2021 AIME I Problems/Problem 9

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}$.

2021 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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2021 AIME I Problems/Problem 9

Contents

Problem

Solution 1

Solution 2(LOC and Trig)

See also