Difference between revisions of "2023 AIME II Problems/Problem 3"
MRENTHUSIASM (talk | contribs) (→Solution 8) |
|||
(21 intermediate revisions by 9 users not shown) | |||
Line 38: | Line 38: | ||
This solution refers to the <b>Diagram</b> section. | This solution refers to the <b>Diagram</b> section. | ||
− | Let | + | Let <math>\angle PAB = \angle PBC = \angle PCA = \theta,</math> from which <math>\angle PAC = 90^\circ-\theta,</math> and <math>\angle APC = 90^\circ.</math> |
Moreover, we have <math>\angle PBA = \angle PCB = 45^\circ-\theta,</math> as shown below: | Moreover, we have <math>\angle PBA = \angle PCB = 45^\circ-\theta,</math> as shown below: | ||
Line 72: | Line 72: | ||
draw(A--B--C--cycle^^P--A^^P--B^^P--C); | draw(A--B--C--cycle^^P--A^^P--B^^P--C); | ||
label("$10$",midpoint(A--P),dir(-30),blue); | label("$10$",midpoint(A--P),dir(-30),blue); | ||
− | |||
− | |||
− | |||
− | |||
label("$\theta$",A,9.5*dir(76),red); | label("$\theta$",A,9.5*dir(76),red); | ||
label("$\theta$",C,9.5*dir(168),red); | label("$\theta$",C,9.5*dir(168),red); | ||
Line 83: | Line 79: | ||
label("$90^\circ-\theta$",A,2*dir(-40),green); | label("$90^\circ-\theta$",A,2*dir(-40),green); | ||
</asy> | </asy> | ||
− | Note that <math>\triangle PAB \sim \triangle PBC</math> by | + | Note that <math>\triangle PAB \sim \triangle PBC</math> by AA Similarity. The ratio of similitude is <math>\frac{PA}{PB} = \frac{PB}{PC} = \frac{AB}{BC},</math> so <math>\frac{10}{PB} = \frac{1}{\sqrt2}</math> and thus <math>PB=10\sqrt2.</math> Similarly, we can figure out that <math>PC=20</math>. |
− | |||
− | Finally, the area of <math>\triangle ABC</math> is <cmath>\frac12\cdot AB\cdot AC = \frac12\cdot | + | Finally, <math>AC=\sqrt{10^2+20^2}=10\sqrt{5}</math>, so the area of <math>\triangle ABC</math> is <cmath>\frac12\cdot AB\cdot AC = \frac12\cdot (10\sqrt{5})^2 = \boxed{250}.</cmath> |
~s214425 | ~s214425 | ||
Line 147: | Line 142: | ||
~Kiran | ~Kiran | ||
− | == | + | ==Solution 5== |
+ | Denote the area of <math>X</math> by <math>[X].</math> As in previous solutions, we see that <math>\angle APC = 90 ^\circ, \triangle BPC \sim \triangle APB</math> with ratio <math>k = \sqrt{2}\implies</math> <cmath>\frac {PC}{PB} = \frac {PB}{PA} = k \implies PC = k^2 \cdot AP = 20 \implies [APC] = \frac {AP \cdot PC}{2} = 100.</cmath> | ||
+ | <cmath>[BPC] = k^2 [APB] = 2 [APB].</cmath> | ||
+ | <cmath>AB = BC, \angle PCA = \angle PAB \implies \frac {[APC]}{[APB]} = \frac {PC}{PA} = 2 \implies</cmath> | ||
+ | <cmath>[ABC] = [APB] + [APC] + [BPC] = [APC] \cdot (\frac {1}{2} + 1 + 2 \cdot \frac {1}{2}) = \frac {5}{2} \cdot [APC] = \boxed{250}.</cmath> | ||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Solution 6== | ||
+ | |||
+ | Denote <math>\angle PCA = \theta</math>. Then, by trig Ceva's: | ||
+ | <cmath>\begin{align*} | ||
+ | \frac{\sin^3(\theta)}{\sin(90-\theta) \cdot \left(\sin(45-\theta)\right)^2} &= 1 \\ | ||
+ | \sin^3(\theta) &= \cos(\theta) \cdot \left(\sin(45) \cos(\theta) - \cos(45) \sin(\theta)\right)^2 \\ | ||
+ | 2\sin^3(\theta) &= \cos(\theta) \cdot \left(\cos(\theta) - \sin(\theta)\right)^2 \\ | ||
+ | 2\sin^2(\theta) &= \cot(\theta) \cdot \left(1 - 2\sin(\theta)\cos(\theta)\right) \\ | ||
+ | 2\sin^2(\theta) &= \cot(\theta) - 2\cos^2(\theta) \\ | ||
+ | \cot(\theta) &= 2 \\ | ||
+ | \sin(\theta) &= \frac{\sqrt{5}}{5}. | ||
+ | \end{align*}</cmath> | ||
+ | Note that <math>\angle APC</math> is a right angle. Therefore: | ||
+ | <cmath>\begin{align*} | ||
+ | \sin(\theta) &= \frac{AP}{AC} \\ | ||
+ | AC &= \frac{10}{\frac{\sqrt{5}}{5}} \\ | ||
+ | &= 10\sqrt{5} \\ | ||
+ | |ABC| &= \frac{AC^2}{2} \\ | ||
+ | &= \boxed{250}. | ||
+ | \end{align*}</cmath> | ||
+ | ~ConcaveTriangle | ||
+ | |||
+ | ==Solution 7== | ||
+ | |||
+ | Notice that point <math>P</math> is one of the two Brocard Points of <math>\triangle ABC</math>. (The angle equalities given in the problem are equivalent to the definition of a Brocard point.) By the Brocard point formula, | ||
+ | <cmath>\begin{align*} \cot(\phi) = \cot(A)+\cot(B)+\cot(C) \end{align*}</cmath> | ||
+ | , where <math>\phi</math> is equal to <math>\angle PAB</math>.(This is also called the Brocard angle of triangle ABC). Because the triangle is an isosceles right triangle, the cotangents are easy to compute: | ||
+ | <cmath>\begin{align*}\cot(\phi) = 0 + 1 + 1 \\ \cot(\phi) = 2\end{align*}</cmath> | ||
+ | By definition, <math>\cot(\phi) = \frac{\cos(\phi)}{\sin(\phi)}</math>. By the Pythagorean identity, <math>\cos(\phi)=\frac{2\sqrt{5}}{5}</math> and <math>\sin(\phi) = \cos(\phi)=\frac{\sqrt{5}}{5}</math>. Consider triangle <math>APB</math>. By the problem condition, <math>\angle PBA = 45-\phi</math>, so <math>\angle BPA = 135^{\circ}</math> | ||
+ | <cmath>\begin{align*}\sin{45-\theta} = \sin{45}\cos{\phi}-\cos{45}\sin{\phi} = \frac{\sqrt{10}}{10}\end{align*}</cmath> | ||
+ | Now, we can use the Law of Sines. | ||
+ | <cmath>\begin{align*} | ||
+ | \frac{AP}{\sin{45-\theta}}&=\frac{AB}{\sin{135}} \\ | ||
+ | 10 \sqrt{10} &= \sqrt{2} AB \\ | ||
+ | AB &= 10 \sqrt{5} | ||
+ | \end{align*}</cmath> | ||
+ | Therefore, the answer is <cmath>[ABC] = \frac{1}{2} (AB)^2 = \boxed{250}.</cmath> | ||
+ | ~ewei12 | ||
+ | |||
+ | ==Solution 8== | ||
+ | |||
+ | Notice that <math>\angle APC = 90^{\circ}</math>, <math>\angle BPA = 135^{\circ}</math>, and <math>\angle CPB = 135^{\circ}</math> (from Solution 4). Now let <math>a = 0</math> and <math>p = 10i</math>. Then by the angle restrictions <math>c = m + 10i</math> and <math>b = -n + (n+10)i</math> for some <math>m, n</math>. Since <math>\angle BAC = 90^{\circ}</math>, <math>ci = b</math>, or <math>(m + 10i)i = -n + (n+10)i</math>. Therefore <math>n = 10</math>, <math>AB = \sqrt{10^2 + 20^2} = \sqrt{500}</math>, and <math>[ABC] = \frac{1}{2} AB^2 = \boxed{250}</math>. | ||
+ | |||
+ | ~aliz | ||
+ | |||
+ | ==Video Solution 1 by SpreadTheMathLove== | ||
https://www.youtube.com/watch?v=APSUN-9Z_AU | https://www.youtube.com/watch?v=APSUN-9Z_AU | ||
+ | |||
+ | ==Video Solution 2 by Piboy== | ||
+ | https://www.youtube.com/watch?v=-WUhMmdXCxU&t=26s&ab_channel=Piboy | ||
+ | |||
+ | ==Video Solution by The Power of Logic(#3 and #4)== | ||
+ | https://youtu.be/dS9K1o4gCA0 | ||
== See also == | == See also == |
Latest revision as of 00:42, 27 June 2024
Contents
Problem
Let be an isosceles triangle with There exists a point inside such that and Find the area of
Diagram
~MRENTHUSIASM
Solution 1
This solution refers to the Diagram section.
Let from which and
Moreover, we have as shown below: Note that by AA Similarity. The ratio of similitude is so and thus Similarly, we can figure out that .
Finally, , so the area of is
~s214425
~MRENTHUSIASM
Solution 2
Since the triangle is a right isosceles triangle, .
Let the common angle be . Note that , thus . From there, we know that .
Note that , so from law of sines we have Dividing by and multiplying across yields From here use the sine subtraction formula, and solve for : Substitute this to find that , thus the area is .
~SAHANWIJETUNGA
Solution 3
Since the triangle is a right isosceles triangle, .
Do some angle chasing yielding:
We have since is a right triangle. Since is a -- triangle, , and .
Note that by a factor of . Thus, , and .
From Pythagorean theorem, so the area of is .
~SAHANWIJETUNGA
Solution 4
Since the triangle is a right isosceles triangle, .
Notice that in triangle , , so . Similar logic shows .
Now, we see that with ratio (as is a -- triangle). Hence, . We use the Law of Cosines to find . Since is a right triangle, the area is .
~Kiran
Solution 5
Denote the area of by As in previous solutions, we see that with ratio vladimir.shelomovskii@gmail.com, vvsss
Solution 6
Denote . Then, by trig Ceva's: Note that is a right angle. Therefore: ~ConcaveTriangle
Solution 7
Notice that point is one of the two Brocard Points of . (The angle equalities given in the problem are equivalent to the definition of a Brocard point.) By the Brocard point formula, , where is equal to .(This is also called the Brocard angle of triangle ABC). Because the triangle is an isosceles right triangle, the cotangents are easy to compute: By definition, . By the Pythagorean identity, and . Consider triangle . By the problem condition, , so Now, we can use the Law of Sines. Therefore, the answer is ~ewei12
Solution 8
Notice that , , and (from Solution 4). Now let and . Then by the angle restrictions and for some . Since , , or . Therefore , , and .
~aliz
Video Solution 1 by SpreadTheMathLove
https://www.youtube.com/watch?v=APSUN-9Z_AU
Video Solution 2 by Piboy
https://www.youtube.com/watch?v=-WUhMmdXCxU&t=26s&ab_channel=Piboy
Video Solution by The Power of Logic(#3 and #4)
See also
2023 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
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.