Difference between revisions of "2020 AIME I Problems/Problem 13"
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pen dotstyle = black; /* point style */ | pen dotstyle = black; /* point style */ | ||
real xmin = -10.645016481888238, xmax = 5.4445786933235505, ymin = 0.7766255516825293, ymax = 9.897545413994122; /* image dimensions */ | real xmin = -10.645016481888238, xmax = 5.4445786933235505, ymin = 0.7766255516825293, ymax = 9.897545413994122; /* image dimensions */ | ||
− | pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen rvwvcq = rgb(0. | + | pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen rvwvcq = rgb(0.8235294117647059,0.0396078431372549,0.07529411764705882); |
draw((-6.837129089839387,8.163360372429347)--(-6.8268938290378,5.895596632024835)--(-4.33118398380513,6.851781504978754)--cycle, linewidth(2) + rvwvcq); | draw((-6.837129089839387,8.163360372429347)--(-6.8268938290378,5.895596632024835)--(-4.33118398380513,6.851781504978754)--cycle, linewidth(2) + rvwvcq); | ||
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Points are defined as shown. It is pretty easy to show that <math>\triangle AFE \sim \triangle AGH</math> by spiral similarity at <math>A</math> by some short angle chasing. Now, note that <math>AD</math> is the altitude of <math>\triangle AFE</math>, as the altitude of <math>AGH</math>. We need to compare these altitudes in order to compare their areas. Note that Stewart's theorem implies that <math>AD/2 = \frac{\sqrt{18}}{2}</math>, the altitude of <math>\triangle AFE</math>. Similarly, the altitude of <math>\triangle AGH</math> is the altitude of <math>\triangle ABC</math>, or <math>\frac{3\sqrt{7}}{2}</math>. However, it's not too hard to see that <math>GB = HC = 1</math>, and therefore <math>[AGH] = [ABC]</math>. From here, we get that the area of <math>\triangle ABC</math> is <math>\frac{15\sqrt{7}}{14} \implies \boxed{036}</math>, by similarity. ~awang11 | Points are defined as shown. It is pretty easy to show that <math>\triangle AFE \sim \triangle AGH</math> by spiral similarity at <math>A</math> by some short angle chasing. Now, note that <math>AD</math> is the altitude of <math>\triangle AFE</math>, as the altitude of <math>AGH</math>. We need to compare these altitudes in order to compare their areas. Note that Stewart's theorem implies that <math>AD/2 = \frac{\sqrt{18}}{2}</math>, the altitude of <math>\triangle AFE</math>. Similarly, the altitude of <math>\triangle AGH</math> is the altitude of <math>\triangle ABC</math>, or <math>\frac{3\sqrt{7}}{2}</math>. However, it's not too hard to see that <math>GB = HC = 1</math>, and therefore <math>[AGH] = [ABC]</math>. From here, we get that the area of <math>\triangle ABC</math> is <math>\frac{15\sqrt{7}}{14} \implies \boxed{036}</math>, by similarity. ~awang11 | ||
− | ==Solution 2(coord bash + basic geometry)== | + | ==Solution 2== |
+ | Let <math>M_A</math>, <math>M_B</math>, <math>M_C</math> be the midpoints of arcs <math>BC</math>, <math>CA</math>, <math>AB</math>. By Fact 5, we know that <math>M_AB = M_AC = M_AI</math>, and so by Ptolmey's theorem, we deduce that <cmath>AB\cdot M_AC + AC\cdot M_AB = BC\cdot M_AA \implies M_AA = 2M_AI.</cmath> | ||
+ | In particular, we have <math>AI = IM_A</math>. | ||
+ | <asy> | ||
+ | defaultpen(fontsize(10pt)); | ||
+ | size(200); | ||
+ | pair A, B, C, D, E, F, I, P, MA, MB, MC; | ||
+ | B = (0,0); | ||
+ | C = (5,0); | ||
+ | A = IP(Circle(B, 4), Circle(C, 6), 0); | ||
+ | I = incenter(A, B, C); | ||
+ | D = extension(A, I, B, C); | ||
+ | P = midpoint(A--D); | ||
+ | E = extension(P, rotate(90, P)*A, B, I); | ||
+ | F = extension(P, rotate(90, P)*A, C, I); | ||
+ | MA = IP(Line(A, I, 20), circumcircle(A, B, C), 1); | ||
+ | MB = IP(Line(B, I, 20), circumcircle(A, B, C), 1); | ||
+ | MC = IP(Line(C, I, 20), circumcircle(A, B, C), 1); | ||
+ | draw(A--B--C--cycle, orange); | ||
+ | draw(circumcircle(A, B, C), red); | ||
+ | draw(A--E--F--cycle, lightblue); | ||
+ | draw(E--D--F, lightblue); | ||
+ | draw(A--MA^^B--MB^^C--MC, heavygreen); | ||
+ | draw(MA--MB--MC--cycle, magenta); | ||
+ | dot("$A$", A, dir(120)); | ||
+ | dot("$B$", B, dir(220)); | ||
+ | dot("$C$", C, dir(320)); | ||
+ | dot("$D$", D, dir(230)); | ||
+ | dot("$E$", E, dir(330)); | ||
+ | dot("$F$", F, dir(250)); | ||
+ | dot("$I$", I, dir(80)); | ||
+ | dot("$M_A$", MA, dir(270)); | ||
+ | dot("$M_B$", MB, dir(60)); | ||
+ | dot("$M_C$", MC, dir(150)); | ||
+ | </asy> | ||
+ | Now the key claim is that: | ||
+ | |||
+ | <b>Claim:</b> <math>\triangle DEF</math> and <math>\triangle M_AM_BM_C</math> are homothetic at <math>I</math> with ratio <math>2</math>. | ||
+ | |||
+ | <i>Proof.</i> First, we show that <math>D</math> is the midpoint of <math>M_AI</math>. Indeed, we have | ||
+ | <cmath>\frac{ID}{DM_A} = \frac{BI}{BM_A}\cdot \frac{\sin\angle IBC}{\sin \angle CBM_A} = \frac{BI}{AI}\cdot\frac{\sin \angle B/2}{\sin \angle A/2} = 1</cmath> | ||
+ | by Ratio lemma and Law of Sines. | ||
+ | |||
+ | Now observe that: | ||
+ | * <math>\overline{M_BM_C}</math> is the perpendicular bisector of <math>\overline{AI}</math>, | ||
+ | * <math>\overline{EF}</math> is the perpendicular bisector of <math>\overline{AD}</math>, and | ||
+ | * <math>ID = AI/2</math>. | ||
+ | |||
+ | Combining these facts gives that <math>\overline{EF}</math> is a midline in <math>\triangle IM_BM_C</math>, which proves the claim. <math>\blacksquare</math> | ||
+ | |||
+ | To finish, we compute <math>[M_AM_BM_C]</math>, noting that <math>[AEF] = [DEF] = \tfrac{1}{4}[M_AM_BM_C]</math>. | ||
+ | |||
+ | By Heron's, we can calculate the circumradius <math>R = 8/\sqrt{7}</math>, and by Law of Cosines, we get | ||
+ | <cmath>\begin{align*}\cos A &= \frac{9}{16}\implies \cos A/2 = \frac{5}{\sqrt{32}} \\ | ||
+ | \cos B &= \frac{1}{8} \implies \cos B/2 = \frac{3}{4} \\ | ||
+ | \cos C &= \frac{3}{4} \implies \cos C/2 = \sqrt{\frac{7}{8}}.\end{align*}</cmath> | ||
+ | Then using <math>[XYZ] = 2R^2\sin X\sin Y\sin Z</math>, we can compute | ||
+ | <cmath>[M_AM_BM_C] = 2\cdot \frac{64}{7}\cdot \frac{5}{\sqrt{32}}\cdot \frac{3}{4}\cdot \frac{\sqrt{7}}{\sqrt{8}} = \frac{30\sqrt{7}}{7}.</cmath> | ||
+ | Thus <math>[AEF] = 15\sqrt{7}/14</math>, which gives a final answer of <math>\boxed{036}</math>. | ||
+ | |||
+ | ~pinetree1 | ||
+ | |||
+ | ==Solution 3(coord bash + basic geometry)== | ||
Let <math>\overline{BC}</math> lie on the x-axis and <math>B</math> be the origin. <math>C</math> is <math>(5,0)</math>. Use Heron's formula to compute the area of triangle <math>ABC</math>. We have <math>s=\frac{15}{2}</math>. and <math>[ABC]=\sqrt{\frac{15 \cdot 7 \cdot 5 \cdot 3}{2^4}}=\frac{15\sqrt{7}}{4}</math>. We now find the altitude, which is <math>\frac{\frac{15\sqrt{7}}{2}}{5}=\frac{3\sqrt{7}}{2}</math>, which is the y-coordinate of <math>A</math>. We now find the x-coordinate of <math>A</math>, which satisfies <math>x^2 + (\frac{3\sqrt{7}}{2})^{2}=16</math>, which gives <math>x=\frac{1}{2}</math> since the triangle is acute. Now using the Angle Bisector Theorem, we have <math>\frac{4}{6}=\frac{BD}{CD}</math> and <math>BD+CD=5</math> to get <math>BD=2</math>. The coordinates of D are <math>(2,0)</math>. | Let <math>\overline{BC}</math> lie on the x-axis and <math>B</math> be the origin. <math>C</math> is <math>(5,0)</math>. Use Heron's formula to compute the area of triangle <math>ABC</math>. We have <math>s=\frac{15}{2}</math>. and <math>[ABC]=\sqrt{\frac{15 \cdot 7 \cdot 5 \cdot 3}{2^4}}=\frac{15\sqrt{7}}{4}</math>. We now find the altitude, which is <math>\frac{\frac{15\sqrt{7}}{2}}{5}=\frac{3\sqrt{7}}{2}</math>, which is the y-coordinate of <math>A</math>. We now find the x-coordinate of <math>A</math>, which satisfies <math>x^2 + (\frac{3\sqrt{7}}{2})^{2}=16</math>, which gives <math>x=\frac{1}{2}</math> since the triangle is acute. Now using the Angle Bisector Theorem, we have <math>\frac{4}{6}=\frac{BD}{CD}</math> and <math>BD+CD=5</math> to get <math>BD=2</math>. The coordinates of D are <math>(2,0)</math>. | ||
Since we want the area of triangle <math>AEF</math>, we will find equations for perpendicular bisector of AD, and the other two angle bisectors. The perpendicular bisector is not too challenging: the midpoint of AD is <math>(\frac{5}{4}, \frac{3\sqrt{7}}{4})</math> and the slope of AD is <math>-\sqrt{7}</math>. The slope of the perpendicular bisector is <math>\frac{1}{\sqrt{7}}</math>. The equation is(in point slope form) <math>y-\frac{3\sqrt{7}}{4}=\frac{1}{\sqrt{7}}(x-\frac{5}{4})</math>. | Since we want the area of triangle <math>AEF</math>, we will find equations for perpendicular bisector of AD, and the other two angle bisectors. The perpendicular bisector is not too challenging: the midpoint of AD is <math>(\frac{5}{4}, \frac{3\sqrt{7}}{4})</math> and the slope of AD is <math>-\sqrt{7}</math>. The slope of the perpendicular bisector is <math>\frac{1}{\sqrt{7}}</math>. The equation is(in point slope form) <math>y-\frac{3\sqrt{7}}{4}=\frac{1}{\sqrt{7}}(x-\frac{5}{4})</math>. | ||
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The area of AEF is equal to <math>\frac{EF \cdot \frac{AD}{2}}{2}</math> since AD is the altitude of that triangle with EF as the base, with <math>\frac{AD}{2}</math> being the height. <math>EF=\frac{5\sqrt{2}}{\sqrt{7}}</math> and <math>AD=3\sqrt{2}</math>, so <math>[AEF]=\frac{15}{2\sqrt{7}}=\frac{15\sqrt{7}}{14}</math> which gives <math>\boxed{036}</math>. NEVER overlook coordinate bash in combination with beginner synthetic techniques.~vvluo | The area of AEF is equal to <math>\frac{EF \cdot \frac{AD}{2}}{2}</math> since AD is the altitude of that triangle with EF as the base, with <math>\frac{AD}{2}</math> being the height. <math>EF=\frac{5\sqrt{2}}{\sqrt{7}}</math> and <math>AD=3\sqrt{2}</math>, so <math>[AEF]=\frac{15}{2\sqrt{7}}=\frac{15\sqrt{7}}{14}</math> which gives <math>\boxed{036}</math>. NEVER overlook coordinate bash in combination with beginner synthetic techniques.~vvluo | ||
− | ==Solution | + | ==Solution 4 (Coordinate Bash + Trig)== |
<asy> | <asy> | ||
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The desired answer is <math>15+7+14=\boxed{036}</math> ~Imayormaynotknowcalculus | The desired answer is <math>15+7+14=\boxed{036}</math> ~Imayormaynotknowcalculus | ||
− | ==Solution | + | ==Solution 5 (Barycentric Coordinates)== |
<asy> | <asy> | ||
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<b>Remark</b>: The area of <math>\triangle{AEF}</math> can also be computed using the <i>Barycentric Area Formula</i>, although it may increase the risk of computational errors; there are also many different ways to proceed once the coordinates are determined. | <b>Remark</b>: The area of <math>\triangle{AEF}</math> can also be computed using the <i>Barycentric Area Formula</i>, although it may increase the risk of computational errors; there are also many different ways to proceed once the coordinates are determined. | ||
− | ==Solution | + | ==Solution 6 (geometry+trig)== |
<asy> | <asy> | ||
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(Added :) ~Math_Genius_164) | (Added :) ~Math_Genius_164) | ||
− | == Solution | + | == Solution 7 == |
[[Image:question13.png|frame|none|###px|]] | [[Image:question13.png|frame|none|###px|]] | ||
First and foremost <math>\big[\triangle{AEF}\big]=\big[\triangle{DEF}\big]</math> as <math>EF</math> is the perpendicular bisector of <math>AD</math>. Now note that quadrilateral <math>ABDF</math> is cyclic, because <math>\angle{ABF}=\angle{FBD}</math> and <math>FA=FD</math>. Similarly quadrilateral <math>AEDC</math> is cyclic, <cmath>\implies \angle{EDA}=\dfrac{C}{2}, \quad \angle{FDA}=\dfrac{B}{2}</cmath> | First and foremost <math>\big[\triangle{AEF}\big]=\big[\triangle{DEF}\big]</math> as <math>EF</math> is the perpendicular bisector of <math>AD</math>. Now note that quadrilateral <math>ABDF</math> is cyclic, because <math>\angle{ABF}=\angle{FBD}</math> and <math>FA=FD</math>. Similarly quadrilateral <math>AEDC</math> is cyclic, <cmath>\implies \angle{EDA}=\dfrac{C}{2}, \quad \angle{FDA}=\dfrac{B}{2}</cmath> | ||
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-tkhalid | -tkhalid | ||
− | ==Solution | + | ==Solution 8 -Trigonometry(only)== |
[[Image:Aime 13.png|frame|none|###px|]] | [[Image:Aime 13.png|frame|none|###px|]] | ||
Trig values we use here: | Trig values we use here: | ||
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Now let <math>AI \cap EF = G</math>. It is easy to angle chase <math>\angle GIE = 90- \frac{B}{2}</math> and <math>\angle GIF = 90- \frac{C}{2}</math>. Since <math>GI = GD - ID = \frac{3\sqrt2}{2}-\sqrt2=\frac{\sqrt2}{2}</math>, we compute that <cmath>EF = EG + FG = \frac{\sqrt2}{2}(\cot B/2 + \cot C/2) = \frac{\sqrt2}{2}(\frac{3}{\sqrt7}+\sqrt7) = \frac{5\sqrt{14}}{7}</cmath> which implies <cmath>[AEF] = AG*EF/2 = \frac{3\sqrt2}{2} * \frac{5\sqrt{14}}{7} / 2 = \frac{15\sqrt7}{14}</cmath> which gives an answer of <math>\boxed{36}</math>. ~Leonard_my_dude~ | Now let <math>AI \cap EF = G</math>. It is easy to angle chase <math>\angle GIE = 90- \frac{B}{2}</math> and <math>\angle GIF = 90- \frac{C}{2}</math>. Since <math>GI = GD - ID = \frac{3\sqrt2}{2}-\sqrt2=\frac{\sqrt2}{2}</math>, we compute that <cmath>EF = EG + FG = \frac{\sqrt2}{2}(\cot B/2 + \cot C/2) = \frac{\sqrt2}{2}(\frac{3}{\sqrt7}+\sqrt7) = \frac{5\sqrt{14}}{7}</cmath> which implies <cmath>[AEF] = AG*EF/2 = \frac{3\sqrt2}{2} * \frac{5\sqrt{14}}{7} / 2 = \frac{15\sqrt7}{14}</cmath> which gives an answer of <math>\boxed{36}</math>. ~Leonard_my_dude~ | ||
− | ==Solution | + | ==Solution 9 (Official MAA 1)== |
Let <math>x = \angle BAD = \angle CAD</math>, <math>y = \angle CBE = \angle ABE</math>, and <math>z = \angle BCF = \angle ACF</math>. Notice that <math>x+y+z = 90^\circ</math>. | Let <math>x = \angle BAD = \angle CAD</math>, <math>y = \angle CBE = \angle ABE</math>, and <math>z = \angle BCF = \angle ACF</math>. Notice that <math>x+y+z = 90^\circ</math>. | ||
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</cmath>The requested sum is <math>15+7+14 = 36</math>. | </cmath>The requested sum is <math>15+7+14 = 36</math>. | ||
− | ==Solution | + | ==Solution 10 (Official MAA 2)== |
Let the point <math>M</math> be the midpoint of <math>\overline{AD}</math>, let <math>I</math> be the incenter of <math>\triangle ABC</math> which is the common point of lines <math>AD</math>, <math>BE</math>, and <math>CF</math>, and let <math>r</math> be the inradius of <math>\triangle ABC</math>. The semiperimeter of <math>\triangle ABC</math> is<cmath>s = \frac{AB + BC + CA}2 = \frac{15}2,</cmath>and Heron's Formula gives the area of <math>\triangle ABC</math> as<cmath>\sqrt{s(s-AB)(s-BC)(s-CA)} = \frac{15\sqrt7}4.</cmath>This area is also <math>rs</math> implying that <math>r = \frac{\sqrt7}2</math>. Stewart's Theorem gives <math>AD =3\sqrt2</math>. Because the ratio of the areas of <math>\triangle IBC</math> and <math>\triangle ABC</math> is <math>\frac{ID}{AD},</math> it follows that<cmath>ID = AD\cdot\frac{\frac{r\cdot BC}2}{\frac{15\sqrt7}4} = \sqrt2.</cmath>Thus <math>IM = MD - ID = \frac{\sqrt2}2</math>. | Let the point <math>M</math> be the midpoint of <math>\overline{AD}</math>, let <math>I</math> be the incenter of <math>\triangle ABC</math> which is the common point of lines <math>AD</math>, <math>BE</math>, and <math>CF</math>, and let <math>r</math> be the inradius of <math>\triangle ABC</math>. The semiperimeter of <math>\triangle ABC</math> is<cmath>s = \frac{AB + BC + CA}2 = \frac{15}2,</cmath>and Heron's Formula gives the area of <math>\triangle ABC</math> as<cmath>\sqrt{s(s-AB)(s-BC)(s-CA)} = \frac{15\sqrt7}4.</cmath>This area is also <math>rs</math> implying that <math>r = \frac{\sqrt7}2</math>. Stewart's Theorem gives <math>AD =3\sqrt2</math>. Because the ratio of the areas of <math>\triangle IBC</math> and <math>\triangle ABC</math> is <math>\frac{ID}{AD},</math> it follows that<cmath>ID = AD\cdot\frac{\frac{r\cdot BC}2}{\frac{15\sqrt7}4} = \sqrt2.</cmath>Thus <math>IM = MD - ID = \frac{\sqrt2}2</math>. | ||
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Note that <math>\angle EFI = 90^{\circ} - \angle FIA = 90^{\circ} - \angle CID = 90^{\circ} - \frac{\angle A}{2} - \frac{\angle C}{2} = \frac{\angle B}{2} = \angle IBC</math>. Thus | Note that <math>\angle EFI = 90^{\circ} - \angle FIA = 90^{\circ} - \angle CID = 90^{\circ} - \frac{\angle A}{2} - \frac{\angle C}{2} = \frac{\angle B}{2} = \angle IBC</math>. Thus | ||
<math>\triangle IBC \sim \triangle IFE</math>. The height of <math>\triangle IBC</math> to <math>I</math> is <math>r=\frac{\sqrt7}2</math>, and the height of <math>\triangle IFE</math> to <math>I</math> is <math>IM=\frac{\sqrt2}2</math>, so <math>EF = BC\cdot \frac{IM}r = \frac{5\sqrt{14}}{7}</math>. The needed area of <math>\triangle AEF</math> is <math>\frac12\cdot EF\cdot \frac{AD}2 = \frac{15\sqrt7}{14}</math>, as above. | <math>\triangle IBC \sim \triangle IFE</math>. The height of <math>\triangle IBC</math> to <math>I</math> is <math>r=\frac{\sqrt7}2</math>, and the height of <math>\triangle IFE</math> to <math>I</math> is <math>IM=\frac{\sqrt2}2</math>, so <math>EF = BC\cdot \frac{IM}r = \frac{5\sqrt{14}}{7}</math>. The needed area of <math>\triangle AEF</math> is <math>\frac12\cdot EF\cdot \frac{AD}2 = \frac{15\sqrt7}{14}</math>, as above. | ||
+ | |||
+ | ==Solution 11 Bash for life== | ||
+ | Firstly, it is easy to find <math>BD=2,CD=3</math> with angle bisector theorem. | ||
+ | |||
+ | Using LOC and some trig formulas we get all those values: | ||
+ | <math>cos\angle{B}=\frac{1}{8},sin\angle{B}=\frac{3\sqrt{7}}{8},cos\angle{\frac{B}{2}}=\frac{3}{4},cos\angle{\frac{{ACB}}{2}}=\frac{\sqrt{14}}{4}</math> | ||
+ | Now we find the coordinates of points <math>A,E,F</math> and we apply shoelace theorem later. Point A's coordinates is <math>(4*\frac{1}{8},4*\frac{3\sqrt{7}}{8})=(\frac{1}{2},\frac{3\sqrt{7}}{2})</math>, Let <math>AJ</math> is perpendicular to <math>BC</math>, <math>tan\angle{JAD}=\frac{2-\frac{1}{2}}{\frac{3\sqrt{7}}{2}}=\frac{\sqrt{7}}{7}</math>, which means the slope of <math>FE</math> is <math>\frac{\sqrt{7}}{7}</math>. Find the coordinate of <math>M</math>, it is easy, <math>(\frac{5}{4},\frac{3\sqrt{7}}{4})</math>, the function <math>EF</math> is <math>y=\frac{\sqrt{7}x}{7}+\frac{4\sqrt{7}}{7}</math>. Now find the intersection of <math>EF,EB</math>, <math>\frac{\sqrt{7}x}{7}+\frac{4\sqrt{7}}{7}=\frac{\sqrt{7}x}{3}</math>, getting that <math>x=3,E(3,\sqrt{7})</math> | ||
+ | Now we look at line segment <math>CF</math>, since <math>cos\angle{\frac{ACB}{2}}=\frac{\sqrt{14}}{4},tan\angle{\frac{ACB}{2}}=\frac{\sqrt{7}}{7}</math>. Since the line passes <math>(5,0)</math>, we can set the equation to get the <math>CF:y=-\frac{\sqrt{7}}{7}+\frac{5\sqrt{7}}{7}</math>, find the intersection of <math>CF,EF</math>,<math>-\frac{\sqrt{7}x}{7}+\frac{5\sqrt{7}}{7}=\frac{\sqrt{7}x}{7}+\frac{4\sqrt{7}}{7}</math>, getting that <math>F:(\frac{1}{2},\frac{9\sqrt{7}}{14})</math> | ||
+ | and in the end we use shoelace theorem with coordinates of <math>A,F,E</math> getting the area <math>\frac{15\sqrt{7}}{14}</math> leads to the final answer <math>\boxed{36}</math> | ||
+ | |||
+ | ~bluesoul | ||
+ | |||
+ | ==Solution 12 (Simple geometry)== | ||
+ | [[File:2020 AIME I 13a.png|350px|right]] | ||
+ | Using the <i><b>Claim</i></b> (below) we get <math>I</math> is orthocenter of <math>\triangle DEF,\angle EDG = \beta,</math> | ||
+ | <math>\angle FDG = \gamma.</math> So area of <math>\triangle DEF</math> is | ||
+ | <cmath>[\triangle DEF] =\frac {DG \cdot FE}{2} = \frac {AD^2}{8} (\tan \beta + \tan \gamma).</cmath> | ||
+ | |||
+ | Semiperimeter of <math>\triangle ABC \hspace{10mm} s = \frac{15}{2},</math> so the bisector | ||
+ | <cmath>AD= \frac{2}{AB + AC}\sqrt{s(s-BC) \cdot AB\cdot AC} = 3\sqrt{2}.</cmath> | ||
+ | |||
+ | We get the inradius by applying Heron's formula | ||
+ | <cmath>r = \sqrt{\frac {(s-AB)(s-BC)(s-AC)}{s}} = \sqrt {\frac {3\cdot 5 \cdot 7}{15 \cdot 4}}=\frac {\sqrt {7}}{2}.</cmath> | ||
+ | |||
+ | We use formulas for inradius and get | ||
+ | <cmath> \tan \beta = \frac{r}{s – AC} = \frac {\sqrt {7}}{3} , \hspace{10mm} \tan \gamma = \frac{r}{s – AB} = \frac {\sqrt {7}}{7}.</cmath> | ||
+ | The area <math>\hspace{30mm} [\triangle DEF] = \frac{15 \sqrt 7}{14}.</math> | ||
+ | [[File:2020 AIME I 13.png|500px|right]] | ||
+ | <i><b>Claim</i></b> | ||
+ | |||
+ | Let <math>I</math> be incenter of <math>\triangle ABC.</math> Then bisector <math>\overline{BI},</math> perpendicular bisector of <math>\overline{AD},</math> and perpendicular dropped to bisector <math>\overline{CI}</math> from point <math>D</math> are concurrent. | ||
+ | |||
+ | <i><b>Proof</i></b> | ||
+ | |||
+ | Denote <math>\angle BAC = 2 \alpha, \angle ABC = 2\beta,\angle ACB = 2 \gamma.</math> | ||
+ | Then <math>\alpha + \beta + \gamma = 90^\circ.</math> | ||
+ | Denote <math>P</math> the intersection point of <math>BC</math> and the tangent line to the circumcircle at point <math>A.</math> | ||
+ | |||
+ | WLOC,<math>\hspace{20mm}\gamma > \beta</math> (case <math>\gamma = \beta</math> is trivial). | ||
+ | |||
+ | <math>\angle PAC = \angle ABC = 2 \beta</math> (this angles are measured by half the arc <math>\overset{\Large\frown} {AC}</math> of the circumcircle). | ||
+ | |||
+ | <cmath>\angle APC = \angle ACD - \angle PAC = 2(\gamma - \beta),</cmath> | ||
+ | <cmath>\angle ADP = 180^o – \angle DAC - \angle ACD = 180^o – \alpha – 2 \gamma = \alpha + 2 \beta = \angle DAP \implies AP = DP.</cmath> Therefore bisector of angle P coincite with the perpendicular bisector of <math>\overline{AD}</math>. | ||
+ | |||
+ | By applying the Law of Sines to <math>\triangle ABP</math> we get <math>\hspace{20mm}\frac {BP}{AP} = \frac {\sin 2 \gamma}{\sin 2 \beta.}</math> | ||
+ | |||
+ | Let <math>E</math> be crosspoint of <math>PG</math> and bisector <math>BI.</math> By applying the Law of Sines to <math>\triangle BEP</math> we get <cmath>\frac {EP}{BP} = \frac {\sin \beta}{\sin(180^o – \beta – (\gamma – \beta))}= \frac {\sin \beta}{\sin \gamma}.</cmath> | ||
+ | |||
+ | Let <math>E'</math> be crosspoint of <math>PG</math> and the perpendicular dropped to bisector <math>\overline{CI}</math> from point <math>D.</math> | ||
+ | <cmath>\frac {E'P}{DP} = \frac {\sin (90^\circ – \gamma)}{\sin(180^o - (90^o – \gamma) – (\gamma - \beta))}= \frac {\cos\gamma}{\cos \beta}.</cmath> | ||
+ | <math>\hspace{50mm}\frac {E'P} {EP} = \frac {E'P}{DP} \cdot \frac {AP}{BP} \cdot \frac {BP}{EP} = \frac {\cos\gamma}{\cos \beta} \cdot \frac {\sin 2\beta}{\sin 2\gamma}\cdot \frac {\sin\gamma}{\sin \beta} = 1 \implies E</math> coincide with <math>E'.</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Solution 13 (Trig)== | ||
+ | Denote <math>I</math> as the incenter of <math>\triangle ABC</math> and <math>M</math> as the intersecting point of <math>AD</math> and <math>EF</math>. | ||
+ | |||
+ | We solve this question based on this equation: <math>[\triangle AEF]=\dfrac{1}{2}\times AM\times (\dfrac{IM}{\tan{\angle EFI}}+\dfrac{IM}{\tan{\angle FEI}})</math>. | ||
+ | |||
+ | |||
+ | Firstly, let's compute the trig value of <math>\angle EFI</math> and <math>\angle FEI</math>. Using the Cosine Law and half-angle formula, we obtain: | ||
+ | <cmath>\cos{\angle B}=\dfrac{1}{8}, \sin{\dfrac{\angle B}{2}}=\dfrac{\sqrt{7}}{4}, \cos{\dfrac{\angle B}{2}}=\dfrac{3}{4}</cmath> | ||
+ | <cmath>\cos{\angle C}=\dfrac{3}{4}, \sin{\dfrac{\angle C}{2}}=\dfrac{1}{\sqrt{8}}, \cos{\dfrac{\angle C}{2}}=\dfrac{\sqrt{7}}{\sqrt{8}}</cmath> | ||
+ | |||
+ | Noticing that by angle chasing, <math>\angle EFI= 90^{\circ}-\angle FIA=90^{\circ}-\angle CID=90^{\circ}-(180^{\circ}-\dfrac{\angle A}{2}-\angle B-\dfrac{\angle C}{2})= \dfrac{\angle B}{2}</math>. | ||
+ | Similarly, <math>\angle FEI=\dfrac{\angle C}{2}</math>. | ||
+ | Thus, <math>\tan{\angle EFI}=\dfrac{\sqrt{7}}{3}</math> and <math>\tan{\angle FEI}=\dfrac{1}{\sqrt{7}}</math> | ||
+ | |||
+ | Then we find <math>AM</math>: by Stewart's theorem, <math>AM=\dfrac{AD}{2}=\dfrac{3\sqrt{2}}{2}</math>. | ||
+ | |||
+ | For <math>MI</math>, first noticing that <math>MI=MD-ID=\dfrac{3\sqrt{2}}{2}-ID</math>. | ||
+ | Then, by Angel Bisector's theorem, we have <math>BD=2, CD=3</math>, and applying Sine Law: | ||
+ | <cmath>\dfrac{ID}{\sin{\dfrac{\angle C}{2}}}=\dfrac{CD}{\sin{\angle CID}}</cmath> | ||
+ | <cmath>\dfrac{ID}{\sin{\dfrac{\angle C}{2}}}=\dfrac{CD}{\cos{\dfrac{\angle B}{2}}}</cmath> | ||
+ | Substituting the known values, we get <math>ID=\sqrt{2}</math>. Thus, <math>MI=\dfrac{\sqrt{2}}{2}</math> | ||
+ | |||
+ | Now, we already find all the unknown values in the very beginning equation. Substituting all the values, we obtain <math>[\triangle AEF]=\dfrac{15\sqrt{7}}{14}</math>, which gives a final answer of <math>\boxed{036}</math>. | ||
+ | |||
+ | ~Ericcc | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/NpB7zpy-yKM | ||
+ | |||
+ | ~MathProblemSolvingSkills.com | ||
+ | |||
+ | |||
==See Also== | ==See Also== | ||
{{AIME box|year=2020|n=I|num-b=12|num-a=14}} | {{AIME box|year=2020|n=I|num-b=12|num-a=14}} | ||
+ | |||
+ | [[Category:Intermediate Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 06:56, 3 August 2024
Contents
- 1 Problem
- 2 Diagram
- 3 Solution 1
- 4 Solution 2
- 5 Solution 3(coord bash + basic geometry)
- 6 Solution 4 (Coordinate Bash + Trig)
- 7 Solution 5 (Barycentric Coordinates)
- 8 Solution 6 (geometry+trig)
- 9 Solution 7
- 10 Solution 8 -Trigonometry(only)
- 11 Solution 9 (Official MAA 1)
- 12 Solution 10 (Official MAA 2)
- 13 Solution 11 Bash for life
- 14 Solution 12 (Simple geometry)
- 15 Solution 13 (Trig)
- 16 Video Solution
- 17 See Also
Problem
Point lies on side
of
so that
bisects
The perpendicular bisector of
intersects the bisectors of
and
in points
and
respectively. Given that
and
the area of
can be written as
where
and
are relatively prime positive integers, and
is a positive integer not divisible by the square of any prime. Find
Diagram
Solution 1
Points are defined as shown. It is pretty easy to show that by spiral similarity at
by some short angle chasing. Now, note that
is the altitude of
, as the altitude of
. We need to compare these altitudes in order to compare their areas. Note that Stewart's theorem implies that
, the altitude of
. Similarly, the altitude of
is the altitude of
, or
. However, it's not too hard to see that
, and therefore
. From here, we get that the area of
is
, by similarity. ~awang11
Solution 2
Let ,
,
be the midpoints of arcs
,
,
. By Fact 5, we know that
, and so by Ptolmey's theorem, we deduce that
In particular, we have
.
Now the key claim is that:
Claim: and
are homothetic at
with ratio
.
Proof. First, we show that is the midpoint of
. Indeed, we have
by Ratio lemma and Law of Sines.
Now observe that:
is the perpendicular bisector of
,
is the perpendicular bisector of
, and
.
Combining these facts gives that is a midline in
, which proves the claim.
To finish, we compute , noting that
.
By Heron's, we can calculate the circumradius , and by Law of Cosines, we get
Then using
, we can compute
Thus
, which gives a final answer of
.
~pinetree1
Solution 3(coord bash + basic geometry)
Let lie on the x-axis and
be the origin.
is
. Use Heron's formula to compute the area of triangle
. We have
. and
. We now find the altitude, which is
, which is the y-coordinate of
. We now find the x-coordinate of
, which satisfies
, which gives
since the triangle is acute. Now using the Angle Bisector Theorem, we have
and
to get
. The coordinates of D are
.
Since we want the area of triangle
, we will find equations for perpendicular bisector of AD, and the other two angle bisectors. The perpendicular bisector is not too challenging: the midpoint of AD is
and the slope of AD is
. The slope of the perpendicular bisector is
. The equation is(in point slope form)
.
The slope of AB, or in trig words, the tangent of
is
.
Finding
and
. Plugging this in to half angle tangent, it gives
as the slope of the angle bisector, since it passes through
, the equation is
.
Similarly, the equation for the angle bisector of
will be
.
For
use the B-angle bisector and the perpendicular bisector of AD equations to intersect at
.
For
use the C-angle bisector and the perpendicular bisector of AD equations to intersect at
.
The area of AEF is equal to
since AD is the altitude of that triangle with EF as the base, with
being the height.
and
, so
which gives
. NEVER overlook coordinate bash in combination with beginner synthetic techniques.~vvluo
Solution 4 (Coordinate Bash + Trig)
Let and
be the line
.
We compute that
, so
.
Thus,
lies on the line
. The length of
at a point
is
, so
.
We now have the coordinates ,
and
.
We also have
by the angle-bisector theorem and
by taking the midpoint.
We have that because
,
by half angle formula.
We also compute , so
.
Now, has slope
, so it's perpendicular bisector has slope
and goes through
.
We find that this line has equation .
As , we have that line
has form
.
Solving for the intersection point of these two lines, we get
and thus
We also have that because ,
has form
.
Intersecting the line and the perpendicular bisector of
yields
.
Solving this, we get and so
.
We now compute .
We also have
.
As , we have
.
The desired answer is ~Imayormaynotknowcalculus
Solution 5 (Barycentric Coordinates)
As usual, we will use homogenized barycentric coordinates.
We have that will have form
. Similarly,
has form
and
has form
.
Since
and
, we also have
.
It remains to determine the equation of the line formed by the perpendicular bisector of
.
This can be found using EFFT. Let a point on
have coordinates
.
We then have that the displacement vector
and that the displacement vector
has form
.
Now, by EFFT, we have
.
This equates to
.
Now, intersecting this with , we have
,
, and
.
This yields
,
, and
, or
.
Similarly, intersecting this with , we have
,
, and
.
Solving this, we obtain
,
, and
, or
.
We finish by invoking the Barycentric Distance Formula twice; our first displacement vector being .
We then have
, thus
.
Our second displacement vector is .
As a result,
, so
.
As , the desired area is
. ~Imayormaynotknowcalculus
Remark: The area of can also be computed using the Barycentric Area Formula, although it may increase the risk of computational errors; there are also many different ways to proceed once the coordinates are determined.
Solution 6 (geometry+trig)
To get the area of , we try to find
and
.
Since is the angle bisector, we can get that
and
. By applying Stewart's Theorem, we can get that
. Therefore
.
Since is the perpendicular bisector of
, we know that
. Since
is the angle bisector of
,
we know that
. By applying the Law of Sines to
and
, we know that
. Since
is not equal to
and therefore these two triangles are not congruent, we know that
and
are supplementary. Then we know that
and
are also supplementary. Given that
, we can get that
is half of
. Similarly, we have
is half of
.
By applying the Law of Cosines, we get , and then
. Similarly, we can get
and
. Based on some trig identities, we can compute that
, and
.
Finally, the area of equals
. Therefore, the final answer is
. ~xamydad
Remark: I didn't figure out how to add segments ,
,
and
. Can someone please help add these segments?
(Added :) ~Math_Genius_164)
Solution 7
First and foremost as
is the perpendicular bisector of
. Now note that quadrilateral
is cyclic, because
and
. Similarly quadrilateral
is cyclic,
Let
,
,
be the
,
, and
excenters of
respectively. Then it follows that
. By angle bisector theorem we have
. Now let the feet of the perpendiculars from
and
to
be
and
resptively. Then by tangents we have
From the previous ratios,
Similarly we can find that
and
and thus
-tkhalid
Solution 8 -Trigonometry(only)
Trig values we use here:
First let the incenter be . Let
be the midpoint of minor arc
on
and let
be the foot of
to
.
We can find using Stewart's Theorem: from Angle Bisector Theorem
and
. Then it is easy to find that
.
Now we trig bash for . Notice that
from the Incenter Excenter Lemma. We obtain that
. To get
we angle chase to get
. Then
gives
. This means
.
Now let . It is easy to angle chase
and
. Since
, we compute that
which implies
which gives an answer of
. ~Leonard_my_dude~
Solution 9 (Official MAA 1)
Let ,
, and
. Notice that
.
In , segment
is the bisector of
, and
lies on the perpendicular bisector of side
. Therefore
is the midpoint of arc
on the circumcircle of
. It follows that
and
. Likewise,
is cyclic,
, and
. Because
is the perpendicular bisector of
, triangles
and
are congruent, implying that
Applying the Law of Sines to and
gives
By the Angle Bisector Theorem,
and
. Combining the above information yields
Applying the Law of Cosines to
gives
,
, and
. By the Half Angle Formulas,
Therefore
The requested sum is
.
Solution 10 (Official MAA 2)
Let the point be the midpoint of
, let
be the incenter of
which is the common point of lines
,
, and
, and let
be the inradius of
. The semiperimeter of
is
and Heron's Formula gives the area of
as
This area is also
implying that
. Stewart's Theorem gives
. Because the ratio of the areas of
and
is
it follows that
Thus
.
Note that . Thus
. The height of
to
is
, and the height of
to
is
, so
. The needed area of
is
, as above.
Solution 11 Bash for life
Firstly, it is easy to find with angle bisector theorem.
Using LOC and some trig formulas we get all those values:
Now we find the coordinates of points
and we apply shoelace theorem later. Point A's coordinates is
, Let
is perpendicular to
,
, which means the slope of
is
. Find the coordinate of
, it is easy,
, the function
is
. Now find the intersection of
,
, getting that
Now we look at line segment
, since
. Since the line passes
, we can set the equation to get the
, find the intersection of
,
, getting that
and in the end we use shoelace theorem with coordinates of
getting the area
leads to the final answer
~bluesoul
Solution 12 (Simple geometry)
Using the Claim (below) we get is orthocenter of
So area of
is
Semiperimeter of so the bisector
We get the inradius by applying Heron's formula
We use formulas for inradius and get
The area
Claim
Let be incenter of
Then bisector
perpendicular bisector of
and perpendicular dropped to bisector
from point
are concurrent.
Proof
Denote
Then
Denote
the intersection point of
and the tangent line to the circumcircle at point
WLOC, (case
is trivial).
(this angles are measured by half the arc
of the circumcircle).
Therefore bisector of angle P coincite with the perpendicular bisector of
.
By applying the Law of Sines to we get
Let be crosspoint of
and bisector
By applying the Law of Sines to
we get
Let be crosspoint of
and the perpendicular dropped to bisector
from point
coincide with
vladimir.shelomovskii@gmail.com, vvsss
Solution 13 (Trig)
Denote as the incenter of
and
as the intersecting point of
and
.
We solve this question based on this equation: .
Firstly, let's compute the trig value of and
. Using the Cosine Law and half-angle formula, we obtain:
Noticing that by angle chasing, .
Similarly,
.
Thus,
and
Then we find : by Stewart's theorem,
.
For , first noticing that
.
Then, by Angel Bisector's theorem, we have
, and applying Sine Law:
Substituting the known values, we get
. Thus,
Now, we already find all the unknown values in the very beginning equation. Substituting all the values, we obtain , which gives a final answer of
.
~Ericcc
Video Solution
~MathProblemSolvingSkills.com
See Also
2020 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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.