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

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==Solution 4 (Cool Reflection, some Pythag and a little Similarity)
  
 
==See also==
 
==See also==

Revision as of 20:34, 14 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 $AB=\frac{5}{6}a$. 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

Solution 3 (Similarity)

Let the foot of the altitude from A to BC be P, to CD be Q, and to BD be R.

Note that all isosceles trapezoids are cyclic quadrilaterals; thus, $A$ is on the circumcircle of $\triangle BCD$ and we have that $PRQ$ is the Simpson Line from $A$. As $\angle QAB = 90^\circ$, we have that $\angle QAR = 90^\circ - \angle RAB =\angle ABR = \angle APR = \angle APQ$, with the last equality coming from cyclic quadrilateral $APBR$. Thus, $\triangle QAR \sim \triangle QPA$ and we have that $\frac{AQ}{AR} = \frac{PQ}{PA}$ or that $\frac{18}{10} = \frac{QP}{15}$, which we can see gives us that $QP = 27$. Further ratios using the same similar triangles gives that $QR = \frac{25}{3}$ and $RP = \frac{56}{3}$.

We also see that quadrilaterals $APBR$ and $ARDQ$ are both cyclic, with diameters of the circumcircles being $AB$ and $AQ$ respectively. The intersection of the circumcircles are the points $A$ and $R$, and we know $DRB$ and $QRP$ are both line segments passing through an intersection of the two circles with one endpoint on each circle. By Fact 5, we know then that there exists a spiral similarity with center A taking $\triangle APQ$ to $\triangle APD$. Because we know a lot about $\triangle APQ$ but very little about $\triangle APD$ and we would like to know more, we wish to find the ratio of similitude between the two triangles.

To do this, we use the one number we have for $\triangle APD$: we know that the altitude from $A$ to $BD$ has length 10. As the two triangles are similar, if we can find the height from $A$ to $PQ$, we can take the ratio of the two heights as the ratio of similitude. To do this, we once again note that $QP = 27$. Using this, we can drop the altitude from $A$ to $QP$ and let it intersect $QP$ at $H$. Then, let $QH = x$ and thus $HP=27-x$. We then have by the Pythagorean Theorem on $\triangle AQH$ and $\triangle APH$: \[15^2 - x^2 = 18^2 - (27-x)^2\] \[225 - x^2 = 324 - (x^2-54x+729)\] \[54x = 630\] \[x=\frac{35}{3}\]

Then, $RH = QH - QR = \frac{35}{3} - \frac{25}{3} = \frac{10}{3}$. This gives us then from right triangle $\triangle ARH$ that $AH = \frac{20\sqrt{2}}{3}$ and thus the ratio of $\triangle APQ$ to $\triangle ABD$ is $\frac{3\sqrt{2}}{4}$. From this, we see then that \[AB = AP * \frac{3\sqrt{2}}{4} = 15 * \frac{3\sqrt{2}}{4} = \frac{45\sqrt{2}}{4}\] and \[AD = AQ * \frac{3\sqrt{2}}{4} = 18 * \frac{3\sqrt{2}}{4} = \frac{27\sqrt{2}}{2}\] The Pythagorean Theorem on $\triangle AQD$ then gives that \[QD = \sqrt{AD^2 - AQ^2} = \sqrt{(\frac{27\sqrt{2}}{2})^2 - 18^2} = \sqrt{\frac{81}{2}} = \frac{9\sqrt{2}}{2}\]

Then, we have the height of trapezoid $ABCD$ is $AQ = 18$, the top base is $AB = \frac{45\sqrt{2}}{4}$, and the bottom base is $CD = \frac{45\sqrt{2}}{4} + 2*\frac{9\sqrt{2}}{2}$. From the equation of a trapezoid, $K = \frac{b_1+b_2}{2} h = \frac{63\sqrt{2}}{4} * 18 = \frac{567\sqrt{2}}{2}$, so the answer is $K\sqrt{2} = \boxed{567}$.

- lvmath

==Solution 4 (Cool Reflection, some Pythag and a little Similarity)

See also

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