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

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~Math_Genius_164
 
~Math_Genius_164
 
==Solution 2(LOC and Trig)==
 
==Solution 2(LOC and Trig)==
−
Call AD and BC <math>a</math>. Draw diagonal  AC and call the foot of the perpendicular from B to AC <math>G</math>. Call the foot of the perpendicular from A to line BC F, and call the foot of the perpindicular from A to DC H. Triangles CBG and CAF are similar, and we get that <math>\frac{10}{15}</math>=<math>\frac{a}{AC}</math> Therefore, <math>AC=1.5a</math>. It then follows that triangles ABF and ADH are similar. Using similar triangles, we can then find that <math>AB=1.2a</math>. Using the Law of Cosine on ABC, We can find that the cosine of angle ABC is <math>-\frac{1}{3}. Since angles ABF and ADH are equivalent and supplementary to angle ABC, we know that the cosine of angle ADH is 1/3. It then follows that </math>a=\frac{27\sqrt{2}}{2}<math>. Then it can be found that the area </math>K<math> is </math>\frac{567\sqrt{2}}{2}<math>. Multiplying this by </math>\sqrt{2}<math>, the answer is </math>\boxed{567}$.
+
Call AD and BC <math>a</math>. Draw diagonal  AC and call the foot of the perpendicular from B to AC <math>G</math>. Call the foot of the perpendicular from A to line BC F, and call the foot of the perpindicular from A to DC H. Triangles CBG and CAF are similar, and we get that <math>\frac{10}{15}</math>=<math>\frac{a}{AC}</math> Therefore, <math>AC=1.5a</math>. It then follows that triangles ABF and ADH are similar. Using similar triangles, we can then find that <math>AB=1.2a</math>. Using the Law of Cosine on ABC, We can find that the cosine of angle ABC is <math>-\frac{1}{3}</math>. Since angles ABF and ADH are equivalent and supplementary to angle ABC, we know that the cosine of angle ADH is 1/3. It then follows that <math>a=\frac{27\sqrt{2}}{2}</math>. Then it can be found that the area <math>K</math> is <math>\frac{567\sqrt{2}}{2}</math>. Multiplying this by <math>\sqrt{2}</math>, the answer is <math>\boxed{567}</math>.
 +
-happykeeper
  
 
==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 12:45, 12 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 1

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

Solution 2(LOC and Trig)

Call AD and BC $a$. Draw diagonal AC and call the foot of the perpendicular from B to AC $G$. Call the foot of the perpendicular from A to line BC F, and call the foot of the perpindicular from A to DC H. Triangles CBG and CAF are similar, and we get that $\frac{10}{15}$=$\frac{a}{AC}$ Therefore, $AC=1.5a$. It then follows that triangles ABF and ADH are similar. Using similar triangles, we can then find that $AB=1.2a$. Using the Law of Cosine on ABC, We can find that the cosine of angle ABC is $-\frac{1}{3}$. Since angles ABF and ADH are equivalent and supplementary to angle ABC, we know that the cosine of angle ADH is 1/3. It then follows that $a=\frac{27\sqrt{2}}{2}$. Then it can be found that the area $K$ is $\frac{567\sqrt{2}}{2}$. Multiplying this by $\sqrt{2}$, the answer is $\boxed{567}$. -happykeeper

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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