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Difference between revisions of "2021 AIME I Problems/Problem 9"

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==Solution==
 
==Solution==
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Construct your isosceles trapezoid. Let, for simplicity, <math>AB = a</math>, <math>AD = BC = b</math>, and <math>CD = c</math>. Extend the sides <math>BC</math> and <math>AD</math> mark the intersection as <math>P</math>. Following what the question states, drop a perpendicular from <math>A</math> to <math>BC</math> labeling the foot as <math>G</math>. Drop another perpendicular from <math>A</math> to <math>CD</math>, calling the foot <math>E</math>. Lastly, drop a perpendicular from <math>A</math> to <math>BD</math>, labeling it <math>F</math>. In addition, drop a perpendicular from <math>B</math> to <math>AC</math> calling its foot <math>F'</math>.
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 +
--DIAGRAM COMING SOON--
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Start out by constructing a triangle <math>ADH</math> congruent to <math>\triangle ABC</math> with its side of length <math>a</math> on line <math>DE</math>. This works because all isosceles triangles are cyclic and as a result, <math>\angle ADC + \angle ABC = 180^\circ</math>.
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Notice that <math>\triangle AGC \sim \triangle BF'C</math> by AA similarity. We are given that <math>AG = 15</math> and by symmetry we can deduce that <math>F'B = 10</math>. As a result, <math>\frac{BF}{AG} = \frac{BC}{AC} = \frac{3}{2}</math>. This gives us that <math>AC = BD = \frac{3}{2} b</math>.
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The question asks us along the lines of finding the area, <math>K</math>, of the trapezoid <math>ABCD</math>. We look at the area of <math>ABC</math> and notice that it can be represented as <math>\frac{1}{2} \cdot AC \cdot 10 = \frac{1}{2} \cdot a \cdot 18</math>. Substituting <math>AC = \frac{3}{2} b</math>, we solve for <math>a</math>, getting <math>a = \frac{5}{6} b</math>.
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Now let us focus on isosceles triangle <math>ACH</math>, where <math>AH = AC = \frac{3}{2} b</math>. Since, <math>AE</math> is an altitude from <math>A</math> to <math>CH</math> of an isosceles triangle, <math>HE</math> must be equal to <math>EC</math>. Since <math>DH = a</math> and <math>DC = c</math>, we can solve to get that <math>DE = \frac{c-a}{2}</math> and <math>EC = \frac{a+c}{2}</math>.
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We must then set up equations using the Pythagorean Theorem, writing everything in terms of <math>a</math>, <math>b</math>, and <math>c</math>. Looking at right triangle <math>AEC</math> we get <cmath>324 + \frac{(a + c)^2}{4} = \frac{9}{4} b^2</cmath> Looking at right triangle <math>AED</math> we get <cmath>b^2 - 324 = \frac{(c-a)^2}{4}</cmath> Now rearranging and solving, we get two equation <cmath>a+c = 3\sqrt{b^2 - 144}</cmath> <cmath>c - a = 2\sqrt{b^2 - 324}</cmath> Those are convenient equations as <math>c+a - (c-a) = 2a = \frac{5}{3} b</math> which gives us <cmath>3\sqrt{b^2 - 324} - 2\sqrt{b^2 - 324} = \frac{5}{3} b</cmath> After some "smart" calculation, we get that <math>b = \frac{27}{\sqrt{2}}</math>.
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Notice that the question asks for <math>K\sqrt{2}</math>, and <math>K = \frac{1}{2} \cdot 18 \cdot (a+c)</math> by applying the trapezoid area formula. Fortunately, this is just <math>27\sqrt{b^2 - 144}</math>, and plugging in the value of <math>b = \frac{27}{\sqrt{2}}</math>, we get that <math>K\sqrt{2} = \boxed{567}</math>.
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~Math_Genius_164
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==See also==
 
==See also==
 
{{AIME box|year=2021|n=I|num-b=8|num-a=10}}
 
{{AIME box|year=2021|n=I|num-b=8|num-a=10}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 21:41, 11 March 2021

Problem

Let $ABCD$ be an isosceles trapezoid with $AD=BC$ and $AB<CD.$ Suppose that the distances from $A$ to the lines $BC,CD,$ and $BD$ are $15,18,$ and $10,$ respectively. Let $K$ be the area of $ABCD.$ Find $\sqrt2 \cdot K.$

Solution

Construct your isosceles trapezoid. Let, for simplicity, $AB = a$, $AD = BC = b$, and $CD = c$. Extend the sides $BC$ and $AD$ mark the intersection as $P$. Following what the question states, drop a perpendicular from $A$ to $BC$ labeling the foot as $G$. Drop another perpendicular from $A$ to $CD$, calling the foot $E$. Lastly, drop a perpendicular from $A$ to $BD$, labeling it $F$. In addition, drop a perpendicular from $B$ to $AC$ calling its foot $F'$.

--DIAGRAM COMING SOON--

Start out by constructing a triangle $ADH$ congruent to $\triangle ABC$ with its side of length $a$ on line $DE$. This works because all isosceles triangles are cyclic and as a result, $\angle ADC + \angle ABC = 180^\circ$.

Notice that $\triangle AGC \sim \triangle BF'C$ by AA similarity. We are given that $AG = 15$ and by symmetry we can deduce that $F'B = 10$. As a result, $\frac{BF}{AG} = \frac{BC}{AC} = \frac{3}{2}$. This gives us that $AC = BD = \frac{3}{2} b$.

The question asks us along the lines of finding the area, $K$, of the trapezoid $ABCD$. We look at the area of $ABC$ and notice that it can be represented as $\frac{1}{2} \cdot AC \cdot 10 = \frac{1}{2} \cdot a \cdot 18$. Substituting $AC = \frac{3}{2} b$, we solve for $a$, getting $a = \frac{5}{6} b$.

Now let us focus on isosceles triangle $ACH$, where $AH = AC = \frac{3}{2} b$. Since, $AE$ is an altitude from $A$ to $CH$ of an isosceles triangle, $HE$ must be equal to $EC$. Since $DH = a$ and $DC = c$, we can solve to get that $DE = \frac{c-a}{2}$ and $EC = \frac{a+c}{2}$.

We must then set up equations using the Pythagorean Theorem, writing everything in terms of $a$, $b$, and $c$. Looking at right triangle $AEC$ we get \[324 + \frac{(a + c)^2}{4} = \frac{9}{4} b^2\] Looking at right triangle $AED$ we get \[b^2 - 324 = \frac{(c-a)^2}{4}\] Now rearranging and solving, we get two equation \[a+c = 3\sqrt{b^2 - 144}\] \[c - a = 2\sqrt{b^2 - 324}\] Those are convenient equations as $c+a - (c-a) = 2a = \frac{5}{3} b$ which gives us \[3\sqrt{b^2 - 324} - 2\sqrt{b^2 - 324} = \frac{5}{3} b\] After some "smart" calculation, we get that $b = \frac{27}{\sqrt{2}}$.

Notice that the question asks for $K\sqrt{2}$, and $K = \frac{1}{2} \cdot 18 \cdot (a+c)$ by applying the trapezoid area formula. Fortunately, this is just $27\sqrt{b^2 - 144}$, and plugging in the value of $b = \frac{27}{\sqrt{2}}$, we get that $K\sqrt{2} = \boxed{567}$.

~Math_Genius_164

See also

2021 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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