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

## Problem

Let $ABCD$ be an isosceles trapezoid with $AD=BC$ and $AB 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 1 (Similar Triangles and Pythagorean Theorem)

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 2 (Similar Triangles and Pythagorean Theorem)

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 $\boxed{567}$.

## Solution 3 (Similar Triangles and Pythagorean Theorem)

Make $AE$ perpendicular to $BC$; $AG$ perpendicular to $BD$; $AF$ perpendicular $DC$.

It's obvious that $\triangle{AEB} \sim \triangle{AFD}$. Let $EB=5x; AB=5y; DF=6x; AD=6y$. Then make $BQ$ perpendicular to $DC$, it's easy to get $BQ=18$.

Since $AB$ parallel to $DC$, $\angle{ABG}=\angle{BDQ}$, so $\triangle{ABG} \sim \triangle{BDQ}$. After drawing the altitude, it's obvious that $FQ=AB=5y$, so $DQ=5y+6x$. According to the property of similar triangles, $AG/BQ=BG/DQ$. So, $\frac{5}{9}=\frac{GB}{(6x+5y)}$, or $GB=\frac{(30x+25y)}{9}$.

Now, we see the $\triangle AEB$, pretty easy to find that $15^2+(5x)^2=(5y)^2$, then we get $x^2+9=y^2$, then express $y$ into $x$ form that $y=\sqrt{x^2+9}$ we put the length of $BG$ back to $\triangle AGB$: $BG^2+100=AB^2$. So, $$\frac{[30x+25\sqrt{(x^2+9)}]^2}{81}+100=(5\sqrt{x^2+9})^2.$$ After calculating, we can have a final equation of $x^2+9=\sqrt{x^2+9}\cdot3x$. It's easy to find $x=\frac{3\sqrt{2}}{4}$ then $y=\frac{9\sqrt{2}}{4}$. So, $$\sqrt{2}\cdot K = \sqrt{2}\cdot(5y+5y+6x+6x)\cdot9=\boxed{567}.$$ ~bluesoul

## Solution 4 (Cyclic Quadrilaterals, Similar Triangles, Pythagorean Theorem)

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 $ARQD$ are both cyclic, with diameters of the circumcircles being $AB$ and $AD$ 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 5 (Cyclic Quadrilaterals, Similar Triangles, Pythagorean Theorem)

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 quadrilateral: \begin{align*} \angle ABE &= 180^\circ - \angle ABC =\angle ADC = \angle ADG, \\ \angle ADG &= \angle ADB =\angle ACB = \angle ACE. \hspace{20mm} \square \end{align*} 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 $\sqrt2\cdot[ABCD]=\boxed{567}$.

## Solution 6 (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 7 (Similar Triangles and Trigonometry)

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

## Solution 8 (Heron's Formula)

Let the points formed by dropping altitudes from $A$ to the lines $BC$, $CD$, and $BD$ be $E$, $F$, and $G$, respectively.

We have $$\triangle ABE \sim \triangle ADF \implies \frac{AD}{18} = \frac{AB}{15} \implies AD = \frac{6}{5}AB$$ and $$BD\cdot10 = 2[ABD] = AB\cdot18 \implies BD = \frac{9}{5}AB.$$

For convenience, let $AB = 5x$. By Heron's formula on $\triangle ABD$, we have sides $5x,6x,9x$ and semiperimeter $10x$, so $$\sqrt{10x\cdot5x\cdot4x\cdot1x} = [ABD] = \frac{AB\cdot18}{2} = 45x \implies 10\sqrt{2}x^2 = 45x \implies x= \frac{45}{10\sqrt{2}},$$ so $AB = 5x = \frac{45}{2\sqrt{2}}$.

Then, $$BE = \sqrt{AB^2 - CA^2} = \sqrt{\left(\frac{45}{2\sqrt{2}}\right)^2 - 15^2} = \sqrt{\frac{225}{8}} = \frac{15}{2\sqrt{2}}$$ and $$\triangle ABE \sim \triangle ADF \implies DF = \frac{6}{5}BE = \frac{6}{5}\cdot\frac{15}{2\sqrt{2}} = \frac{18}{2\sqrt{2}}.$$

Finally, recalling that $ABCD$ is isosceles, $$K = [ABCD] = \frac{18}{2}(AB + (AB + 2DF)) = 18(AB + DF) = 18\left(\frac{45}{2\sqrt{2}} + \frac{18}{2\sqrt{2}}\right) = \frac{567}{\sqrt{2}},$$ so $\sqrt{2}\cdot K = \boxed{567}$.