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

Revision as of 22:55, 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 Simson 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 Solution by advanture)

First, draw the diagram. Then, notice that since $ABCD$ is isosceles, $\Delta ABD \cong \Delta BAC$ , and the length of the altitude from $B$ to $AC$ is also $10$ . Let the foot of this altitude be $F$ , and let the foot of the altitude from $A$ to $BC$ be denoted as $E$ . Then, $\Delta BCF \sim \Delta ACE$ . So, $\frac{BC}{AC} = \frac{BF}{AE} = \frac{2}{3}$ . Now, notice that $[ABC] = \frac{10 \times AC} {2} = \frac{AB \times 18}{2} \implies AC = \frac{9 \times AB}{5}$ , where $[ABC]$ denotes the area of triangle $ABC$ . Letting $AB = x$ , this equality becomes $AC = \frac{9x}{5}$ . Also, from $\frac{BC}{AC} = \frac{2}{3}$ , we have $BC = \frac{6x}{5}$ . Now, by the Pythagorean theorem on triangles $ABF$ and $CBF$ , we have $AF = \sqrt{x^{2}-100}$ and $CF = \sqrt{ \left( \frac{6x}{5} \right) ^{2}-100}$ . Notice that $AC = AF + CF$ , so $\frac{9x}{5} = \sqrt{x^{2}-100} + \sqrt{ \left( \frac{6x}{5} \right) ^{2}-100}$ . Squaring both sides of the equation once, moving $x^{2}-100$ and $\left( \frac{6x}{5} \right) ^{2}-100$ to the right, dividing both sides by $2$ , and squaring the equation once more, we are left with $\frac{32x^{4}}{25} = 324x^{2}$ . Dividing both sides by $x^{2}$ (since we know $x$ is positive), we are left with $\frac{32x^{2}}{25} = 324$ . Solving for $x$ gives us $x = \frac{45}{2\sqrt{2}}$ .

Now, let the foot of the perpendicular from $A$ to $CD$ be $G$ . Then let $DG = y$ . Let the foot of the perpendicular from $B$ to $CD$ be $H$ . Then, $CH$ is also equal to $y$ . Notice that $ABHG$ is a rectangle, so $GH = x$ . Now, we have $CG = GH + CH = x + y$ . By the Pythagorean theorem applied to $\Delta AGC$ , we have $(x+y)^{2}+18^{2}= \left( \frac{9x}{5} \right) ^{2}$ . We know that $\frac{9x}{5} = \frac{9}{5} \cdot \frac{45}{2\sqrt{2}} = \frac{81}{2\sqrt{2}}$ , so we can plug this into this equation. Solving for $x+y$ , we get $x+y=\frac{63}{2\sqrt{2}}$ .

Finally, to find $[ABCD]$ , we use the formula for the area of a trapezoid: $K = [ABCD] = \frac{b_{1}+b_{2}}{2} \cdot h = \frac{AB+CD}{2} \cdot 18 = \frac{x+(CG+DG)}{2} \cdot 18 = \frac{2x+2y}{2} \cdot 18 = (x+y) \cdot 18 = \frac{63}{2\sqrt{2}} \cdot 18 = \frac{567}{\sqrt{2}}$ . The problem asks us for $K \cdot \sqrt{2}$ , which comes out to be $\boxed{567}$ .

~advanture

@@ Line 27: / Line 27: @@
 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, <math>A</math> is on the circumcircle of <math>\triangle BCD</math> and we have that <math>PRQ</math> is the Simpson Line from <math>A</math>. As <math>\angle QAB = 90^\circ</math>, we have that <math>\angle QAR = 90^\circ - \angle RAB =\angle ABR = \angle APR = \angle APQ</math>, with the last equality coming from cyclic quadrilateral <math>APBR</math>. Thus, <math>\triangle QAR \sim \triangle QPA</math> and we have that <math>\frac{AQ}{AR} = \frac{PQ}{PA}</math> or that <math>\frac{18}{10} = \frac{QP}{15}</math>, which we can see gives us that <math>QP = 27</math>. Further ratios using the same similar triangles gives that <math>QR = \frac{25}{3}</math> and <math>RP = \frac{56}{3}</math>.
+Note that all isosceles trapezoids are cyclic quadrilaterals; thus, <math>A</math> is on the circumcircle of <math>\triangle BCD</math> and we have that <math>PRQ</math> is the Simson Line from <math>A</math>. As <math>\angle QAB = 90^\circ</math>, we have that <math>\angle QAR = 90^\circ - \angle RAB =\angle ABR = \angle APR = \angle APQ</math>, with the last equality coming from cyclic quadrilateral <math>APBR</math>. Thus, <math>\triangle QAR \sim \triangle QPA</math> and we have that <math>\frac{AQ}{AR} = \frac{PQ}{PA}</math> or that <math>\frac{18}{10} = \frac{QP}{15}</math>, which we can see gives us that <math>QP = 27</math>. Further ratios using the same similar triangles gives that <math>QR = \frac{25}{3}</math> and <math>RP = \frac{56}{3}</math>.
 We also see that quadrilaterals <math>APBR</math> and <math>ARDQ</math> are both cyclic, with diameters of the circumcircles being <math>AB</math> and <math>AQ</math> respectively. The intersection of the circumcircles are the points <math>A</math> and <math>R</math>, and we know <math>DRB</math> and <math>QRP</math> 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 <math>\triangle APQ</math> to <math>\triangle APD</math>. Because we know a lot about <math>\triangle APQ</math> but very little about <math>\triangle APD</math> and we would like to know more, we wish to find the ratio of similitude between the two triangles.

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
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