Difference between revisions of "2021 AIME I Problems/Problem 9"
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~MRENTHUSIASM | ~MRENTHUSIASM | ||
− | ==Solution 1== | + | ==Solution 1 (Similar Triangles and Pythagorean Theorem)== |
− | + | Let <math>\overline{AE}, \overline{AF},</math> and <math>\overline{AG}</math> be the perpendiculars from <math>A</math> to <math>\overleftrightarrow{BC}, \overleftrightarrow{CD},</math> and <math>\overleftrightarrow{BD},</math> respectively. Next, let <math>H</math> be the intersection of <math>\overline{AF}</math> and <math>\overline{BD}.</math> | |
− | + | We set <math>AB=x</math> and <math>AH=y,</math> 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 <math>HF=18-y</math> by segment subtraction, and <math>BG=\sqrt{x^2-10^2}</math> and <math>HG=\sqrt{y^2-10^2}</math> by the Pythagorean Theorem. | ||
+ | |||
+ | Since <math>\angle ABG</math> and <math>\angle HAG</math> are both complementary to <math>\angle AHB,</math> we have <math>\angle ABG = \angle HAG,</math> from which <math>\triangle ABG \sim \triangle HAG</math> by AA. It follows that <math>\frac{BG}{AG}=\frac{AG}{HG},</math> so <math>BG\cdot HG=AG^2,</math> or <cmath>\sqrt{x^2-10^2}\cdot\sqrt{y^2-10^2}=10^2. \hspace{10mm} (1)</cmath> | ||
+ | Since <math>\angle AHB = \angle FHD</math> by vertical angles, we have <math>\triangle AHB \sim \triangle FHD</math> by AA, with the ratio of similitude <math>\frac{AH}{FH}=\frac{BA}{DF}.</math> It follows that <math>DF=BA\cdot\frac{FH}{AH}=x\cdot\frac{18-y}{y}.</math> | ||
+ | |||
+ | Since <math>\angle EBA = \angle ECD = \angle FDA</math> by angle chasing, we have <math>\triangle EBA \sim \triangle FDA</math> by AA, with the ratio of similitude <math>\frac{EA}{FA}=\frac{BA}{DA}.</math> It follows that <math>DA=BA\cdot\frac{FA}{EA}=x\cdot\frac{18}{15}=\frac{6}{5}x.</math> | ||
− | + | By the Pythagorean Theorem on right <math>\triangle ADF,</math> we have <math>DF^2+AF^2=AD^2,</math> or <cmath>\left(x\cdot\frac{18-y}{y}\right)^2+18^2=\left(\frac{6}{5}x\right)^2. \hspace{7mm} (2)</cmath> | |
+ | Solving this system of equations (<math>(1)</math> and <math>(2)</math>), we get <math>x=\frac{45\sqrt2}{4}</math> and <math>y=\frac{90}{7},</math> so <math>AB=x=\frac{45\sqrt2}{4}</math> and <math>CD=AB+2DF=x+2\left(x\cdot\frac{18-y}{y}\right)=\frac{81\sqrt2}{4}.</math> Finally, the area of <math>ABCD</math> is <cmath>K=\frac{AB+CD}{2}\cdot AF=\frac{567\sqrt2}{2},</cmath> from which <math>\sqrt2 \cdot K=\boxed{567}.</math> | ||
− | + | ~MRENTHUSIASM | |
− | + | <u><b>Remark</b></u> | |
− | + | Instead of solving the system of equations <math>(1)</math> and <math>(2),</math> which can be time consuming, by noting that <math>\triangle ACF \sim \triangle ABG</math> by AA, we could find out <math>\frac{AB}{AG} = \frac{AC}{AF}</math>, which gives <math>AC = \frac{9}{5}x</math>. We also know that <math>EB = \sqrt{x^2 - 15^2}</math> by Pythagorean Theorem on <math>\triangle ABE</math>. From <math>BC = AD = \frac{6}{5}x,</math> we apply the Pythagorean Theorem to <math>\triangle ACE</math> and obtain | |
+ | <cmath>AC^2 = (EB+BC)^2 + AE^2.</cmath> | ||
+ | Substituting, we get | ||
+ | <cmath>\frac{81}{25}x^2 = \left(\sqrt{x^2 -225}+\frac{6}{5}x\right)^2+225 \iff x = 3\sqrt{x^2 - 15^2},</cmath> | ||
+ | from which <math>x = \frac{45\sqrt{2}}{4}.</math> | ||
− | + | ~Chupdogs | |
+ | |||
+ | ==Solution 2 (Similar Triangles and Pythagorean Theorem)== | ||
+ | |||
+ | First, draw the diagram. Then, notice that since <math>ABCD</math> is isosceles, <math>\Delta ABD \cong \Delta BAC</math>, and the length of the altitude from <math>B</math> to <math>AC</math> is also <math>10</math>. Let the foot of this altitude be <math>F</math>, and let the foot of the altitude from <math>A</math> to <math>BC</math> be denoted as <math>E</math>. Then, <math>\Delta BCF \sim \Delta ACE</math>. So, <math>\frac{BC}{AC} = \frac{BF}{AE} = \frac{2}{3}</math>. Now, notice that <math>[ABC] = \frac{10 \cdot AC} {2} = \frac{AB \cdot 18}{2} \implies AC = \frac{9 \cdot AB}{5}</math>, where <math>[ABC]</math> denotes the area of triangle <math>ABC</math>. Letting <math>AB = x</math>, this equality becomes <math>AC = \frac{9x}{5}</math>. Also, from <math>\frac{BC}{AC} = \frac{2}{3}</math>, we have <math>BC = \frac{6x}{5}</math>. Now, by the Pythagorean theorem on triangles <math>ABF</math> and <math>CBF</math>, we have <math>AF = \sqrt{x^{2}-100}</math> and <math>CF = \sqrt{ \left( \frac{6x}{5} \right) ^{2}-100}</math>. Notice that <math>AC = AF + CF</math>, so <math>\frac{9x}{5} = \sqrt{x^{2}-100} + \sqrt{ \left( \frac{6x}{5} \right) ^{2}-100}</math>. Squaring both sides of the equation once, moving <math>x^{2}-100</math> and <math> \left( \frac{6x}{5} \right) ^{2}-100</math> to the right, dividing both sides by <math>2</math>, and squaring the equation once more, we are left with <math>\frac{32x^{4}}{25} = 324x^{2}</math>. Dividing both sides by <math>x^{2}</math> (since we know <math>x</math> is positive), we are left with <math>\frac{32x^{2}}{25} = 324</math>. Solving for <math>x</math> gives us <math>x = \frac{45}{2\sqrt{2}}</math>. | ||
+ | |||
+ | Now, let the foot of the perpendicular from <math>A</math> to <math>CD</math> be <math>G</math>. Then let <math>DG = y</math>. Let the foot of the perpendicular from <math>B</math> to <math>CD</math> be <math>H</math>. Then, <math>CH</math> is also equal to <math>y</math>. Notice that <math>ABHG</math> is a rectangle, so <math>GH = x</math>. Now, we have <math>CG = GH + CH = x + y</math>. By the Pythagorean theorem applied to <math>\Delta AGC</math>, we have <math>(x+y)^{2}+18^{2}= \left( \frac{9x}{5} \right) ^{2}</math>. We know that <math>\frac{9x}{5} = \frac{9}{5} \cdot \frac{45}{2\sqrt{2}} = \frac{81}{2\sqrt{2}}</math>, so we can plug this into this equation. Solving for <math>x+y</math>, we get <math>x+y=\frac{63}{2\sqrt{2}}</math>. | ||
+ | |||
+ | Finally, to find <math>[ABCD]</math>, we use the formula for the area of a trapezoid: <math>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}}</math>. The problem asks us for <math>K \cdot \sqrt{2}</math>, which comes out to be <math>\boxed{567}</math>. | ||
+ | |||
+ | ~advanture | ||
− | + | ==Solution 3 (Similar Triangles and Pythagorean Theorem)== | |
+ | Make <math>AE</math> perpendicular to <math>BC</math>; <math>AG</math> perpendicular to <math>BD</math>; <math>AF</math> perpendicular <math>DC</math>. | ||
− | + | It's obvious that <math>\triangle{AEB} \sim \triangle{AFD}</math>. Let <math>EB=5x; AB=5y; DF=6x; AD=6y</math>. Then make <math>BQ</math> perpendicular to <math>DC</math>, it's easy to get <math>BQ=18</math>. | |
− | |||
− | + | Since <math>AB</math> parallel to <math>DC</math>, <math>\angle{ABG}=\angle{BDQ}</math>, so <math>\triangle{ABG} \sim \triangle{BDQ}</math>. After drawing the altitude, it's obvious that <math>FQ=AB=5y</math>, so <math>DQ=5y+6x</math>. According to the property of similar triangles, <math>AG/BQ=BG/DQ</math>. So, <math>\frac{5}{9}=\frac{GB}{(6x+5y)}</math>, or <math>GB=\frac{(30x+25y)}{9}</math>. | |
− | + | Now, we see the <math>\triangle AEB</math>, pretty easy to find that <math>15^2+(5x)^2=(5y)^2</math>, then we get <math>x^2+9=y^2</math>, then express <math>y</math> into <math>x</math> form that <math>y=\sqrt{x^2+9}</math> | |
+ | we put the length of <math>BG</math> back to <math>\triangle AGB</math>: <math>BG^2+100=AB^2</math>. So, <cmath>\frac{[30x+25\sqrt{(x^2+9)}]^2}{81}+100=(5\sqrt{x^2+9})^2.</cmath> | ||
+ | After calculating, we can have a final equation of <math>x^2+9=\sqrt{x^2+9}\cdot3x</math>. It's easy to find <math>x=\frac{3\sqrt{2}}{4}</math> then <math>y=\frac{9\sqrt{2}}{4}</math>. So, <cmath>\sqrt{2}\cdot K = \sqrt{2}\cdot(5y+5y+6x+6x)\cdot9=\boxed{567}.</cmath> | ||
+ | ~bluesoul | ||
− | ==Solution | + | ==Solution 4 (Cyclic Quadrilaterals, Similar Triangles, Pythagorean Theorem)== |
Let the foot of the altitude from <math>A</math> to <math>BC</math> be <math>P</math>, to <math>CD</math> be <math>Q</math>, and to <math>BD</math> be <math>R</math>. | Let the foot of the altitude from <math>A</math> to <math>BC</math> be <math>P</math>, to <math>CD</math> be <math>Q</math>, and to <math>BD</math> be <math>R</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>. | 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> | + | We also see that quadrilaterals <math>APBR</math> and <math>ARQD</math> are both cyclic, with diameters of the circumcircles being <math>AB</math> and <math>AD</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 <math>A</math> 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. |
To do this, we use the one number we have for <math>\triangle APD</math>: we know that the altitude from <math>A</math> to <math>BD</math> has length <math>10</math>. As the two triangles are similar, if we can find the height from <math>A</math> to <math>PQ</math>, we can take the ratio of the two heights as the ratio of similitude. To do this, we once again note that <math>QP = 27</math>. Using this, we can drop the altitude from <math>A</math> to <math>QP</math> and let it intersect <math>QP</math> at <math>H</math>. Then, let <math>QH = x</math> and thus <math>HP=27-x</math>. We then have by the Pythagorean Theorem on <math>\triangle AQH</math> and <math>\triangle APH</math>: | To do this, we use the one number we have for <math>\triangle APD</math>: we know that the altitude from <math>A</math> to <math>BD</math> has length <math>10</math>. As the two triangles are similar, if we can find the height from <math>A</math> to <math>PQ</math>, we can take the ratio of the two heights as the ratio of similitude. To do this, we once again note that <math>QP = 27</math>. Using this, we can drop the altitude from <math>A</math> to <math>QP</math> and let it intersect <math>QP</math> at <math>H</math>. Then, let <math>QH = x</math> and thus <math>HP=27-x</math>. We then have by the Pythagorean Theorem on <math>\triangle AQH</math> and <math>\triangle APH</math>: | ||
Line 76: | Line 133: | ||
Then, we have the height of trapezoid <math>ABCD</math> is <math>AQ = 18</math>, the top base is <math>AB = \frac{45\sqrt{2}}{4}</math>, and the bottom base is <math>CD = \frac{45\sqrt{2}}{4} + 2\cdot\frac{9\sqrt{2}}{2}</math>. From the equation of a trapezoid, <math>K = \frac{b_1+b_2}{2} \cdot h = \frac{63\sqrt{2}}{4} \cdot 18 = \frac{567\sqrt{2}}{2}</math>, so the answer is <math>K\sqrt{2} = \boxed{567}</math>. | Then, we have the height of trapezoid <math>ABCD</math> is <math>AQ = 18</math>, the top base is <math>AB = \frac{45\sqrt{2}}{4}</math>, and the bottom base is <math>CD = \frac{45\sqrt{2}}{4} + 2\cdot\frac{9\sqrt{2}}{2}</math>. From the equation of a trapezoid, <math>K = \frac{b_1+b_2}{2} \cdot h = \frac{63\sqrt{2}}{4} \cdot 18 = \frac{567\sqrt{2}}{2}</math>, so the answer is <math>K\sqrt{2} = \boxed{567}</math>. | ||
− | + | ~lvmath | |
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− | ==Solution 5 ( | + | ==Solution 5 (Cyclic Quadrilaterals, Similar Triangles, Pythagorean Theorem)== |
Let <math>E,F,</math> and <math>G</math> be the feet of the altitudes from <math>A</math> to <math>BC,CD,</math> and <math>DB</math>, respectively. | Let <math>E,F,</math> and <math>G</math> be the feet of the altitudes from <math>A</math> to <math>BC,CD,</math> and <math>DB</math>, respectively. | ||
Line 94: | Line 141: | ||
Claim: We have <math>2</math> pairs of similar right triangles: <math>\triangle AEB \sim \triangle AFD</math> and <math>\triangle AGD \sim \triangle AEC</math>. | Claim: We have <math>2</math> pairs of similar right triangles: <math>\triangle AEB \sim \triangle AFD</math> and <math>\triangle AGD \sim \triangle AEC</math>. | ||
− | Proof: Note that <math>ABCD</math> is cyclic. We need one more angle, and we get this from this cyclic | + | Proof: Note that <math>ABCD</math> is cyclic. We need one more angle, and we get this from this cyclic quadrilateral: |
− | + | <cmath>\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*}</cmath> | ||
Let <math>AD=a</math>. We obtain from the similarities <math>AB = \frac{5a}{6}</math> and <math>AC=BD=\frac{3a}{2}</math>. | Let <math>AD=a</math>. We obtain from the similarities <math>AB = \frac{5a}{6}</math> and <math>AC=BD=\frac{3a}{2}</math>. | ||
Line 105: | Line 154: | ||
Applying the Pythagorean theorem on <math>\triangle ADF</math>, we get <math>324=a^2 - \frac{a^2}{9}=\frac{8a^2}{9}</math>. | Applying the Pythagorean theorem on <math>\triangle ADF</math>, we get <math>324=a^2 - \frac{a^2}{9}=\frac{8a^2}{9}</math>. | ||
− | Thus, <math>a=\frac{27}{\sqrt{2}}</math>, and <math>[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}}</math>, yielding <math>\boxed{567}</math>. | + | Thus, <math>a=\frac{27}{\sqrt{2}}</math>, and <math>[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}}</math>, yielding <math>\sqrt2\cdot[ABCD]=\boxed{567}</math>. |
− | ==Solution 6 ( | + | ==Solution 6 (Similar Triangles and Trigonometry)== |
− | Let <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>. |
+ | |||
+ | Note that <math>\triangle CBG\sim\triangle CAF</math>, and we get that <math>\frac{10}{15}=\frac{a}{AC}</math>. Therefore, <math>AC=\frac32 a</math>. It then follows that <math>\triangle ABF\sim\triangle ADH</math>. Using similar triangles, we can then find that <math>AB=\frac{5}{6}a</math>. Using the Law of Cosines on <math>\triangle ABC</math>, We can find that the <math>\cos\angle ABC=-\frac{1}{3}</math>. Since <math>\angle ABF=\angle ADH</math>, and each is supplementary to <math>\angle ABC</math>, we know that the <math>\cos\angle ADH=\frac{1}{3}</math>. It then follows that <math>a=\frac{27\sqrt{2}}{2}</math>. Then it can be found that the area <math>K</math> is <math>\frac{567\sqrt{2}}{2}</math>. Multiplying this by <math>\sqrt{2}</math>, the answer is <math>\boxed{567}</math>. | ||
+ | |||
+ | ~happykeeper | ||
+ | |||
+ | ==Solution 7 (Similar Triangles and Trigonometry)== | ||
+ | Draw the distances in terms of <math>B</math>, as shown in the diagram. By similar triangles, <math>\triangle{AEC}\sim\triangle{BIC}</math>. As a result, let <math>AB=u</math>, then <math>BC=AD=\frac{6}{5}u</math> and <math>2AC=3BC</math>. The triangle <math>ABC</math> is <math>6-5-9</math> which <math>\cos(\angle{ABC})=-\frac{1}{3}</math>. By angle subtraction, <math>\cos(180-\theta)=-\cos\theta</math>. Therefore, <math>AB=\frac{45}{2\sqrt{2}}=\frac{45\sqrt{2}}{4}</math> and <math>AD=BC=\frac{27}{\sqrt{2}}</math>. By trapezoid area formula, the area of <math>ABCD</math> is equal to <math>(AB+DF)\cdot 18=567\cdot \frac{\sqrt{2}}{2}</math> which <math>\sqrt{2}\cdot k=\boxed{567}</math>. | ||
+ | |||
+ | ~math2718281828459 | ||
− | + | ==Solution 8 (Heron's Formula)== | |
<asy> | <asy> | ||
− | |||
size(250); | size(250); | ||
pair A, B, C, D, E, F, G, H; | pair A, B, C, D, E, F, G, H; | ||
Line 127: | Line 184: | ||
draw(rightanglemark(A,F,C),red); | draw(rightanglemark(A,F,C),red); | ||
draw(rightanglemark(A,G,D),red); | draw(rightanglemark(A,G,D),red); | ||
+ | filldraw(A--D--F--cycle,yellow,black+linewidth(1.5)); | ||
+ | filldraw(A--B--E--cycle,yellow,black+linewidth(1.5)); | ||
dot("$A$",A,1.5*NW,linewidth(4)); | dot("$A$",A,1.5*NW,linewidth(4)); | ||
dot("$B$",B,1.5*NE,linewidth(4)); | dot("$B$",B,1.5*NE,linewidth(4)); | ||
dot("$C$",C,1.5*SE,linewidth(4)); | dot("$C$",C,1.5*SE,linewidth(4)); | ||
dot("$D$",D,1.5*SW,linewidth(4)); | dot("$D$",D,1.5*SW,linewidth(4)); | ||
− | dot( | + | dot(E,linewidth(4)); |
− | + | dot(F,linewidth(4)); | |
− | + | dot(G,linewidth(4)); | |
− | + | label("$E$",E,NE); | |
+ | label("$F$",F, S); | ||
+ | label("$G$",G,SE); | ||
draw(A--B--C--D--cycle^^B--D^^B--E); | draw(A--B--C--D--cycle^^B--D^^B--E); | ||
draw(A--E^^A--F^^A--G,dashed); | draw(A--E^^A--F^^A--G,dashed); | ||
label("$10$",midpoint(A--G),1.5*(1,0)); | label("$10$",midpoint(A--G),1.5*(1,0)); | ||
label("$15$",midpoint(A--E),1.5*N); | label("$15$",midpoint(A--E),1.5*N); | ||
+ | label("$5x$",midpoint(A--B),S); | ||
+ | label("$6x$",midpoint(A--D),1.5*(-1,0)); | ||
Label L = Label("$18$", align=(0,0), position=MidPoint, filltype=Fill(0,3,white)); | 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)); | draw(C+(5,0)--(81sqrt(2)/8,18)+(5,0), L=L, arrow=Arrows(),bar=Bars(15)); | ||
− | |||
− | |||
</asy> | </asy> | ||
− | + | Let the points formed by dropping altitudes from <math>A</math> to the lines <math>BC</math>, <math>CD</math>, and <math>BD</math> be <math>E</math>, <math>F</math>, and <math>G</math>, respectively. | |
+ | |||
+ | We have | ||
+ | <cmath>\triangle ABE \sim \triangle ADF \implies \frac{AD}{18} = \frac{AB}{15} \implies AD = \frac{6}{5}AB</cmath> | ||
+ | and | ||
+ | <cmath>BD\cdot10 = 2[ABD] = AB\cdot18 \implies BD = \frac{9}{5}AB.</cmath> | ||
+ | For convenience, let <math>AB = 5x</math>. By Heron's formula on <math>\triangle ABD</math>, we have sides <math>5x,6x,9x</math> and semiperimeter <math>10x</math>, so | ||
+ | <cmath>\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}},</cmath> | ||
+ | so <math>AB = 5x = \frac{45}{2\sqrt{2}}</math>. | ||
+ | |||
+ | Then, | ||
+ | <cmath>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}}</cmath> | ||
+ | and | ||
+ | <cmath>\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}}.</cmath> | ||
+ | Finally, recalling that <math>ABCD</math> is isosceles, | ||
+ | <cmath>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}},</cmath> | ||
+ | so <math>\sqrt{2}\cdot K = \boxed{567}</math>. | ||
+ | |||
+ | ~[[User:emerald_block|emerald_block]] | ||
+ | |||
+ | ==Solution 9 (Three Heights)== | ||
+ | [[File:2021 AIME I 9.jpg|450px|right]] | ||
+ | |||
+ | Let <math>\overline{AE}, \overline{AF},</math> and <math>\overline{AG}</math> be the perpendiculars from <math>A</math> to <math>{BC}, {CD},</math> and <math>{BD},</math> respectively. | ||
+ | <math>AE = 15, AF = 18, AG =10</math>. | ||
+ | Denote by <math>G'</math> the base of the perpendicular from <math>B</math> to <math>AC, H</math> be the base of the perpendicular from <math>C</math> to <math>AB</math>. Denote <math>\theta = \angle{CBH}.</math> | ||
+ | It is clear that <cmath>BG' = AG, CH = AF, \triangle CBH \ =\triangle ADF,</cmath> the area of <math>ABCD</math> is equal to the area of the rectangle <math>AFCH.</math> | ||
+ | |||
+ | The problem is reduced to finding <math>AH</math>. | ||
+ | |||
+ | In triangle <math>ABC</math> all altitudes are known: | ||
+ | <cmath>AB : BC : AC = \frac{1}{CH}\ : \frac{1}{AE}\ : \frac{1}{BG'}\ =</cmath> | ||
+ | <cmath>= \frac{1}{AF}\ : \frac{1}{AE}\ : \frac{1}{AG}\ = 5 : 6 : 9.</cmath> | ||
+ | We apply the Law of Cosines to <math>\triangle ABC</math> and get<math>:</math> | ||
+ | <cmath>\begin{align*} 2\cdot AB\cdot BC \cdot \cos\theta = AC^2 – AB^2 – BC^2, \end{align*}</cmath> | ||
+ | <cmath>\begin{align*} 2\cdot 5\cdot 6\cdot \cos\theta = 60 \cos\theta = 9^2 – 5^2 – 6^2 = 20, \cos\theta =\frac{1}{3}. \end{align*}</cmath> | ||
+ | <cmath>\begin{align*} BH = BC \cos\theta = \frac{BC}{3}.\end{align*}</cmath> | ||
+ | We apply the Pythagorean Law to <math>\triangle HBC</math> and get<math>:</math> | ||
+ | <cmath>\begin{align*} HC^2 = 18^2 = BC^2 – BH^2 = 9\cdot BH^2 – BH^2 = 8 BH^2.\end{align*}</cmath> | ||
+ | <cmath>\begin{align*} BH = \frac{9}{\sqrt2}, AH = (\frac{5}{2} + 1)\cdot BH = \frac{63}{2\cdot \sqrt2}. \end{align*}</cmath> | ||
+ | Required area is | ||
+ | <cmath>\begin{align*} K = \frac{63}{2\cdot \sqrt{2}} \cdot 18 = \frac{567}{\sqrt{2}} \implies \sqrt{2} K=\boxed{567}. \end{align*}</cmath> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Solution 10 (Area)== | ||
+ | Let <math>F</math> be on <math>DC</math> such that <math>AF \| DC</math>. Let <math>G</math> be on <math>BD</math> such that <math>AG \| BD</math>. | ||
+ | |||
+ | Let <math>m</math> be the length of <math>AB</math>. Let <math>n</math> be the length of <math>AD</math>. | ||
+ | |||
+ | The area of <math>\triangle ABD</math> can be expressed in three ways: <math>\frac{1}{2}(15)(BC) = \frac{1}{2}(15)(n)</math>, <math>\frac{1}{2}(18)(m)</math>, and <math>\frac{1}{2}(10)(BD)</math>. | ||
+ | |||
+ | <cmath> | ||
+ | \frac{1}{2}(15)(n) = \frac{1}{2}(18)(m) | ||
+ | </cmath> | ||
+ | <cmath> | ||
+ | 15n = 18m | ||
+ | </cmath> | ||
+ | <cmath> | ||
+ | 5n = 6m | ||
+ | </cmath> | ||
+ | <cmath> | ||
+ | n = \frac{6}{5}m | ||
+ | </cmath> | ||
+ | |||
+ | Now, <math>BD = BG + GD = \sqrt{m^2-100} + \sqrt{n^2-100}</math>. We can substitute in <math>n = \frac{6}{5}m</math> to get | ||
− | + | <math>BD = \sqrt{m^2-100} + \sqrt{(\frac{6}{5}m)^2-100}</math>. | |
− | |||
− | + | We have | |
+ | <cmath> | ||
+ | \frac{1}{2}(10)\left(\sqrt{m^2-100} + \sqrt{(\frac{6}{5}m)^2-100}\right) = \frac{1}{2}(18)(m) | ||
+ | </cmath> | ||
+ | After a fairly straightforward algebraic bash, we get <math>m = \frac{45\sqrt{2}}{4}</math>, and <math>n = (\frac{6}{5})(\frac{45\sqrt{2}}{4}) = \frac{27\sqrt{2}}{2}</math>. By the Pythagorean Theorem on <math>\triangle ADF</math>, <math>DF^2 = n^2 - 18^2 = \frac{729}{2} - 324 = \frac{81}{2}</math>, and <math>DF = \frac{9\sqrt{2}}{2}</math>. | ||
− | + | Thus, <math>DC = 2DF + AB = 9\sqrt{2}+\frac{45\sqrt{2}}{4} = \frac{81\sqrt{2}}{4}</math>. Therefore, <math>K = \frac{1}{2}(\frac{45\sqrt{2}}{4}+\frac{81\sqrt{2}}{4}) \cdot 18 = \frac{63\sqrt{2}}{2} \cdot 18 = \frac{567\sqrt{2}}{2}</math>. The requested answer is <math>K \cdot \sqrt{2} = \boxed{567}</math>. | |
− | |||
− | ~ | + | ~ adam_zheng |
− | ==Solution | + | ==Video Solution== |
− | + | https://youtu.be/uItEKVj-tF8 | |
− | ~ | + | ~Mathproblemsolvingskills.com |
==Video Solution== | ==Video Solution== |
Latest revision as of 18:16, 25 January 2024
Contents
- 1 Problem
- 2 Diagram
- 3 Solution 1 (Similar Triangles and Pythagorean Theorem)
- 4 Solution 2 (Similar Triangles and Pythagorean Theorem)
- 5 Solution 3 (Similar Triangles and Pythagorean Theorem)
- 6 Solution 4 (Cyclic Quadrilaterals, Similar Triangles, Pythagorean Theorem)
- 7 Solution 5 (Cyclic Quadrilaterals, Similar Triangles, Pythagorean Theorem)
- 8 Solution 6 (Similar Triangles and Trigonometry)
- 9 Solution 7 (Similar Triangles and Trigonometry)
- 10 Solution 8 (Heron's Formula)
- 11 Solution 9 (Three Heights)
- 12 Solution 10 (Area)
- 13 Video Solution
- 14 Video Solution
- 15 See Also
Problem
Let be an isosceles trapezoid with
and
Suppose that the distances from
to the lines
and
are
and
respectively. Let
be the area of
Find
Diagram
~MRENTHUSIASM
Solution 1 (Similar Triangles and Pythagorean Theorem)
Let and
be the perpendiculars from
to
and
respectively. Next, let
be the intersection of
and
We set and
as shown below.
From here, we obtain
by segment subtraction, and
and
by the Pythagorean Theorem.
Since and
are both complementary to
we have
from which
by AA. It follows that
so
or
Since
by vertical angles, we have
by AA, with the ratio of similitude
It follows that
Since by angle chasing, we have
by AA, with the ratio of similitude
It follows that
By the Pythagorean Theorem on right we have
or
Solving this system of equations (
and
), we get
and
so
and
Finally, the area of
is
from which
~MRENTHUSIASM
Remark
Instead of solving the system of equations and
which can be time consuming, by noting that
by AA, we could find out
, which gives
. We also know that
by Pythagorean Theorem on
. From
we apply the Pythagorean Theorem to
and obtain
Substituting, we get
from which
~Chupdogs
Solution 2 (Similar Triangles and Pythagorean Theorem)
First, draw the diagram. Then, notice that since is isosceles,
, and the length of the altitude from
to
is also
. Let the foot of this altitude be
, and let the foot of the altitude from
to
be denoted as
. Then,
. So,
. Now, notice that
, where
denotes the area of triangle
. Letting
, this equality becomes
. Also, from
, we have
. Now, by the Pythagorean theorem on triangles
and
, we have
and
. Notice that
, so
. Squaring both sides of the equation once, moving
and
to the right, dividing both sides by
, and squaring the equation once more, we are left with
. Dividing both sides by
(since we know
is positive), we are left with
. Solving for
gives us
.
Now, let the foot of the perpendicular from to
be
. Then let
. Let the foot of the perpendicular from
to
be
. Then,
is also equal to
. Notice that
is a rectangle, so
. Now, we have
. By the Pythagorean theorem applied to
, we have
. We know that
, so we can plug this into this equation. Solving for
, we get
.
Finally, to find , we use the formula for the area of a trapezoid:
. The problem asks us for
, which comes out to be
.
~advanture
Solution 3 (Similar Triangles and Pythagorean Theorem)
Make perpendicular to
;
perpendicular to
;
perpendicular
.
It's obvious that . Let
. Then make
perpendicular to
, it's easy to get
.
Since parallel to
,
, so
. After drawing the altitude, it's obvious that
, so
. According to the property of similar triangles,
. So,
, or
.
Now, we see the , pretty easy to find that
, then we get
, then express
into
form that
we put the length of
back to
:
. So,
After calculating, we can have a final equation of
. It's easy to find
then
. So,
~bluesoul
Solution 4 (Cyclic Quadrilaterals, Similar Triangles, Pythagorean Theorem)
Let the foot of the altitude from to
be
, to
be
, and to
be
.
Note that all isosceles trapezoids are cyclic quadrilaterals; thus, is on the circumcircle of
and we have that
is the Simson Line from
. As
, we have that
, with the last equality coming from cyclic quadrilateral
. Thus,
and we have that
or that
, which we can see gives us that
. Further ratios using the same similar triangles gives that
and
.
We also see that quadrilaterals and
are both cyclic, with diameters of the circumcircles being
and
respectively. The intersection of the circumcircles are the points
and
, and we know
and
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
taking
to
. Because we know a lot about
but very little about
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 : we know that the altitude from
to
has length
. As the two triangles are similar, if we can find the height from
to
, we can take the ratio of the two heights as the ratio of similitude. To do this, we once again note that
. Using this, we can drop the altitude from
to
and let it intersect
at
. Then, let
and thus
. We then have by the Pythagorean Theorem on
and
:
Then,
. This gives us then from right triangle
that
and thus the ratio of
to
is
. From this, we see then that
and
The Pythagorean Theorem on
then gives that
Then, we have the height of trapezoid
is
, the top base is
, and the bottom base is
. From the equation of a trapezoid,
, so the answer is
.
~lvmath
Solution 5 (Cyclic Quadrilaterals, Similar Triangles, Pythagorean Theorem)
Let and
be the feet of the altitudes from
to
and
, respectively.
Claim: We have pairs of similar right triangles:
and
.
Proof: Note that is cyclic. We need one more angle, and we get this from this cyclic quadrilateral:
Let
. We obtain from the similarities
and
.
By Ptolemy, , so
.
We obtain , so
.
Applying the Pythagorean theorem on , we get
.
Thus, , and
, yielding
.
Solution 6 (Similar Triangles and Trigonometry)
Let . Draw diagonal
and let
be the foot of the perpendicular from
to
,
be the foot of the perpendicular from
to line
, and
be the foot of the perpendicular from
to
.
Note that , and we get that
. Therefore,
. It then follows that
. Using similar triangles, we can then find that
. Using the Law of Cosines on
, We can find that the
. Since
, and each is supplementary to
, we know that the
. It then follows that
. Then it can be found that the area
is
. Multiplying this by
, the answer is
.
~happykeeper
Solution 7 (Similar Triangles and Trigonometry)
Draw the distances in terms of , as shown in the diagram. By similar triangles,
. As a result, let
, then
and
. The triangle
is
which
. By angle subtraction,
. Therefore,
and
. By trapezoid area formula, the area of
is equal to
which
.
~math2718281828459
Solution 8 (Heron's Formula)
Let the points formed by dropping altitudes from
to the lines
,
, and
be
,
, and
, respectively.
We have
and
For convenience, let
. By Heron's formula on
, we have sides
and semiperimeter
, so
so
.
Then,
and
Finally, recalling that
is isosceles,
so
.
Solution 9 (Three Heights)
Let and
be the perpendiculars from
to
and
respectively.
.
Denote by
the base of the perpendicular from
to
be the base of the perpendicular from
to
. Denote
It is clear that
the area of
is equal to the area of the rectangle
The problem is reduced to finding .
In triangle all altitudes are known:
We apply the Law of Cosines to
and get
We apply the Pythagorean Law to
and get
Required area is
vladimir.shelomovskii@gmail.com, vvsss
Solution 10 (Area)
Let be on
such that
. Let
be on
such that
.
Let be the length of
. Let
be the length of
.
The area of can be expressed in three ways:
,
, and
.
Now, . We can substitute in
to get
.
We have
After a fairly straightforward algebraic bash, we get
, and
. By the Pythagorean Theorem on
,
, and
.
Thus, . Therefore,
. The requested answer is
.
~ adam_zheng
Video Solution
~Mathproblemsolvingskills.com
Video Solution
https://www.youtube.com/watch?v=6rLnl8z7lnM
See Also
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 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.