Difference between revisions of "2009 AMC 10A Problems/Problem 10"
(New page: == Problem == Triangle <math>ABC</math> has a right angle at <math>B</math>. Point <math>D</math> is the foot of the altitude from <math>B</math>, <math>AD=3</math>, and <math>DC=4</math>...) |
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\mathrm{(E)}\ 42 | \mathrm{(E)}\ 42 | ||
</math> | </math> | ||
+ | [[Category: Introductory Geometry Problems]] | ||
− | == Solution == | + | == Solution 1== |
− | It is a well-known fact that in any right triangle <math>ABC</math> with the right angle at <math>B</math> and <math>D</math> the foot of the altitude from <math>B</math> onto <math>AC</math> we have <math>BD^2 = AD\cdot CD</math>. (See below for a proof.) Then <math>BD = \sqrt{ 3\cdot 4 } = 2\sqrt 3</math>, and the area of the triangle <math>ABC</math> is <math>\frac{AC\cdot BD}2 = \boxed{ | + | It is a well-known fact that in any right triangle <math>ABC</math> with the right angle at <math>B</math> and <math>D</math> the foot of the altitude from <math>B</math> onto <math>AC</math> we have <math>BD^2 = AD\cdot CD</math>. (See below for a proof.) Then <math>BD = \sqrt{ 3\cdot 4 } = 2\sqrt 3</math>, and the area of the triangle <math>ABC</math> is <math>\frac{AC\cdot BD}2 = 7\sqrt3\Rightarrow\boxed{\text{(B)}}</math>. |
− | Proof: Consider the [[Pythagorean theorem]] for each of the triangles <math>ABC</math>, <math>ABD</math>, and <math>CBD</math>. We get: | + | ''Proof'': Consider the [[Pythagorean theorem]] for each of the triangles <math>ABC</math>, <math>ABD</math>, and <math>CBD</math>. We get: |
# <math>AB^2 + BC^2 = AC^2 = (AD+DC)^2 = AD^2 + DC^2 + 2 \cdot AD \cdot DC</math>. | # <math>AB^2 + BC^2 = AC^2 = (AD+DC)^2 = AD^2 + DC^2 + 2 \cdot AD \cdot DC</math>. | ||
# <math>AB^2 = AD^2 + BD^2</math> | # <math>AB^2 = AD^2 + BD^2</math> | ||
# <math>BC^2 = BD^2 + CD^2</math> | # <math>BC^2 = BD^2 + CD^2</math> | ||
− | Substituting equations 2 and 3 into the left hand side of equation 1, we get <math>BD^2 = AD \cdot DC</math>, | + | Substituting equations 2 and 3 into the left hand side of equation 1, we get <math>BD^2 = AD \cdot DC</math>. |
+ | |||
+ | Alternatively, note that <math>\triangle ABD \sim \triangle BCD \Longrightarrow \frac{AD}{BD} = \frac{BD}{CD}</math>. | ||
+ | |||
+ | == Solution 2 == | ||
+ | |||
+ | For those looking for a dumber solution, we can use Pythagoras and manipulation of area formulas to solve the problem. | ||
+ | |||
+ | Assume the length of <math>BD</math> is equal to <math>h</math>. Then, by Pythagoras, we have, | ||
+ | |||
+ | <cmath>AB^2 = h^2 + 9 \Rightarrow AB = \sqrt{h^2 + 9}</cmath> | ||
+ | <cmath>BC^2 = h^2 + 16 \Rightarrow BC = \sqrt{h^2 + 16}</cmath> | ||
+ | |||
+ | Then, by area formulas, we know: | ||
+ | |||
+ | <cmath>\frac{1}{2}\sqrt{(h^2+9)(h^2+16)} = \frac{1}{2}(7)(h)</cmath> | ||
+ | |||
+ | Squaring and solving the above equation yields our solution that <math>h^2 = 12 \Rightarrow h = 2\sqrt{3}.</math> Since the area of the triangle is half of this quantity multiplied by the base, we have | ||
+ | <cmath>\text{area} = \frac{1}{2}(7)(2\sqrt{3})\Rightarrow \boxed{7\sqrt{3}}</cmath> | ||
+ | |||
+ | == Solution 3 (Power of a point)== | ||
+ | |||
+ | Draw the circumcircle <math>\omega</math> of the <math>\Delta ABC</math>. Because <math>\Delta ABC</math> is a right angle triangle, AC is the diameter of the circumcircle. By applying [[Power of a Point Theorem]], we can have <math>BD=DE</math> and <math>AD\cdot CD=BD^2</math> <math>\Rightarrow BD=\sqrt{3\times 4}=2\sqrt{3}</math>. Then we have <math>S_{[ABC]}=\frac{1}{2}(7)(2\sqrt{3})=\boxed{7\sqrt{3}}</math> | ||
+ | |||
+ | |||
+ | <asy> | ||
+ | unitsize(6mm); | ||
+ | defaultpen(linewidth(.8pt)+fontsize(8pt)); | ||
+ | dotfactor=4; | ||
+ | |||
+ | pair B=(0,0), C=(sqrt(28),0), A=(0,sqrt(21)), E=(6*sqrt(28)/7,8*sqrt(21)/7); | ||
+ | pair D=foot(B,A,C); | ||
+ | pair[] ps={B,C,A,D}; | ||
+ | |||
+ | |||
+ | filldraw(Circle((sqrt(28)/2,sqrt(21)/2),sqrt(49)/2),white,black); | ||
+ | draw(A--B--C--cycle); | ||
+ | draw(B--D); | ||
+ | draw(E--D); | ||
+ | draw(rightanglemark(B,D,C)); | ||
+ | |||
+ | dot(ps); | ||
+ | label("$A$",A,NW); | ||
+ | label("$B$",B,SW); | ||
+ | label("$E$",E,NE); | ||
+ | label("$C$",C,SE); | ||
+ | label("$D$",D,NE); | ||
+ | label("$3$",midpoint(A--D),NE); | ||
+ | label("$4$",midpoint(D--C),NE); | ||
+ | </asy> | ||
+ | ~Bran_Qin | ||
+ | |||
+ | ===Solution 4 (Fakesolve)/Answer Choices=== | ||
+ | The area of the triangle is <math>\frac{7x}{2}</math>. We want to fine what <math>x</math> is. Now, we try each answer choice. | ||
+ | |||
+ | <math>A): 4 \sqrt{3}</math>. I am too lazy to go over this, but we immediately see that this is very improbably due to the area being <math>\frac{7x}{2}</math>. This does not work. | ||
+ | <math>B): 7 \sqrt{3}</math>. This is promising. This means that <math>x = 2\sqrt{3}</math>. Now, applying Pythagorean Theorem, we have the vertical sides is <math>\sqrt {21}</math> and the horizontal side is <math>\sqrt {28}</math>. Multiplying these and dividing by <math>2</math> indeed gives us <math>7 \sqrt {3}</math> as desired. Therefore, the answer is <math>\boxed{7 \sqrt{3}}</math> | ||
+ | |||
+ | ~Arcticturn | ||
+ | |||
+ | ===Solution 5=== | ||
+ | Let <math>\overline{BD} = x</math>, and <math>\angle ABD = \theta</math>. Since <math>m\angle ABC = 90, DBC = 90 - \theta</math>. <math>\tan{\theta} = \frac{3}{x}</math>, and <math>\tan{90-\theta} = \frac{4}{x}</math>. Using trig identities, we can simplify the second one to <math>\cot{\theta} = \frac{4}{x}</math>. Let <math>y = \tan{\theta}</math>. We get <math>xy = 3</math>, and using the fact that <math>\cot{a} = \frac{1}{\tan{a}}</math>, <math>\frac{x}{y} = 4</math>. Solving for x and y(using substitution), we get <math>y^2 = \frac{3}{4}</math>. We could substitute back in and solve, but there is an easier way. Squaring <math>xy = 3</math>, we get <math>x^2y^2 = 9</math>. Since we know <math>y^2 = \frac{3}{4}</math>, <math>x = \sqrt{12}</math>. Using the Pythagorean Theorem in the original triangle, we get the legs of the triangle to be <math>\sqrt{9+x^2}</math> and <math>\sqrt{16+x^2}</math>. Putting in <math>x</math>, we get the equation for the area, which is <math>\frac{\sqrt{28} \cdot \sqrt{21}}{2} = \frac{14\sqrt{2}}{2} = \boxed{\textbf{(B) } 7\sqrt{3}}</math> | ||
+ | |||
+ | ~idk12345678 | ||
+ | |||
+ | == Video Solution by OmegaLearn== | ||
+ | https://youtu.be/4_x1sgcQCp4?t=1195 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
== See Also == | == See Also == | ||
{{AMC10 box|year=2009|ab=A|num-b=9|num-a=11}} | {{AMC10 box|year=2009|ab=A|num-b=9|num-a=11}} | ||
+ | {{MAA Notice}} |
Revision as of 17:57, 30 April 2024
Contents
[hide]Problem
Triangle has a right angle at . Point is the foot of the altitude from , , and . What is the area of ?
Solution 1
It is a well-known fact that in any right triangle with the right angle at and the foot of the altitude from onto we have . (See below for a proof.) Then , and the area of the triangle is .
Proof: Consider the Pythagorean theorem for each of the triangles , , and . We get:
- .
Substituting equations 2 and 3 into the left hand side of equation 1, we get .
Alternatively, note that .
Solution 2
For those looking for a dumber solution, we can use Pythagoras and manipulation of area formulas to solve the problem.
Assume the length of is equal to . Then, by Pythagoras, we have,
Then, by area formulas, we know:
Squaring and solving the above equation yields our solution that Since the area of the triangle is half of this quantity multiplied by the base, we have
Solution 3 (Power of a point)
Draw the circumcircle of the . Because is a right angle triangle, AC is the diameter of the circumcircle. By applying Power of a Point Theorem, we can have and . Then we have
~Bran_Qin
Solution 4 (Fakesolve)/Answer Choices
The area of the triangle is . We want to fine what is. Now, we try each answer choice.
. I am too lazy to go over this, but we immediately see that this is very improbably due to the area being . This does not work. . This is promising. This means that . Now, applying Pythagorean Theorem, we have the vertical sides is and the horizontal side is . Multiplying these and dividing by indeed gives us as desired. Therefore, the answer is
~Arcticturn
Solution 5
Let , and . Since . , and . Using trig identities, we can simplify the second one to . Let . We get , and using the fact that , . Solving for x and y(using substitution), we get . We could substitute back in and solve, but there is an easier way. Squaring , we get . Since we know , . Using the Pythagorean Theorem in the original triangle, we get the legs of the triangle to be and . Putting in , we get the equation for the area, which is
~idk12345678
Video Solution by OmegaLearn
https://youtu.be/4_x1sgcQCp4?t=1195
~ pi_is_3.14
See Also
2009 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.