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

(Added a new solution)
(Added third solution)
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</math>
 
</math>
  
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== Solution ==
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== Solution 1 ==
  
 
The broken flagpole forms a right triangle with legs <math>1</math> and <math>x</math>, and hypotenuse <math>5-x</math>. The [[Pythagorean theorem]] now states that <math>1^2 + x^2 = (5-x)^2</math>, hence <math>10x = 24</math>, and <math>x=\boxed{2.4}</math>.
 
The broken flagpole forms a right triangle with legs <math>1</math> and <math>x</math>, and hypotenuse <math>5-x</math>. The [[Pythagorean theorem]] now states that <math>1^2 + x^2 = (5-x)^2</math>, hence <math>10x = 24</math>, and <math>x=\boxed{2.4}</math>.
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== Solution #2 ==
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== Solution 2 ==
  
 
A right triangle is formed with the bottom of the flagpole, the snapped part, and the ground. One leg is of length <math>1</math> and the other is length <math>x</math>. By the [[Pythagorean theorem]], we know that <math>\sqrt{x^2+1^2}</math> must be the length of the snapped part of the flagpole. Observe that all the answer choices are rational. If <math>x</math> is rational, <math>5-x</math>, which is the snapped part, must also be rational. Therefore, <math>1, x, 5-x</math> must form a scaled Pythagorean triple. We know that <math>10, 24, 26</math> is a Pythagorean triple, so the corresponding answer must be <math>1, 2.4, 2.6</math>. Adding together the <math>x</math> and the snapped part, this does indeed equal <math>5</math>, so our solution is done.
 
A right triangle is formed with the bottom of the flagpole, the snapped part, and the ground. One leg is of length <math>1</math> and the other is length <math>x</math>. By the [[Pythagorean theorem]], we know that <math>\sqrt{x^2+1^2}</math> must be the length of the snapped part of the flagpole. Observe that all the answer choices are rational. If <math>x</math> is rational, <math>5-x</math>, which is the snapped part, must also be rational. Therefore, <math>1, x, 5-x</math> must form a scaled Pythagorean triple. We know that <math>10, 24, 26</math> is a Pythagorean triple, so the corresponding answer must be <math>1, 2.4, 2.6</math>. Adding together the <math>x</math> and the snapped part, this does indeed equal <math>5</math>, so our solution is done.
 +
a
 +
== Solution 3 ==
 +
 +
<asy>
 +
size(300);
 +
pair A, B, C, D, E, x;
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A =(0, 5);
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B = (0, 0);
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C = (3, 0);
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D = (0, 1.6);
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E = (A+C)/2;
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x = (B+D)/2;
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 +
draw(MP("A",A, (0, 1))--MP("B",B,(-1,-1))--MP("C",C,(1, -1))--cycle);
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draw(C--MP("D",D, (-1, 0))--MP("E",E,(1, 1)));
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MP("x",x, (-1, 0));
 +
 +
draw(rightanglemark(A, B, C));
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draw(rightanglemark(A, E, D));
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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.
 +
 +
Draw the altitude <math>DE \perp AC</math>. Since <math>\triangle ADC</math> is isosceles, we know that <math>AE = EC</math>. Also note that <math>\triangle AED \sim \triangle ABC</math>. Therefore,
 +
<cmath>\begin{align*}
 +
AD &= AE \times \frac{AD}{AE} \\
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&= \frac{AC}{2} \times \frac{AC}{AB} \\
 +
&= \frac{AC^2}{2 AB} \\
 +
&= \frac{AB^2 + BC^2}{2 AB}
 +
\end{align*}</cmath>
 +
 +
Since <math>AB = 5</math> and <math>BC = 1</math>, we have that <math>AD = \frac{5^2 + 1^2}{2 \cdot 5} = 2.6</math>, and thus <math>x = AB - AD = \boxed{2.4}</math>.
  
 
== See Also ==
 
== See Also ==

Revision as of 07:24, 25 June 2018

Problem

A flagpole is originally $5$ meters tall. A hurricane snaps the flagpole at a point $x$ meters above the ground so that the upper part, still attached to the stump, touches the ground $1$ meter away from the base. What is $x$?

$\text{(A) } 2.0 \qquad \text{(B) } 2.1 \qquad \text{(C) } 2.2 \qquad \text{(D) } 2.3 \qquad \text{(E) } 2.4$

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 Pythagorean theorem, 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. a

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("B",B,(-1,-1))--MP("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 $AB$ 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 $x = AB - AD = \boxed{2.4}$.

See Also

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

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