problem_id int64 1 978 | question stringlengths 86 2.11k | source stringlengths 19 76 | solution stringlengths 94 14.7k | asymptote_code stringlengths 44 17.8k | solution_image_url stringclasses 16
values |
|---|---|---|---|---|---|
957 | Equilateral $\triangle ABC$ with side length $14$ is rotated about its center by angle $\theta$, where $0 < \theta < 60^{\circ}$, to form $\triangle DEF$. See the figure. The area of hexagon $ADBECF$ is $91\sqrt{3}$. What is $\tan\theta$?
$\textbf{(A)}~\frac{3}{4}\qquad\textbf{(B)}~\frac{5\sqrt{3}}{11}\qquad\textbf{(... | 2024 AMC 12B Problem 19 | Let the circumcenter of the circle inscribing this polygon be $O$. The area of the equilateral triangle is $\frac{\sqrt{3}}{4}*196=49\sqrt{3}$. The area of one of the three smaller triangles, say $\triangle{DBE}$ is $14\sqrt{3}$. Let $BH$ be the altitude of $\triangle{DBE}$, so if we extend $BH$ to point $M$ where $MO\... | // Block 1
// Modified Asymptote diagram with correct right angle mark and alpha label.
import olympiad;
defaultpen(fontsize(13));
size(200);
// Circumradius R = 14*sqrt(3)/3
real R = 14*sqrt(3)/3;
// Define original points with scaling
pair O = (0,0);
pair A = R * dir(225);
pair B = R * dir(-15);
pair C = R * dir(1... | [] |
957 | Equilateral $\triangle ABC$ with side length $14$ is rotated about its center by angle $\theta$, where $0 < \theta < 60^{\circ}$, to form $\triangle DEF$. See the figure. The area of hexagon $ADBECF$ is $91\sqrt{3}$. What is $\tan\theta$?
$\textbf{(A)}~\frac{3}{4}\qquad\textbf{(B)}~\frac{5\sqrt{3}}{11}\qquad\textbf{(... | 2024 AMC 12B Problem 19 | \[[\triangle ABD] = \frac{[ADBECF]-[\triangle ABC]}{3} = \frac{91\sqrt{3}-49\sqrt{3}}{3} = 14\sqrt{3}\]
Let $M$ be the intersection of $AB$ and $DE$. Since $DMB$ is isosceles and $\angle AMD = \theta$, we have $\angle ABD = \theta/2$. Also, all of the hexagon's internal angles are equal, so $\angle ADB = 120^\circ$.
... | // Block 1
defaultpen(fontsize(13));
size(220);
// Base points and rotation
pair O = (0,0);
pair A = dir(225);
pair B = dir(-15);
pair D = rotate(38.21, O)*A;
pair E = rotate(38.21, O)*B;
// Intersection point of AB and DE
pair M = extension(A, B, D, E);
// Triangle and segments
defaultpen(fontsize(13));
size(220);
... | [] |
958 | The measures of the smallest angles of three different right triangles sum to $90^\circ$. All three triangles have side lengths that are primitive Pythagorean triples. Two of them are $3-4-5$ and $5-12-13$. What is the perimeter of the third triangle?
$\textbf{(A) } 40 \qquad\textbf{(B) } 126 \qquad\textbf{(C) } 154 \... | 2024 AMC 12B Problem 21 | We can start by scaling the $3-4-5$ right triangle up by a factor of $3$ and "gluing" it to the $5-12-13$ triangle's longer leg. Let $\alpha$, $\beta$, and $\theta$ be the smallest angles in the $3-4-5$, $5-12-13$ and unknown triangle respectively. We can construct the following diagram of the two known triangles.
We... | // Block 1
pair A = (0,0);
pair B = (5,0);
pair C = (14,0);
pair D = (5,12);
draw(A--C--D--cycle);
draw(B--D);
draw(rightanglemark(A,B,D,20));
label("5", A--B, S);
label("9", B--C, S);
label("15", C--D, NE);
label("13", D--A, NW);
label("12", B--D, E);
label("$\beta$", D, 6*dir(257));
label("$\alpha$", D, 4*dir(287... | [] |
959 | A semicircle has diameter $\overline{AB}$ and chord $\overline{CD}$ of length $16$ parallel to $\overline{AB}$. A smaller semicircle
with diameter on $\overline{AB}$ and tangent to $\overline{CD}$ is cut from the larger semicircle, as shown below.
What is the area of the resulting figure, shown shaded?
$\textbf{(A)... | 2025 AMC 10A Problem 10 | The problem doesn't define where the chord is within the circle. So, let's place the chord such that when lines are drawn from its ends to the center of the larger semicircle, an equilateral triangle with side length $16$ is formed.
In this definition, the radius of the larger semicircle is $16,$ giving it an area ... | // Block 1
import graph; unitsize(14mm);
defaultpen(linewidth(.8pt)+fontsize(10pt));
dotfactor=4;
fill(Arc((0,0),3,0,180)--cycle,palered);
fill(Arc((0,0),2.598,0,180)--cycle,white);
draw(Arc((0,0),3,0,180),black);
draw(Arc((0,0),2.598,0,180),black);
draw((-2.598,0) -- (2.598,0), white);
draw((-1.5,2.598) -- (1.5,... | [] |
959 | A semicircle has diameter $\overline{AB}$ and chord $\overline{CD}$ of length $16$ parallel to $\overline{AB}$. A smaller semicircle
with diameter on $\overline{AB}$ and tangent to $\overline{CD}$ is cut from the larger semicircle, as shown below.
What is the area of the resulting figure, shown shaded?
$\textbf{(A)... | 2025 AMC 10A Problem 10 | We can move this small semicircle to the middle of the big semicircle. Let $r$ be the radius of the small semicircle. By Pythagorean Theorem, the line draw from the midpoint of the diameter of the big semicircle to the big semicircle at the height of the small semicircle (see the diagram) has length $\sqrt{r^2 + 64}$. ... | // Block 1
import graph;
unitsize(14mm);
defaultpen(linewidth(.8pt)+fontsize(10pt));
dotfactor=4;
real R = 3;
pair A = (-R,0), B = (R,0);
fill(Arc((0,0),R,0,180)--cycle,palered);
draw(Arc((0,0),R,0,180),black);
real h = 0.75;
pair C = (-sqrt(R^2 - h^2), h);
pair D = ( sqrt(R^2 - h^2), h);
draw(C--D,black);
label("$C$",... | [] |
960 | In the figure below, $ABEF$ is a rectangle, $\overline{AD}\perp\overline{DE}$, $AF=7$, $AB=1$, and $AD=5$.
What is the area of $\triangle ABC$?
$\textbf{(A) } \frac{3}{8} \qquad\textbf{(B) } \frac{4}{9} \qquad\textbf{(C) } \frac{1}{8}\sqrt{13} \qquad\textbf{(D) } \frac{7}{15} \qquad\textbf{(E) } \frac{1}{8}\sqrt{15}$ | 2025 AMC 10A Problem 15 | As shown in the figure, establish a coordinate system with point A as the origin.
Suppose $C = (a, 1)$, then the equation of $AC$ is $y = \dfrac{1}{a}x$. Since $DE \perp AC$, the slope of $DE$ is $-a$, the equation of $DE$ is $y - 1 = -a(x - 7) \to ax + y - 7a - 1 = 0$.
Notice that the length of $AD$ is the distanc... | // Block 1
unitsize(1cm);
pair _A, _B, _C, _D, _E, _F;
_A = (0, 0);
_B = (0, 1);
_E = (7, 1);
_F = (7, 0);
_C = (3/4, 1);
_D = extension(_A, _C, _E, _E + rotate(90)*(_A - _C));
fill(_A -- _B -- _C -- cycle, gray);
draw(_A -- _B -- _E -- _D -- _A -- _F -- _E);
draw((-1, 0) -- (8, 0), Arrows);
draw((0, -1) -- (0, 5), Arr... | [] |
960 | In the figure below, $ABEF$ is a rectangle, $\overline{AD}\perp\overline{DE}$, $AF=7$, $AB=1$, and $AD=5$.
What is the area of $\triangle ABC$?
$\textbf{(A) } \frac{3}{8} \qquad\textbf{(B) } \frac{4}{9} \qquad\textbf{(C) } \frac{1}{8}\sqrt{13} \qquad\textbf{(D) } \frac{7}{15} \qquad\textbf{(E) } \frac{1}{8}\sqrt{15}$ | 2025 AMC 10A Problem 15 | Start by assigning the variable $x$ to the length of side $\overline{AC}$. Since $\overline{AD}=5$, $\overline {CD}=5-x$. Additionally, the diagonal $\overline{AE}=5\sqrt{2}$ by the Pythagorean Theorem. Notice that with this diagonal, we've created a right triangle $\triangle ADE$.
Now, you could manually use the ... | // Block 1
unitsize(1cm);
pen small=fontsize(9pt);
pair A, B, C, D, E, F;
draw(brace((5,5),(5,0)),1bp+black);
label(" $ $ $5$",brace((5,5),(5,0)), E);
A = (5, 5);
B = (5.6, 4.2);
C = (5, 3.75);
D = (5, 0);
E = (0, 0);
F = (-0.6, 0.8);
fill(A--B--C--cycle, palered);
dot(A);
dot(B);
dot(C);
dot(D);
dot(E);
dot(F);
label(... | [] |
961 | A silo (right circular cylinder) with diameter 20 meters stands in a field. MacDonald is located 20 meters west and 15 meters south of the center of the silo. McGregor is located 20 meters east and $g > 0$ meters south of the center of the silo. The line of sight between MacDonald and McGregor is tangent to the silo. T... | 2025 AMC 10A Problem 20 | Let the silo center be $O$, let the point MacDonald is situated at be $A$, and let the point $20$ meters west of the silo center be $B$. $ABO$ is then a right triangle with side lengths $15, 20,$ and $25$.
Let the point $20$ meters east of the silo center be $C$, and let the point McGregor is at be $D$ with $CD=g>0... | // Block 1
import olympiad;
size(340);
// numeric setup for reliable AoPS rendering
real R = 10;
pair O = (0,0);
pair B = (-20,0);
pair A = (-20,-15);
pair C = (20,0);
real g = 5.55; // approximate numeric version of (20√21−75)/3
pair D = (20,-g);
pair E = (2.30,-9.74);
pair F = intersectionpoint(A--(A+4*(D-A)), B--(B... | [] |
961 | A silo (right circular cylinder) with diameter 20 meters stands in a field. MacDonald is located 20 meters west and 15 meters south of the center of the silo. McGregor is located 20 meters east and $g > 0$ meters south of the center of the silo. The line of sight between MacDonald and McGregor is tangent to the silo. T... | 2025 AMC 10A Problem 20 | We use the point names given in Solution 1.
By Pythagorean Theorem (twice), we have \[AE=\sqrt{AO^2-OE^2}=\sqrt{AB^2+BO^2-10^2}=\sqrt{525}=5\sqrt{21}.\] Similarly, we find that \[DE=\sqrt{DO^2-OE^2}=\sqrt{CD^2+CO^2-10^2}=\sqrt{300+g^2}.\] Note that the altitude from $D$ to $AB$ (call the foot $H$) has length $40$. Als... | // Block 1
import olympiad;
size(340);
// numeric setup for reliable AoPS rendering
real R = 10;
pair O = (0,0);
pair B = (-20,0);
pair A = (-20,-15);
pair C = (20,0);
real g = 5.55; // approximate numeric version of (20√21−75)/3
pair D = (20,-g);
pair E = (2.30,-9.74);
pair F = intersectionpoint(A--(A+4*(D-A)), B--(B... | [] |
961 | A silo (right circular cylinder) with diameter 20 meters stands in a field. MacDonald is located 20 meters west and 15 meters south of the center of the silo. McGregor is located 20 meters east and $g > 0$ meters south of the center of the silo. The line of sight between MacDonald and McGregor is tangent to the silo. T... | 2025 AMC 10A Problem 20 | First, taking the Pythagorean Theorem on $\Delta BOD$ and $\Delta AOE$ gives $AO = 25, AE = 5\sqrt{21}$. By cosine angular addition, $\cos{(\angle BOD + \angle AOE)} = \cos{\angle BOE} = \frac{8-3\sqrt21}{25}$, so $\cos{\angle EOC} = \frac{3\sqrt21-8}{25}$.
Dropping the altitude from $E$ to $BC$ and labelling the foo... | // Block 1
import olympiad;
size(340);
// numeric setup for reliable AoPS rendering
real R = 10;
pair O = (0,0);
pair B = (-20,0);
pair A = (-20,-15);
pair C = (20,0);
real g = 5.55; // approximate numeric version of (20√21−75)/3
pair D = (20,-g);
pair E = (2.30,-9.74);
pair F = intersectionpoint(A--(A+4*(D-A)), B--(B... | [] |
961 | A silo (right circular cylinder) with diameter 20 meters stands in a field. MacDonald is located 20 meters west and 15 meters south of the center of the silo. McGregor is located 20 meters east and $g > 0$ meters south of the center of the silo. The line of sight between MacDonald and McGregor is tangent to the silo. T... | 2025 AMC 10A Problem 20 | $F$ is the midpoint of $AD$, $DG \perp AB$. Notice that $\triangle OEF \sim \triangle DGA$, we have
\[\frac{OE}{OF} = \frac{DG}{DA}\]
Given the length of $CD$ is $g$, $OF$ is the midline, $CD = BG$, therefore $GA = 15 - g$, $DA = \sqrt{40^2 + (15 - g)^2}$, $OF = \dfrac{15 + g}{2}$
\[\frac{OE}{OF} = \frac{DG}{GA}\]
\[\c... | // Block 1
import geometry;
unitsize(7);
pair _O = (0,0);
pair _B = (-20,0);
pair _A = (-20,-15);
pair _C = (20,0);
pair _D = (20, -(20sqrt(21)-75)/3);
pair _E = extension(_A, _D, _O, _O + rotate(90)*(_A - _D));
pair _F = (_A + _D)/2;
pair _G = (-20, _D.y);
draw(arc(_O, 10, 0, -180));
draw(_A -- _D -- _C -- _B -- _A ^... | [] |
962 | Triangle $\triangle ABC$ has side lengths $AB = 80$, $BC = 45$, and $AC = 75$. The bisector of $\angle B$ and the altitude to side $\overline{AB}$ intersect at point $P$. What is $BP$?
$\textbf{(A)}~18\qquad\textbf{(B)}~19\qquad\textbf{(C)}~20\qquad\textbf{(D)}~21\qquad\textbf{(E)}~22$ | 2025 AMC 10A Problem 23 | Preface: When we have a problem as such, involving a simple diagram with minimal instructions, I use a method I named "rulerbash". Rulerbash should only be used in specific cases and as a last resort, mainly in the event of a time crunch or a difficult problem.
Start with the longest side, drawing a line with a lengt... | // Block 1
unitsize(0.25cm); pair A=(0,0), B=(8,0); path cA=circle(A,7.5), cB=circle(B,4.5); pair C=intersectionpoint(cA,cB); draw(A--B); draw(cA); draw(cB); draw(A--C--B--cycle); dot(A); dot(B); dot(C); label("A",A,S, fontsize(8)); label("B",B,S, fontsize(8)); label("C",C,N, fontsize(8)); label("8 cm",(A+B)/2, S, f... | [] |
962 | Triangle $\triangle ABC$ has side lengths $AB = 80$, $BC = 45$, and $AC = 75$. The bisector of $\angle B$ and the altitude to side $\overline{AB}$ intersect at point $P$. What is $BP$?
$\textbf{(A)}~18\qquad\textbf{(B)}~19\qquad\textbf{(C)}~20\qquad\textbf{(D)}~21\qquad\textbf{(E)}~22$ | 2025 AMC 10A Problem 23 | Due to Angle Bisector Theorem
\[\frac{CE}{AE} = \frac{BC}{BA} = \frac{45}{80} = \frac{9}{16}\]
\[CE = AC \times \frac{9}{25} = 27\]
\[AE = AC \times \frac{16}{25} = 48\]
Notice that $\dfrac{CE}{CB} = \dfrac{CB}{CA} = \dfrac{3}{5}$, $\triangle CBE \sim \triangle CAB$, $\angle A = \angle CBE = \angle EBF$, so $\triangle ... | /* Figure drawn by reda*/ import geometry; unitsize(2.5); pair _A = (0, 0); pair _B = (80, 0); pair _C = (75*(75^2+80^2-45^2)/(2*75*80), 75*sqrt(1-((75^2+80^2-45^2)/(2*75*80))^2)); pair _D = (_C.x, 0); pair _E = (_A * 45 + _C * 80)/(45 + 80); pair _F = (_E.x, 0); pair _P = extension(_C, _D, _B, _E); draw(_B -- _A -- _... | [] |
963 | A point $P$ is chosen at random inside square $ABCD$. The probability that $\overline{AP}$ is neither the shortest nor the longest side of $\triangle APB$ can be written as $\frac{a + b \pi - c \sqrt{d}}{e}$, where $a, b, c, d,$ and $e$ are positive integers, $\text{gcd}(a, b, c, e) = 1$, and $d$ is not divisible by th... | 2025 AMC 10A Problem 25 | Say WLOG that $AB$ is the top side of the square, and the square is of side length 1. Let us say that the midpoint of $AB$ is $M$, while the midpoint of $CD$ is $Q$. Drawing a vertical line to split the square in half, we notice that if $P$ is to the left of the line, $AP < BP$, and if P is to the right of the line, $A... | // Block 1
size(200);
pair A = (0,1), B = (1,1), C = (1,0), D = (0,0);
pair M = (0.5,1), Q = (0.5,0);
draw((0.5,0)--(0.5,1));
draw(A--B--C--D--cycle);
draw(arc(A,1,270,360));
pair N = intersectionpoint(arc(A,1,270,360), (0.5,0)--(0.5,1));
dot(N); label("$N$", N, NE);
dot(M); label("$M$", M, N);
dot(Q); label("$Q$... | [] |
963 | A point $P$ is chosen at random inside square $ABCD$. The probability that $\overline{AP}$ is neither the shortest nor the longest side of $\triangle APB$ can be written as $\frac{a + b \pi - c \sqrt{d}}{e}$, where $a, b, c, d,$ and $e$ are positive integers, $\text{gcd}(a, b, c, e) = 1$, and $d$ is not divisible by th... | 2025 AMC 10A Problem 25 | Assume the sides of this square is 1, hence we only need to find the area of the desired regions. From Solution 1, it is easy to see that the regions are the bottom left region c and the top right region b, hence we must compute $b+c$. Also, define $k$ to be the circular segment. We have two equations right off the ba... | // Block 1
size(200);
import olympiad;
pair A = (0,1), B = (1,1), C = (1,0), D = (0,0);
pair M = (0.5,1), Q = (0.5,0);
draw((0.5,0)--(0.5,1));
draw(A--B--C--D--cycle);
draw(arc(A,1,270,360));
pair N = intersectionpoint(arc(A,1,270,360), (0.5,0)--(0.5,1));
dot(N); label("$N$", N, NE);
draw(arc(B,1,180,270)); ... | [] |
964 | Pentagon $ABCDE$ is inscribed in a circle, and $\angle BEC = \angle CED = 30^\circ$. Let line $AC$ and line $BD$ intersect at point $F$, and suppose that $AB = 9$ and $AD = 24$. What is $BF$?
$\textbf{(A) } \frac{57}{11} \qquad\textbf{(B) } \frac{59}{11} \qquad\textbf{(C) } \frac{60}{11} \qquad\textbf{(D) } \frac{61}{... | 2025 AMC 12A Problem 8 | We will scale down the diagram by a factor of $3$ so that $AB = 3$ and $AD = 8.$ Since $\angle BEC = 30^{\circ},$ it follows that $\angle BAC = \angle BDC = 30^{\circ}$ as they all subtend the same arc. Similarly, since $\angle CED = 30^{\circ},$ it follows that $\angle CAD = \angle CBD = 30^{\circ}$ as well.
We obtai... | // Block 1
/* Made by MRENTHUSIASM */
size(200);
real r = 7*sqrt(3);
pair O, A, B, C, D, E, F;
O = origin;
B = r*dir(30);
C = r*dir(-30);
D = r*dir(-90);
E = r*dir(180);
A = intersectionpoints(Circle(O,r),Circle(B,9))[0];
F = intersectionpoint(A--C,B--D);
draw(Circle(O,r)^^B--C--D--E--A^^B--E--C--F);
draw(A--D--B--cyc... | [] |
965 | Let $w$ be the complex number $2+i$, where $i=\sqrt{-1}$. What real number $r$ has the property that $r$, $w$, and $w^2$ are three collinear points in the complex plane?
$\textbf{(A)}~\frac34\qquad\textbf{(B)}~1\qquad\textbf{(C)}~\frac75\qquad\textbf{(D)}~\frac32\qquad\textbf{(E)}~\frac53$ | 2025 AMC 12A Problem 9 | Recall that the slope of a line is $m=\tan\phi,$ where $\phi$ is the angle formed by the line and the positive $x$-axis.
Note that $|w|=\sqrt{2^2+1^2}=\sqrt{5}.$ In polar coordinates, let $w=\sqrt{5}\operatorname{cis}\theta.$ It follows that $\tan\theta=\frac12.$
By De Moivre's Theorem, we have $w^2=5\operatorname{ci... | // Block 1
/* Made by MRENTHUSIASM */
size(200);
int xMin = -2;
int xMax = 5;
int yMin = -2;
int yMax = 5;
//Draws the horizontal gridlines
void horizontalLines()
{
for (int i = yMin+1; i < yMax; ++i)
{
draw((xMin,i)--(xMax,i), mediumgray+linewidth(0.4));
}
}
//Draws the vertical gridlines
void verticalLi... | [] |
966 | The orthocenter of a triangle is the concurrent intersection of the three (possibly extended) altitudes. What is the sum of the coordinates of the orthocenter of the triangle whose vertices are $A(2,31), B(8,27),$ and $C(18,27)$?
$\textbf{(A)}~5\qquad\textbf{(B)}~17\qquad\textbf{(C)}~10+4\sqrt{17} +2\sqrt{13}\qquad\te... | 2025 AMC 12A Problem 11 | Let the intersection of the heights and their corresponding sides be $H_1$ and $H_2$, respectively, and let the intersection of line $AH_1$ and $H_2B$ be $E$. Since $\angle EH_2A = \angle EH_1B = 90^{\circ}$, $\angle EBH_1 = 90^{\circ}- \angle BEH_1 = \angle EAH_2$, we have
\[\tan{\angle EBH_1} = \tan{\angle EAH_2} = ... | // Block 1
size(200);
pair A = (2,31);
pair B = (8, 27);
pair C = (18, 27);
pair H1 = (2, 27);
pair H2 = (8.5882352941176, 29.3529411764706);
pair D = (2,3);
label("$A$", A, NW);
label("$B$", (8.5, 27), S);
label("$C$", C, SE);
label("$H_1$", H1, SW);
label("$H_2$", H2, NE);
label("$E$", E, S);
draw(A--B--C--cycle);
... | [] |
967 | Triangle $\triangle ABC$ has side lengths $AB = 80$, $BC = 45$, and $AC = 75$. The bisector of $\angle B$ and the altitude to side $\overline{AB}$ intersect at point $P$. What is $BP$?
$\textbf{(A)}~18\qquad\textbf{(B)}~19\qquad\textbf{(C)}~20\qquad\textbf{(D)}~21\qquad\textbf{(E)}~22$ | 2025 AMC 12A Problem 16 | Preface: When we have a problem as such, involving a simple diagram with minimal instructions, I use a method I named "rulerbash". Rulerbash should only be used in specific cases and as a last resort, mainly in the event of a time crunch or a difficult problem.
Start with the longest side, drawing a line with a lengt... | // Block 1
unitsize(0.25cm);
pair A=(0,0), B=(8,0);
path cA=circle(A,7.5), cB=circle(B,4.5);
pair C=intersectionpoint(cA,cB);
draw(A--B);
draw(cA);
draw(cB);
draw(A--C--B--cycle);
dot(A); dot(B); dot(C);
label("A",A,S, fontsize(8));
label("B",B,S, fontsize(8));
label("C",C,N, fontsize(8));
label("8 cm",(A+B)/2, S, f... | [] |
967 | Triangle $\triangle ABC$ has side lengths $AB = 80$, $BC = 45$, and $AC = 75$. The bisector of $\angle B$ and the altitude to side $\overline{AB}$ intersect at point $P$. What is $BP$?
$\textbf{(A)}~18\qquad\textbf{(B)}~19\qquad\textbf{(C)}~20\qquad\textbf{(D)}~21\qquad\textbf{(E)}~22$ | 2025 AMC 12A Problem 16 | Due to Angle Bisector Theorem
\[\frac{CE}{AE} = \frac{BC}{BA} = \frac{45}{80} = \frac{9}{16}\]
\[CE = AC \times \frac{9}{25} = 27\]
\[AE = AC \times \frac{16}{25} = 48\]
Notice that $\dfrac{CE}{CB} = \dfrac{CB}{CA} = \dfrac{3}{5}$, $\triangle CBE \sim \triangle CAB$, $\angle A = \angle CBE = \angle EBF$, so $\triangle ... | // Block 1
/* Figure drawn by reda*/
import geometry;
unitsize(2.5);
pair _A = (0, 0);
pair _B = (80, 0);
pair _C = (75*(75^2+80^2-45^2)/(2*75*80), 75*sqrt(1-((75^2+80^2-45^2)/(2*75*80))^2));
pair _D = (_C.x, 0);
pair _E = (_A * 45 + _C * 80)/(45 + 80);
pair _F = (_E.x, 0);
pair _P = extension(_C, _D, _B, _E);
draw(_B... | [] |
968 | A circle of radius $r$ is surrounded by $12$ circles of radius $1,$ externally tangent to the central circle and sequentially tangent to each other, as shown. Then $r$ can be written as $\sqrt a + \sqrt b + c,$ where $a, b, c$ are integers. What is $a+b+c?$
$\textbf{(A) } 3 \qquad \textbf{(B) } 5 \qquad \textbf{(C) ... | 2025 AMC 12A Problem 24 | We connect the centers of the smaller circles at $0^\circ$ and $90^\circ$ as follows:
Note that connecting the centers of the 12 smaller circles forms a regular 12-gon. A regular 12-gon has angles
\[\frac{10\cdot 180}{12} = 150^\circ,\]
so $\angle OAB = \angle ODC = 75^\circ.$ Furthermore, since $OA=OD$ and $\angl... | // Block 1
import graph;
size(250);
real R = -1 + sqrt(2) + sqrt(6);
real r = 1;
pair O = (0,0);
label("$O$", 0, W);
pair C0 = (R + r) * dir(0);
label("$A$", C0, E);
pair C15 = (R + r) * dir(15);
label("2", C15, NE);
pair C30 = (R + r) * dir(30);
label("$B$", C30, E);
pair C45 = (R + r) * dir(45);
label("2", C45, NE);
... | [] |
969 | In $\triangle ABC$, $AB = 10$, $AC = 18$, and $\angle B = 130^\circ$. Let $O$ be the center of the circle containing points $A, B, C$. What is the degree measure of $\angle CAO$?
$\textbf{(A)}\ 20^\circ \qquad \textbf{(B)}\ 30^\circ \qquad \textbf{(C)}\ 40^\circ \qquad \textbf{(D)}\ 50^\circ \qquad \textbf{(E)}\ 60^\... | 2025 AMC 10B Problem 5 | Since $\angle ABC = 130^\circ$, by the Inscribed Angle Theorem, it intercepts the major arc $AC$, so that arc has measure $260^\circ$. Point $O$ is the circumcenter, and is therefore outside of triangle ABC. Therefore, the minor arc $AC$ has measure $360^\circ - 260^\circ = 100^\circ$. The central angle $\angle AOC$ su... | // Block 1
size(220);
// Choose coordinates so that angle B is obtuse (~130°)
pair B = (0,0);
pair C = (3,0);
pair A = 3*dir(130);
// Circumcenter of triangle ABC (outside since angle at B is obtuse)
pair O = circumcenter(A,B,C);
// Draw circumcircle
draw(circle(O, abs(O - A)), gray(0.7));
// Draw triangle ABC
draw... | [] |
970 | The figure below shows an equilateral triangle, a rhombus with a $60^\circ$ angle, and a regular hexagon, each of them containing some mutually tangent congruent disks. Let $T, R,$ and $H,$ respectively, denote the ratio in each case of the total area of the disks to the area of the enclosing polygon. Which of the foll... | 2025 AMC 10B Problem 12 | To solve this problem, we can dissect the different shapes into equilateral triangles. We can keep figure $T$ the same, and then divide the rhombus $R$ into two equilateral triangles (as a result of one of the angles equaling $60^\circ$), with one circle in each that is the incircle to its triangle. We can do the same ... | // Block 1
/* Figure by reda */
unitsize(15);
real r = 1 + sqrt(3);
draw(r*dir(90) -- r*dir(210) -- r*dir(330) -- cycle);
filldraw(circle(dir(90), sqrt(3)/2), rgb("#CCC9E6"));
filldraw(circle(dir(210), sqrt(3)/2), rgb("#CCC9E6"));
filldraw(circle(dir(330), sqrt(3)/2), rgb("#CCC9E6"));
draw(dir(90)+dir(-30)*sqrt(3) --... | [] |
971 | The altitude to the hypotenuse of a $30^\circ{-}60^\circ{-}90^\circ$ is divided into two segments of lengths $x<y$ by the median to the shortest side of the triangle. What is the ratio $\tfrac{x}{x+y}$?
$\textbf{(A) } \dfrac{3}{7} \qquad \textbf{(B) } \dfrac{\sqrt3}{4} \qquad \textbf{(C) } \dfrac{4}{9} \qquad \textbf{... | 2025 AMC 10B Problem 13 | Without loss of generality, let $\triangle ABC$ have side-lengths $AB=2, BC=2\sqrt{3},$ and $AC=4.$ Let $D$ be the foot of the perpendicular from $B$ to $\overline{AC}, \ E$ be the midpoint of $\overline{AB},$ and $F$ be the intersection of $\overline{CE}$ and $\overline{BD}.$ Note that $\triangle ADB$ and $\triangle B... | /* Made by MRENTHUSIASM */ size(250); pair A, B, C, D, E, F; A = (0,1); B = origin; C = (sqrt(3),0); D = foot(B,A,C); E = midpoint(A--B); F = intersectionpoint(B--D,C--E); markscalefactor=0.008; draw(rightanglemark(A,B,C)^^rightanglemark(B,D,C),red+linewidth(1.25)); draw(A--B--C--cycle,linewidth(1.25)); draw(B--D^^C--... | [] |
971 | The altitude to the hypotenuse of a $30^\circ{-}60^\circ{-}90^\circ$ is divided into two segments of lengths $x<y$ by the median to the shortest side of the triangle. What is the ratio $\tfrac{x}{x+y}$?
$\textbf{(A) } \dfrac{3}{7} \qquad \textbf{(B) } \dfrac{\sqrt3}{4} \qquad \textbf{(C) } \dfrac{4}{9} \qquad \textbf{... | 2025 AMC 10B Problem 13 | This solution refers to the diagram in Solution 1, with the definitions slightly modified.
Let $AB=2$. Label the foot of the median $M$ and the foot of the altitude $H$. In $\triangle ABH$, we construct the midsegment $\overline{MP}$, which is parallel to $\overline{BH}$ and half as long as $\overline{BH}$. In additio... | /* Made by MRENTHUSIASM */ size(250); pair A, B, C, H, M, F, P; A = (0,1); B = origin; C = (sqrt(3),0); H = foot(B,A,C); M = midpoint(A--B); F = intersectionpoint(B--H,C--M); P = midpoint(A--H); markscalefactor=0.008; draw(rightanglemark(A,B,C)^^rightanglemark(B,H,C),red+linewidth(1.25)); draw(A--B--C--cycle,linewidt... | [] |
971 | The altitude to the hypotenuse of a $30^\circ{-}60^\circ{-}90^\circ$ is divided into two segments of lengths $x<y$ by the median to the shortest side of the triangle. What is the ratio $\tfrac{x}{x+y}$?
$\textbf{(A) } \dfrac{3}{7} \qquad \textbf{(B) } \dfrac{\sqrt3}{4} \qquad \textbf{(C) } \dfrac{4}{9} \qquad \textbf{... | 2025 AMC 10B Problem 13 | This solution refers to the definitions and the diagram in Solution 1.
Let $M$ be the midpoint of $\overline{AC},$ so $DM=1.$ Let $G$ be the intersection of $\overline{BM}$ and $\overline{CE},$ so $G$ is the centroid of $\triangle ABC.$ In addition, we construct $\overline{DE},$ as shown below:
In any right triangle,... | /* Made by MRENTHUSIASM */ size(250); pair A, B, C, D, E, F, M, G; A = (0,1); B = origin; C = (sqrt(3),0); D = foot(B,A,C); E = midpoint(A--B); F = intersectionpoint(B--D,C--E); M = midpoint(A--C); G = centroid(A,B,C); markscalefactor=0.008; draw(rightanglemark(A,B,C)^^rightanglemark(B,D,C),red+linewidth(1.25)); draw(... | [] |
971 | The altitude to the hypotenuse of a $30^\circ{-}60^\circ{-}90^\circ$ is divided into two segments of lengths $x<y$ by the median to the shortest side of the triangle. What is the ratio $\tfrac{x}{x+y}$?
$\textbf{(A) } \dfrac{3}{7} \qquad \textbf{(B) } \dfrac{\sqrt3}{4} \qquad \textbf{(C) } \dfrac{4}{9} \qquad \textbf{... | 2025 AMC 10B Problem 13 | This solution refers to the definitions and the diagram in Solution 1.
Since $E$ is the midpoint of $\overline{AB}$, we have $AE=1$, and because $\angle A=60^\circ$ and $AD=1$, $\triangle AED$ is equilateral. Its perpendicular bisector meets $\overline{AE}$ at $H$, giving $AH=HE=\tfrac12$ and $DH=\tfrac{\sqrt3}{2}$.
... | size(325); pair A, B, C, D, E, F, G, H; A = (0,2); B = origin; C = (2*sqrt(3),0); D = foot(B,A,C); E = midpoint(A--B); F = intersectionpoint(B--D,C--E); // G and H for the parallel/perpendicular construction G = (-sqrt(3), 1.5); H = (0, 1.5); markscalefactor=0.015; // Draw figure draw(rightanglemark(A,B,C)^^rightan... | [] |
971 | The altitude to the hypotenuse of a $30^\circ{-}60^\circ{-}90^\circ$ is divided into two segments of lengths $x<y$ by the median to the shortest side of the triangle. What is the ratio $\tfrac{x}{x+y}$?
$\textbf{(A) } \dfrac{3}{7} \qquad \textbf{(B) } \dfrac{\sqrt3}{4} \qquad \textbf{(C) } \dfrac{4}{9} \qquad \textbf{... | 2025 AMC 10B Problem 13 | WLOG $\overline{AB} = 2$ as the problem statement concerns only lengths ratios.
Draw $\overline{EM}$ connecting the median of $\triangle ABE$ to right angle $\angle E$, we know that by inscribing right $\triangle ABE$ in a circle and by Thales theorem, $\overline{ME} = \overline{MA} = \overline{MB}$ the circumradius.
$... | unitsize(0.3mm); /* Generated by Cloud's Excalidraw to Asymptote */ draw((-561, -520)--(-553, -515)--(-548, -523)--(-556, -528)--cycle, rgb(0.94, 0.54, 0)+linewidth(1)+solid);draw((-700, -590)--(-691, -590)--(-691, -599)--(-700, -599)--cycle, rgb(0.94, 0.54, 0)+linewidth(1)+solid);draw((-567, -510)--(-559, -505)--(-554... | [] |
972 | A container has a $1\times 1$ square bottom, a $3\times 3$ open square top, and four congruent trapezoidal sides, as shown. Starting when the container is empty, a hose that runs water at a constant rate takes $35$ minutes to fill the container up to the midline of the trapezoids.
How many more minutes will it take... | 2025 AMC 10B Problem 19 | Extend the bottom of the container to form an upside down square pyramid. Then take a diagonal cross section as shown below.
You can notice that there are several similar triangles, so we write several equations.
\[\frac{x_3}{x_1+x_2+x_3}=\frac{\sqrt{2}}{3\sqrt{2}}\]
\[\frac{x_3}{x_2+x_3}=\frac{\sqrt{2}}{2\sqrt{2}}\... | size(220); // Outer (largest) upside-down isosceles triangle pair A = (0,-4); // bottom vertex pair B = (-3,0); // left base point pair C = (3,0); // right base point draw(A--B--C--cycle); // Middle triangle (scaled toward A) real k2 = 0.6; pair B2 = A + k2*(B - A); pair C2 = A + k2*(C - A); draw(A--... | [] |
973 | Four congruent semicircles are inscribed in a square of side length $1$ so that their diameters are on the sides of the square, one endpoint of each diameter is at a vertex of the square, and adjacent semicircles are tangent to each other. A small circle centered at the center of the square is tangent to each of the fo... | 2025 AMC 10B Problem 20 | Let the radius of the large semicircles be $r$, and let the diameter of the inner be $D$. Draw lines as follows:
By the Pythagorean Theorem, $r^2+(1-r)^2=(2r)^2$, so $2r^2+2r-1=0$. Hence $r=\frac{\sqrt{3}-1}{2}$ (we take the positive solution).
Now apply the Pythagorean Theorem on the quadrilateral’s altitude down t... | // Block 1
import geometry;
size(6cm);
real r = (sqrt(3)-1)/2, R = (sqrt(2) - 1)*(sqrt(3)-1)/2;
draw((0,0) -- (1,0) -- (1, 1) -- (0,1) -- cycle);
draw(arc((r, 0), r, 0, 180));
draw(arc((1, r), r, 90, 270));
draw(arc((1 - r, 1), r, 180, 360));
draw(arc((0, 1 - r), r, 270, 450));
draw(circle((1/2,1/2),R));
draw((0.36602... | [] |
973 | Four congruent semicircles are inscribed in a square of side length $1$ so that their diameters are on the sides of the square, one endpoint of each diameter is at a vertex of the square, and adjacent semicircles are tangent to each other. A small circle centered at the center of the square is tangent to each of the fo... | 2025 AMC 10B Problem 20 | We have the diagram:
We label the vertices of the square $A$,$B$,$C$, and $D$, the centers of the semicircle $E$, $F$, $G$, and $H$, and the center of the small circle $O$. Lets call the radius of the big semicircle $x$ and the radius of the small circle $r$.
Looking at $\triangle HDG$, we can see it is a right t... | // Block 1
import geometry;
size(6cm);
real r = (sqrt(3)-1)/2, R = (sqrt(2) - 1)*(sqrt(3)-1)/2;
draw((0,0) -- (1,0) -- (1, 1) -- (0,1) -- cycle);
draw(arc((r, 0), r, 0, 180));
draw(arc((1, r), r, 90, 270));
draw(arc((1 - r, 1), r, 180, 360));
draw(arc((0, 1 - r), r, 270, 450));
draw(circle((1/2,1/2),R));
draw((0.36602... | [] |
973 | Four congruent semicircles are inscribed in a square of side length $1$ so that their diameters are on the sides of the square, one endpoint of each diameter is at a vertex of the square, and adjacent semicircles are tangent to each other. A small circle centered at the center of the square is tangent to each of the fo... | 2025 AMC 10B Problem 20 | We realize that we can connect the radii of the semicircles to form an inner square (let $r$ denote the radius of the semicircles and $d$ denote their diameter)
To find the length of the inner square, we apply the Pythagorean Theorem
\[(1 - r)^{2}+r^{2}= (2r)^{2}\]
We now substitute $r = \frac{d}{2}$
\[(1 - \frac{d}{2... | // Block 1
import geometry;
size(6cm);
real r = (sqrt(3)-1)/2, R = (sqrt(2) - 1)*(sqrt(3)-1)/2;
draw((0,0) -- (1,0) -- (1, 1) -- (0,1) -- cycle);
draw((1,0.366025403784) -- (0.366025403784, 0));
draw((0.366025403784, 0) -- (0, 0.63397459621));
draw((1,0.366025403784) -- (0.63397459621, 1));
draw((0.63397459621, 1) -- ... | [] |
974 | Square $ABCD$ has sides of length $4$. Points $P$ and $Q$ lie on $\overline{AD}$ and $\overline{CD}$, respectively, with $AP=\frac{8}{5}$ and $DQ=\frac{10}{3}$. A path begins along the segment from $P$ to $Q$ and continues by reflecting against the sides of $ABCD$ (with congruent incoming and outgoing angles). If the p... | 2025 AMC 10B Problem 25 | The square can be placed with $A=(0,0),B=(4,0),C=(4,4),D=(0,4)$. Then \[P=(0,\frac{8}{5}),Q=(\frac{10}{3},4).\]
Straight line reflection is equivalent to passing through. The straight line from $P$ to $Q$ is
\[y = \frac{18}{25}x + \frac{8}{5}\]
\[25y = 18x + 40 \equiv 4 \pmod{18}\]
Notice that $25 \times 2 \equiv 7 ... | // Block 1
unitsize(20);
pair _A, _B, _C, _D, _P, _Q;
_A = (0, 0);
_B = (0, 4);
_C = (4, 4);
_D = (4, 0);
_P = (0, 8/5);
_Q = (4, 10/3);
draw(_A -- _B -- _C -- _D -- _A ^^ _P -- _Q);
draw(extension(_P, _Q, _B, _C) -- _Q -- 2*_C - extension(_P, _Q, _B, _C), dashed);
label("$A$", _A, SW);
label("$B$", _B, NW);
label("$C$... | [] |
975 | The altitude to the hypotenuse of a $30^\circ{-}60^\circ{-}90^\circ$ is divided into two segments of lengths $x<y$ by the median to the shortest side of the triangle. What is the ratio $\tfrac{x}{x+y}$?
$\textbf{(A) } \dfrac{3}{7} \qquad \textbf{(B) } \dfrac{\sqrt3}{4} \qquad \textbf{(C) } \dfrac{4}{9} \qquad \textbf{... | 2025 AMC 12B Problem 10 | Without loss of generality, let $\triangle ABC$ have side-lengths $AB=2, BC=2\sqrt{3},$ and $AC=4.$ Let $D$ be the foot of the perpendicular from $B$ to $\overline{AC}, \ E$ be the midpoint of $\overline{AB},$ and $F$ be the intersection of $\overline{CE}$ and $\overline{BD}.$ Note that $\triangle ADB$ and $\triangle B... | // Block 1
/* Made by MRENTHUSIASM */
size(250);
pair A, B, C, D, E, F;
A = (0,1);
B = origin;
C = (sqrt(3),0);
D = foot(B,A,C);
E = midpoint(A--B);
F = intersectionpoint(B--D,C--E);
markscalefactor=0.008;
draw(rightanglemark(A,B,C)^^rightanglemark(B,D,C),red+linewidth(1.25));
draw(A--B--C--cycle,linewidth(1.25));
dra... | [] |
975 | The altitude to the hypotenuse of a $30^\circ{-}60^\circ{-}90^\circ$ is divided into two segments of lengths $x<y$ by the median to the shortest side of the triangle. What is the ratio $\tfrac{x}{x+y}$?
$\textbf{(A) } \dfrac{3}{7} \qquad \textbf{(B) } \dfrac{\sqrt3}{4} \qquad \textbf{(C) } \dfrac{4}{9} \qquad \textbf{... | 2025 AMC 12B Problem 10 | This solution refers to the diagram in Solution 1, with the definitions slightly modified.
Let $AB=2$. Label the foot of the median $M$ and the foot of the altitude $H$. In $\triangle ABH$, we construct the midsegment $\overline{MP}$, which is parallel to $\overline{BH}$ and half as long as $\overline{BH}$. In additio... | // Block 1
/* Made by MRENTHUSIASM */
size(250);
pair A, B, C, H, M, F, P;
A = (0,1);
B = origin;
C = (sqrt(3),0);
H = foot(B,A,C);
M = midpoint(A--B);
F = intersectionpoint(B--H,C--M);
P = midpoint(A--H);
markscalefactor=0.008;
draw(rightanglemark(A,B,C)^^rightanglemark(B,H,C),red+linewidth(1.25));
draw(A--B--C--cyc... | [] |
975 | The altitude to the hypotenuse of a $30^\circ{-}60^\circ{-}90^\circ$ is divided into two segments of lengths $x<y$ by the median to the shortest side of the triangle. What is the ratio $\tfrac{x}{x+y}$?
$\textbf{(A) } \dfrac{3}{7} \qquad \textbf{(B) } \dfrac{\sqrt3}{4} \qquad \textbf{(C) } \dfrac{4}{9} \qquad \textbf{... | 2025 AMC 12B Problem 10 | This solution refers to the definitions and the diagram in Solution 1.
Let $M$ be the midpoint of $\overline{AC},$ so $DM=1.$ Let $G$ be the intersection of $\overline{BM}$ and $\overline{CE},$ so $G$ is the centroid of $\triangle ABC.$ In addition, we construct $\overline{DE},$ as shown below:
In any right triangle,... | // Block 1
/* Made by MRENTHUSIASM */
size(250);
pair A, B, C, D, E, F, M, G;
A = (0,1);
B = origin;
C = (sqrt(3),0);
D = foot(B,A,C);
E = midpoint(A--B);
F = intersectionpoint(B--D,C--E);
M = midpoint(A--C);
G = centroid(A,B,C);
markscalefactor=0.008;
draw(rightanglemark(A,B,C)^^rightanglemark(B,D,C),red+linewidth(1.... | [] |
975 | The altitude to the hypotenuse of a $30^\circ{-}60^\circ{-}90^\circ$ is divided into two segments of lengths $x<y$ by the median to the shortest side of the triangle. What is the ratio $\tfrac{x}{x+y}$?
$\textbf{(A) } \dfrac{3}{7} \qquad \textbf{(B) } \dfrac{\sqrt3}{4} \qquad \textbf{(C) } \dfrac{4}{9} \qquad \textbf{... | 2025 AMC 12B Problem 10 | This solution refers to the definitions and the diagram in Solution 1.
Since $E$ is the midpoint of $\overline{AB}$, we have $AE=1$, and because $\angle A=60^\circ$ and $AD=1$, $\triangle AED$ is equilateral. Its perpendicular bisector meets $\overline{AE}$ at $H$, giving $AH=HE=\tfrac12$ and $DH=\tfrac{\sqrt3}{2}$.
... | // Block 1
size(325);
pair A, B, C, D, E, F, G, H;
A = (0,2);
B = origin;
C = (2*sqrt(3),0);
D = foot(B,A,C);
E = midpoint(A--B);
F = intersectionpoint(B--D,C--E);
// G and H for the parallel/perpendicular construction
G = (-sqrt(3), 1.5);
H = (0, 1.5);
markscalefactor=0.015;
// Draw figure
draw(rightanglemark(A,B,... | [] |
975 | The altitude to the hypotenuse of a $30^\circ{-}60^\circ{-}90^\circ$ is divided into two segments of lengths $x<y$ by the median to the shortest side of the triangle. What is the ratio $\tfrac{x}{x+y}$?
$\textbf{(A) } \dfrac{3}{7} \qquad \textbf{(B) } \dfrac{\sqrt3}{4} \qquad \textbf{(C) } \dfrac{4}{9} \qquad \textbf{... | 2025 AMC 12B Problem 10 | WLOG $\overline{AB} = 2$ as the problem statement concerns only lengths ratios.
Draw $\overline{EM}$ connecting the median of $\triangle ABE$ to right angle $\angle E$, we know that by inscribing right $\triangle ABE$ in a circle and by Thales theorem, $\overline{ME} = \overline{MA} = \overline{MB}$ the circumradius.
$... | // Block 1
unitsize(0.3mm);
/* Generated by Cloud's Excalidraw to Asymptote */
draw((-561, -520)--(-553, -515)--(-548, -523)--(-556, -528)--cycle, rgb(0.94, 0.54, 0)+linewidth(1)+solid);draw((-700, -590)--(-691, -590)--(-691, -599)--(-700, -599)--cycle, rgb(0.94, 0.54, 0)+linewidth(1)+solid);draw((-567, -510)--(-559, -... | [] |
976 | A circle has been divided into 6 sectors of different sizes. Then 2 of the sectors are painted red, 2 painted green, and 2 painted blue so that no two neighboring sectors are painted the same color. One such coloring is shown below.
How many different colorings are possible?
$\textbf{(A)}~12\qquad\textbf{(B)}~16\... | 2025 AMC 12B Problem 13 | You can tackle the problem by forgetting about the different sizes at first. This gets you 4 combinations:
B-R-G-B-R-G
B-R-G-B-G-R
B-R-B-G-R-G
B-G-B-R-G-R
Now we can multiply by 6 since we have 6 different pieces. $6\cdot 4 =\boxed{\textbf{(D)}~24}$.
~Taha Jazaeri | // Block 1
size(300);
pair base = (0,0);
real r = 1;
// Define colors
pen B = blue;
pen R = red;
pen G = green;
// Patterns (each row = one circle)
// B,R,G,B,R,G etc.
pen[][] patterns = {
{B,R,G,B,R,G},
{B,R,G,B,G,R},
{B,R,B,G,R,G},
{B,G,B,R,G,R}
};
for (int c = 0; c < 4; ++c) {
pair O = base + (3*c, 0)... | [] |
977 | A container has a $1\times 1$ square bottom, a $3\times 3$ open square top, and four congruent trapezoidal sides, as shown. Starting when the container is empty, a hose that runs water at a constant rate takes $35$ minutes to fill the container up to the midline of the trapezoids.
How many more minutes will it take... | 2025 AMC 12B Problem 15 | Extend the bottom of the container to form an upside down square pyramid. Then take a diagonal cross section as shown below.
You can notice that there are several similar triangles, so we write several equations.
\[\frac{x_3}{x_1+x_2+x_3}=\frac{\sqrt{2}}{3\sqrt{2}}\]
\[\frac{x_3}{x_2+x_3}=\frac{\sqrt{2}}{2\sqrt{2}}\... | // Block 1
size(220);
// Outer (largest) upside-down isosceles triangle
pair A = (0,-4); // bottom vertex
pair B = (-3,0); // left base point
pair C = (3,0); // right base point
draw(A--B--C--cycle);
// Middle triangle (scaled toward A)
real k2 = 0.6;
pair B2 = A + k2*(B - A);
pair C2 = A + k2*(C - A... | [] |
978 | Two non-congruent triangles have the same area. Each triangle has sides of length $8$ and $9$, and the third side of each triangle has integer length. What is the sum of the lengths of the third sides?
$\textbf{(A) } 20 \qquad \textbf{(B) } 22 \qquad \textbf{(C) } 24 \qquad \textbf{(D) } 26 \qquad \textbf{(E) } 28$ | 2025 AMC 12B Problem 21 | ~Diagram by imosilver
Let the angles between the $8$ and $9$ sides be $\theta_1$ and $\theta_2$. Relating the two triangles' areas, we get
\[\frac{1}{2} \cdot 8 \cdot 9 \cdot \sin{\theta_1} = \frac{1}{2} \cdot 8 \cdot 9 \cdot \sin{\theta_2}\]
\[\sin{\theta_1} = \sin{\theta_2}\]
For this to be true, $\theta_1 + \theta_... | // Block 1
size(250);
pair A = (0,0);
pair C1 = (-9,0);
pair C2 = (9,0);
pair B = (-4.0/3.0, 4*sqrt(35)/3);
pen pen1 = blue+1.2;
pen pen2 = red+1.2;
draw(B--A--C1--cycle, pen1);
draw(B--A--C2--cycle, pen2);
dot("$A$", A, S);
dot("$C_1$", C1, SW);
dot("$C_2$", C2, SE);
dot("$B$", B, N);
label("$8$", (A+B)/2, W);
la... | [] |
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