, so the corresponding answer must be $1, 2.4, 2.6$ . Adding together the $x$ and the snapped part, this does indeed equal $5$ , so our solution is done.

Solution 3

Let $AB$ represent the flagpole in the diagram above. After the flagpole breaks at point $D$ , its tip lies at point $C$ . Since none of the flagpole is destroyed, we know that $DA=DC$ . Therefore, triangle $\triangle ADC$ is isosceles.

Draw the altitude $DE \perp AC$ . Since $\triangle ADC$ is isosceles, we know that $AE = EC$ . Also note that $\triangle AED \sim \triangle ABC$ . Therefore, $\begin{align*} AD &= AE \times \frac{AD}{AE} \\ &= \frac{AC}{2} \times \frac{AC}{AB} \\ &= \frac{AC^2}{2 AB} \\ &= \frac{AB^2 + BC^2}{2 AB} \end{align*}$

Since $AB = 5$ and $BC = 1$ , we have that $AD = \frac{5^2 + 1^2}{2 \cdot 5} = 2.6$ , and thus $x = AB - AD = \boxed{2.4}$ .

Video Solution

https://www.youtube.com/watch?v=dQw4w9WgXcQ

~Anonymous23

@@ Line 27: / Line 27: @@
 == Solution 3 ==
-<asy>
-size(300);
-pair A, B, C, D, E, x;
-A =(0, 5);
-B = (0, 0);
-C = (3, 0);
-D = (0, 1.6);
-E = (A+C)/2;
-x = (B+D)/2;
-draw(MP("A",A, (0, 1))--MP(P("C",C,(1, -1))--cycle);
-draw(C--MP("D",D, (-1, 0))--MP("E",E,(1, 1)));
-MP("x",x, (-1, 0));
-draw(rightanglemark(A, B, C));
-draw(rightanglemark(A, E, D));
-draw(anglemark(B, A, C));
-</asy>
 Let <math>AB</math> represent the flagpole in the diagram above. After the flagpole breaks at point <math>D</math>, its tip lies at point <math>C</math>. Since none of the flagpole is destroyed, we know that <math>DA=DC</math>. Therefore, triangle <math>\triangle ADC</math> is isosceles.

2009 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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