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

(Solution 2 (LOC and Trig))
(Solution 2 (Trigonometry))
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~MRENTHUSIASM
 
~MRENTHUSIASM
  
==Solution 2 (Trigonometry)==
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==Solution 2 (Similar Triangles and Trigonometry)==
 
Let <math>AD=BC=a</math>. Draw diagonal <math>AC</math> and let <math>G</math> be the foot of the perpendicular from <math>B</math> to <math>AC</math>, <math>F</math> be the foot of the perpendicular from <math>A</math> to line <math>BC</math>, and <math>H</math> be the foot of the perpendicular from <math>A</math> to <math>DC</math>.
 
Let <math>AD=BC=a</math>. Draw diagonal <math>AC</math> and let <math>G</math> be the foot of the perpendicular from <math>B</math> to <math>AC</math>, <math>F</math> be the foot of the perpendicular from <math>A</math> to line <math>BC</math>, and <math>H</math> be the foot of the perpendicular from <math>A</math> to <math>DC</math>.
  

Revision as of 12:51, 13 December 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 $\sqrt2 \cdot K.$

Diagram

[asy] /* Made by MRENTHUSIASM */ size(250); pair A, B, C, D, E, F, G, H; A = (-45sqrt(2)/8,18); B = (45sqrt(2)/8,18); C = (81sqrt(2)/8,0); D = (-81sqrt(2)/8,0); E = foot(A,C,B); F = foot(A,C,D); G = foot(A,B,D); H = intersectionpoint(A--F,B--D); markscalefactor=0.1; draw(rightanglemark(A,E,B),red); draw(rightanglemark(A,F,C),red); draw(rightanglemark(A,G,D),red); dot("$A$",A,1.5*NW,linewidth(4)); dot("$B$",B,1.5*NE,linewidth(4)); dot("$C$",C,1.5*SE,linewidth(4)); dot("$D$",D,1.5*SW,linewidth(4)); dot(E,linewidth(4)); dot(F,linewidth(4)); dot(G,linewidth(4)); draw(A--B--C--D--cycle^^B--D^^B--E); draw(A--E^^A--F^^A--G,dashed); label("$10$",midpoint(A--G),1.5*(1,0)); label("$15$",midpoint(A--E),1.5*N); Label L = Label("$18$", align=(0,0), position=MidPoint, filltype=Fill(0,3,white)); draw(C+(5,0)--(81sqrt(2)/8,18)+(5,0), L=L, arrow=Arrows(),bar=Bars(15)); [/asy] ~MRENTHUSIASM

Solution 2 (Similar Triangles and Trigonometry)

Let $AD=BC=a$. Draw diagonal $AC$ and let $G$ be the foot of the perpendicular from $B$ to $AC$, $F$ be the foot of the perpendicular from $A$ to line $BC$, and $H$ be the foot of the perpendicular from $A$ to $DC$.

Note that $\triangle CBG\sim\triangle CAF$, and we get that $\frac{10}{15}=\frac{a}{AC}$. Therefore, $AC=\frac32 a$. It then follows that $\triangle ABF\sim\triangle ADH$. Using similar triangles, we can then find that $AB=\frac{5}{6}a$. Using the Law of Cosines on $\triangle ABC$, We can find that the $\cos\angle ABC=-\frac{1}{3}$. Since $\angle ABF=\angle ADH$, and each is supplementary to $\angle ABC$, we know that the $\cos\angle ADH=\frac{1}{3}$. 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 $\triangle APH$: \begin{align*} 15^2 - x^2 &= 18^2 - (27-x)^2 \\ 225 - x^2 &= 324 - (x^2-54x+729) \\ 54x &= 630 \\ x &= \frac{35}{3}. \end{align*} 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 \[AB = AP \cdot \frac{3\sqrt{2}}{4} = 15 \cdot \frac{3\sqrt{2}}{4} = \frac{45\sqrt{2}}{4}\] and \[AD = AQ \cdot \frac{3\sqrt{2}}{4} = 18 \cdot \frac{3\sqrt{2}}{4} = \frac{27\sqrt{2}}{2}.\] The Pythagorean Theorem on $\triangle AQD$ then gives that \[QD = \sqrt{AD^2 - AQ^2} = \sqrt{\left(\frac{27\sqrt{2}}{2}\right)^2 - 18^2} = \sqrt{\frac{81}{2}} = \frac{9\sqrt{2}}{2}.\] 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 $CD = \frac{45\sqrt{2}}{4} + 2\cdot\frac{9\sqrt{2}}{2}$. From the equation of a trapezoid, $K = \frac{b_1+b_2}{2} \cdot h = \frac{63\sqrt{2}}{4} \cdot 18 = \frac{567\sqrt{2}}{2}$, 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 $[ABC] = \frac{10 \cdot AC} {2} = \frac{AB \cdot 18}{2} \implies AC = \frac{9 \cdot AB}{5}$, 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

Solution 5 (Compact Similarity Solution)

Let $E,F,$ and $G$ be the feet of the altitudes from $A$ to $BC,CD,$ and $DB$, respectively.

Claim: We have $2$ pairs of similar right triangles: $\triangle AEB \sim \triangle AFD$ and $\triangle AGD \sim \triangle AEC$.

Proof: Note that $ABCD$ is cyclic. We need one more angle, and we get this from this cyclic quad: \[\angle ABE = 180^\circ - \angle ABC =\angle ADC = \angle ADG\] \[\angle ADG = \angle ADB =\angle ACB = \angle ACE \square\]

Let $AD=a$. We obtain from the similarities $AB = \frac{5a}{6}$ and $AC=BD=\frac{3a}{2}$.

By Ptolemy, $\left(\frac{3a}{2}\right)^2 = a^2 + \frac{5a}{6} \cdot CD$, so $\frac{5a^2}{4} = \frac{5a}{6} \cdot CD$.

We obtain $CD=\frac{3a}{2}$, so $DF=\frac{CD-AB}{2}=\frac{a}{3}$.

Applying the Pythagorean theorem on $\triangle ADF$, we get $324=a^2 - \frac{a^2}{9}=\frac{8a^2}{9}$.

Thus, $a=\frac{27}{\sqrt{2}}$, and $[ABCD]=\frac{AB+CD}{2} \cdot 18 = \frac{\frac{5a}{6} +\frac{9a}{6}}{2} \cdot 18 = 18 \cdot \frac{7}{6} \cdot \frac{27}{\sqrt{2}} = \frac{567}{\sqrt{2}}$, yielding $\boxed{567}$.

Solution 6 (Two Variables, Two Equations)

Let $\overline{AE}, \overline{AF},$ and $\overline{AG}$ be the perpendiculars from $A$ to $\overleftrightarrow{BC}, \overleftrightarrow{CD},$ and $\overleftrightarrow{BD},$ respectively. Next, let $H$ be the intersection of $\overline{AF}$ and $\overline{BD}.$

We set $AB=x$ and $AH=y,$ as shown below. [asy] /* Made by MRENTHUSIASM */ size(250); pair A, B, C, D, E, F, G, H; A = (-45sqrt(2)/8,18); B = (45sqrt(2)/8,18); C = (81sqrt(2)/8,0); D = (-81sqrt(2)/8,0); E = foot(A,C,B); F = foot(A,C,D); G = foot(A,B,D); H = intersectionpoint(A--F,B--D); markscalefactor=0.1; draw(rightanglemark(A,E,B),red); draw(rightanglemark(A,F,C),red); draw(rightanglemark(A,G,D),red); dot("$A$",A,1.5*NW,linewidth(4)); dot("$B$",B,1.5*NE,linewidth(4)); dot("$C$",C,1.5*SE,linewidth(4)); dot("$D$",D,1.5*SW,linewidth(4)); dot("$E$",E,1.5*dir(E),linewidth(4)); dot("$F$",F,1.5*S,linewidth(4)); dot("$G$",G,SE,linewidth(4)); dot("$H$",H,SE,linewidth(4)); draw(A--B--C--D--cycle^^B--D^^B--E); draw(A--E^^A--F^^A--G,dashed); label("$10$",midpoint(A--G),1.5*(1,0)); label("$15$",midpoint(A--E),1.5*N); Label L = Label("$18$", align=(0,0), position=MidPoint, filltype=Fill(0,3,white)); draw(C+(5,0)--(81sqrt(2)/8,18)+(5,0), L=L, arrow=Arrows(),bar=Bars(15)); label("$x$",midpoint(A--B),N); label("$y$",midpoint(A--H),W); [/asy] From here, we obtain $HF=18-y$ by segment subtraction, and $BG=\sqrt{x^2-10^2}$ and $HG=\sqrt{y^2-10^2}$ by the Pythagorean Theorem.

Since $\angle ABG$ and $\angle HAG$ are both complementary to $\angle AHB,$ we have $\angle ABG = \angle HAG,$ from which $\triangle ABG \sim \triangle HAG$ by AA. It follows that $\frac{BG}{AG}=\frac{AG}{HG},$ so $BG\cdot HG=AG^2,$ or \[\sqrt{x^2-10^2}\cdot\sqrt{y^2-10^2}=10^2. \hspace{10mm} (1)\] Since $\angle AHB = \angle FHD$ by vertical angles, we have $\triangle AHB \sim \triangle FHD$ by AA, with the ratio of similitude $\frac{AH}{FH}=\frac{BA}{DF}.$ It follows that $DF=BA\cdot\frac{FH}{AH}=x\cdot\frac{18-y}{y}.$

Since $\angle EBA = \angle ECD = \angle FDA$ by angle chasing, we have $\triangle EBA \sim \triangle FDA$ by AA, with the ratio of similitude $\frac{EA}{FA}=\frac{BA}{DA}.$ It follows that $DA=BA\cdot\frac{FA}{EA}=x\cdot\frac{18}{15}=\frac{6}{5}x.$

By the Pythagorean Theorem on right $\triangle ADF,$ we have $DF^2+AF^2=AD^2,$ or \[\left(x\cdot\frac{18-y}{y}\right)^2+18^2=\left(\frac{6}{5}x\right)^2. \hspace{7mm} (2)\] Solving this system of equations ($(1)$ and $(2)$), we get $x=\frac{45\sqrt2}{4}$ and $y=\frac{90}{7},$ so $AB=x=\frac{45\sqrt2}{4}$ and $CD=AB+2DF=x+2\left(x\cdot\frac{18-y}{y}\right)=\frac{81\sqrt2}{4}.$ Finally, the area of $ABCD$ is \[K=\frac{AB+CD}{2}\cdot AF=\frac{567\sqrt2}{2},\] from which $\sqrt2 \cdot K=\boxed{567}.$

~MRENTHUSIASM

Solution 7 (Similar Triangles)

Draw the distances in terms of $B$, as shown in the diagram. By similar triangles, $\triangle{AEC}\sim\triangle{BIC}$. As a result, let $AB=u$, then $BC=AD=\frac{6}{5}u$ and $2AC=3BC$. The triangle $ABC$ is $6-5-9$ which $\cos(\angle{ABC})=-\frac{1}{3}$. By angle subtraction, $\cos(180-\theta)=-\cos\theta$. Therefore, $AB=\frac{45}{2\sqrt{2}}=\frac{45\sqrt{2}}{4}$ and $AD=BC=\frac{27}{\sqrt{2}}$. By trapezoid area formula, the area of $ABCD$ is equal to $(AB+DF)\cdot 18=567\cdot \frac{\sqrt{2}}{2}$ which $\sqrt{2}\cdot k=\boxed{567}$.

~math2718281828459

Video Solution

https://www.youtube.com/watch?v=6rLnl8z7lnM

See Also

2021 AIME I (ProblemsAnswer KeyResources)
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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