{"year": "2008", "tier": "T4", "problem_label": "1", "problem_type": "Geometry", "exam": "HMMT", "problem": "How many different values can $\\angle A B C$ take, where $A, B, C$ are distinct vertices of a cube?", "solution": "5 . In a unit cube, there are 3 types of triangles, with side lengths $(1,1, \\sqrt{2}),(1, \\sqrt{2}, \\sqrt{3})$ and $(\\sqrt{2}, \\sqrt{2}, \\sqrt{2})$. Together they generate 5 different angle values.", "metadata": {"resource_path": "HarvardMIT/segmented/en-112-2008-feb-geo-solutions.jsonl", "problem_match": "\n1. [3]", "solution_match": "\nAnswer: "}}
{"year": "2008", "tier": "T4", "problem_label": "2", "problem_type": "Geometry", "exam": "HMMT", "problem": "Let $A B C$ be an equilateral triangle. Let $\\Omega$ be its incircle (circle inscribed in the triangle) and let $\\omega$ be a circle tangent externally to $\\Omega$ as well as to sides $A B$ and $A C$. Determine the ratio of the radius of $\\Omega$ to the radius of $\\omega$.", "solution": "$\\quad 3$ Label the diagram as shown below, where $\\Omega$ and $\\omega$ also denote the center of the corresponding circles. Note that $A M$ is a median and $\\Omega$ is the centroid of the equilateral triangle. So $A M=3 M \\Omega$. Since $M \\Omega=N \\Omega$, it follows that $A M / A N=3$, and triangle $A B C$ is the image of triangle $A B^{\\prime} C^{\\prime}$ after a scaling by a factor of 3 , and so the two incircles must also be related by a scale factor of 3 .\n![](https://cdn.mathpix.com/cropped/2025_01_24_684c7884f8823da5eea3g-1.jpg?height=289&width=329&top_left_y=1048&top_left_x=936)", "metadata": {"resource_path": "HarvardMIT/segmented/en-112-2008-feb-geo-solutions.jsonl", "problem_match": "\n2. [3]", "solution_match": "\nAnswer: "}}
{"year": "2008", "tier": "T4", "problem_label": "3", "problem_type": "Geometry", "exam": "HMMT", "problem": "Let $A B C$ be a triangle with $\\angle B A C=90^{\\circ}$. A circle is tangent to the sides $A B$ and $A C$ at $X$ and $Y$ respectively, such that the points on the circle diametrically opposite $X$ and $Y$ both lie on the side $B C$. Given that $A B=6$, find the area of the portion of the circle that lies outside the triangle.\n![](https://cdn.mathpix.com/cropped/2025_01_24_684c7884f8823da5eea3g-1.jpg?height=253&width=253&top_left_y=1543&top_left_x=974)", "solution": "$\\pi-2$ Let $O$ be the center of the circle, and $r$ its radius, and let $X^{\\prime}$ and $Y^{\\prime}$ be the points diametrically opposite $X$ and $Y$, respectively. We have $O X^{\\prime}=O Y^{\\prime}=r$, and $\\angle X^{\\prime} O Y^{\\prime}=90^{\\circ}$. Since triangles $X^{\\prime} O Y^{\\prime}$ and $B A C$ are similar, we see that $A B=A C$. Let $X^{\\prime \\prime}$ be the projection of $Y^{\\prime}$ onto $A B$. Since $X^{\\prime \\prime} B Y^{\\prime}$ is similar to $A B C$, and $X^{\\prime \\prime} Y^{\\prime}=r$, we have $X^{\\prime \\prime} B=r$. It follows that $A B=3 r$, so $r=2$.\n![](https://cdn.mathpix.com/cropped/2025_01_24_684c7884f8823da5eea3g-1.jpg?height=386&width=391&top_left_y=2070&top_left_x=908)\n\nThen, the desired area is the area of the quarter circle minus that of the triangle $X^{\\prime} O Y^{\\prime}$. And the answer is $\\frac{1}{4} \\pi r^{2}-\\frac{1}{2} r^{2}=\\pi-2$.", "metadata": {"resource_path": "HarvardMIT/segmented/en-112-2008-feb-geo-solutions.jsonl", "problem_match": "\n3. [4]", "solution_match": "\nAnswer: "}}
{"year": "2008", "tier": "T4", "problem_label": "4", "problem_type": "Geometry", "exam": "HMMT", "problem": "In a triangle $A B C$, take point $D$ on $B C$ such that $D B=14, D A=13, D C=4$, and the circumcircle of $A D B$ is congruent to the circumcircle of $A D C$. What is the area of triangle $A B C$ ?", "solution": "108\n![](https://cdn.mathpix.com/cropped/2025_01_24_684c7884f8823da5eea3g-2.jpg?height=332&width=438&top_left_y=539&top_left_x=887)\n\nThe fact that the two circumcircles are congruent means that the chord $A D$ must subtend the same angle in both circles. That is, $\\angle A B C=\\angle A C B$, so $A B C$ is isosceles. Drop the perpendicular $M$ from $A$ to $B C$; we know $M C=9$ and so $M D=5$ and by Pythagoras on $A M D, A M=12$. Therefore, the area of $A B C$ is $\\frac{1}{2}(A M)(B C)=\\frac{1}{2}(12)(18)=108$.", "metadata": {"resource_path": "HarvardMIT/segmented/en-112-2008-feb-geo-solutions.jsonl", "problem_match": "\n4. [4]", "solution_match": "\nAnswer: "}}
{"year": "2008", "tier": "T4", "problem_label": "5", "problem_type": "Geometry", "exam": "HMMT", "problem": "A piece of paper is folded in half. A second fold is made such that the angle marked below has measure $\\phi\\left(0^{\\circ}<\\phi<90^{\\circ}\\right)$, and a cut is made as shown below.\n![](https://cdn.mathpix.com/cropped/2025_01_24_684c7884f8823da5eea3g-2.jpg?height=185&width=633&top_left_y=1225&top_left_x=784)\n\nWhen the piece of paper is unfolded, the resulting hole is a polygon. Let $O$ be one of its vertices. Suppose that all the other vertices of the hole lie on a circle centered at $O$, and also that $\\angle X O Y=144^{\\circ}$, where $X$ and $Y$ are the the vertices of the hole adjacent to $O$. Find the value(s) of $\\phi$ (in degrees).", "solution": "$81^{\\circ}$ Try actually folding a piece of paper. We see that the cut out area is a kite, as shown below. The fold was made on $A C$, and then $B E$ and $D E$. Since $D C$ was folded onto $D A$, we have $\\angle A D E=\\angle C D E$.\n![](https://cdn.mathpix.com/cropped/2025_01_24_684c7884f8823da5eea3g-2.jpg?height=419&width=268&top_left_y=1750&top_left_x=969)\n\nEither $A$ or $C$ is the center of the circle. If it's $A$, then $\\angle B A D=144^{\\circ}$, so $\\angle C A D=72^{\\circ}$. Using $C A=D A$, we see that $\\angle A C D=\\angle A D C=54^{\\circ}$. So $\\angle E D A=27^{\\circ}$, and thus $\\phi=72^{\\circ}+27^{\\circ}=99^{\\circ}$, which is inadmissible, as $\\phi<90^{\\circ}$.\nSo $C$ is the center of the circle. Then, $\\angle C A D=\\angle C D A=54^{\\circ}, \\angle A D E=27^{\\circ}$, and $\\phi=54^{\\circ}+27^{\\circ}=81^{\\circ}$.", "metadata": {"resource_path": "HarvardMIT/segmented/en-112-2008-feb-geo-solutions.jsonl", "problem_match": "\n5. [5]", "solution_match": "\nAnswer: "}}
{"year": "2008", "tier": "T4", "problem_label": "6", "problem_type": "Geometry", "exam": "HMMT", "problem": "Let $A B C$ be a triangle with $\\angle A=45^{\\circ}$. Let $P$ be a point on side $B C$ with $P B=3$ and $P C=5$. Let $O$ be the circumcenter of $A B C$. Determine the length $O P$.", "solution": "$\\sqrt{\\sqrt{17}}$ Using extended Sine law, we find the circumradius of $A B C$ to be $R=\\frac{B C}{2 \\sin A}=4 \\sqrt{2}$. By considering the power of point $P$, we find that $R^{2}-O P^{2}=P B \\cdot P C=15$. So $O P=\\sqrt{R^{2}-15}=$ $\\sqrt{16 \\cdot 2-15}=\\sqrt{17}$.", "metadata": {"resource_path": "HarvardMIT/segmented/en-112-2008-feb-geo-solutions.jsonl", "problem_match": "\n6. [5]", "solution_match": "\nAnswer: "}}
{"year": "2008", "tier": "T4", "problem_label": "7", "problem_type": "Geometry", "exam": "HMMT", "problem": "Let $C_{1}$ and $C_{2}$ be externally tangent circles with radius 2 and 3 , respectively. Let $C_{3}$ be a circle internally tangent to both $C_{1}$ and $C_{2}$ at points $A$ and $B$, respectively. The tangents to $C_{3}$ at $A$ and $B$ meet at $T$, and $T A=4$. Determine the radius of $C_{3}$.", "solution": "8 Let $D$ be the point of tangency between $C_{1}$ and $C_{2}$. We see that $T$ is the radical center of the three circles, and so it must lie on the radical axis of $C_{1}$ and $C_{2}$, which happens to be their common tangent $T D$. So $T D=4$.\n![](https://cdn.mathpix.com/cropped/2025_01_24_684c7884f8823da5eea3g-3.jpg?height=654&width=757&top_left_y=711&top_left_x=725)\n\nWe have\n\n$$\n\\tan \\frac{\\angle A T D}{2}=\\frac{2}{T D}=\\frac{1}{2}, \\quad \\text { and } \\quad \\tan \\frac{\\angle B T D}{2}=\\frac{3}{T D}=\\frac{3}{4} .\n$$\n\nThus, the radius of $C_{3}$ equals to\n\n$$\n\\begin{aligned}\nT A \\tan \\frac{\\angle A T B}{2} & =4 \\tan \\left(\\frac{\\angle A T D+\\angle B T D}{2}\\right) \\\\\n& =4 \\cdot \\frac{\\tan \\frac{\\angle A T D}{2}+\\tan \\frac{\\angle B T D}{2}}{1-\\tan \\frac{\\angle A T D}{2} \\tan \\frac{\\angle B T D}{2}} \\\\\n& =4 \\cdot \\frac{\\frac{1}{2}+\\frac{3}{4}}{1-\\frac{1}{2} \\cdot \\frac{3}{4}} \\\\\n& =8 .\n\\end{aligned}\n$$", "metadata": {"resource_path": "HarvardMIT/segmented/en-112-2008-feb-geo-solutions.jsonl", "problem_match": "\n7. [6]", "solution_match": "\nAnswer: "}}
{"year": "2008", "tier": "T4", "problem_label": "8", "problem_type": "Geometry", "exam": "HMMT", "problem": "Let $A B C$ be an equilateral triangle with side length 2 , and let $\\Gamma$ be a circle with radius $\\frac{1}{2}$ centered at the center of the equilateral triangle. Determine the length of the shortest path that starts somewhere on $\\Gamma$, visits all three sides of $A B C$, and ends somewhere on $\\Gamma$ (not necessarily at the starting point). Express your answer in the form of $\\sqrt{p}-q$, where $p$ and $q$ are rational numbers written as reduced fractions.", "solution": "$\\sqrt{\\frac{28}{3}}-1$ Suppose that the path visits sides $A B, B C, C A$ in this order. Construct points $A^{\\prime}, B^{\\prime}, C^{\\prime}$ so that $C^{\\prime}$ is the reflection of $C$ across $A B, A^{\\prime}$ is the reflection of $A$ across $B C^{\\prime}$, and $B^{\\prime}$ is the reflection of $B$ across $A^{\\prime} C^{\\prime}$. Finally, let $\\Gamma^{\\prime}$ be the circle with radius $\\frac{1}{2}$ centered at the center of $A^{\\prime} B^{\\prime} C^{\\prime}$. Note that $\\Gamma^{\\prime}$ is the image of $\\Gamma$ after the three reflections: $A B, B C^{\\prime}, C^{\\prime} A^{\\prime}$.\n![](https://cdn.mathpix.com/cropped/2025_01_24_684c7884f8823da5eea3g-4.jpg?height=302&width=640&top_left_y=234&top_left_x=786)\n\nWhen the path hits $A B$, let us reflect the rest of the path across $A B$ and follow this reflected path. When we hit $B C^{\\prime}$, let us reflect the rest of the path across $B C^{\\prime}$, and follow the new path. And when we hit $A^{\\prime} C^{\\prime}$, reflect the rest of the path across $A^{\\prime} C^{\\prime}$ and follow the new path. We must eventually end up at $\\Gamma^{\\prime}$.\nIt is easy to see that the shortest path connecting some point on $\\Gamma$ to some point on $\\Gamma^{\\prime}$ lies on the line connecting the centers of the two circles. We can easily find the distance between the two centers to be $\\sqrt{3^{2}+\\left(\\frac{1}{\\sqrt{3}}\\right)^{2}}=\\sqrt{\\frac{28}{3}}$. Therefore, the length of the shortest path connecting $\\Gamma$ to $\\Gamma^{\\prime}$ has length $\\sqrt{\\frac{28}{3}}-1$. By reflecting this path three times back into $A B C$, we get a path that satisfies our conditions.", "metadata": {"resource_path": "HarvardMIT/segmented/en-112-2008-feb-geo-solutions.jsonl", "problem_match": "\n8. [6]", "solution_match": "\nAnswer: "}}
{"year": "2008", "tier": "T4", "problem_label": "9", "problem_type": "Geometry", "exam": "HMMT", "problem": "Let $A B C$ be a triangle, and $I$ its incenter. Let the incircle of $A B C$ touch side $B C$ at $D$, and let lines $B I$ and $C I$ meet the circle with diameter $A I$ at points $P$ and $Q$, respectively. Given $B I=$ $6, C I=5, D I=3$, determine the value of $(D P / D Q)^{2}$.", "solution": "<br> $\\qquad$\n\n![](https://cdn.mathpix.com/cropped/2025_01_24_684c7884f8823da5eea3g-4.jpg?height=435&width=386&top_left_y=1273&top_left_x=910)\n\nLet the incircle touch sides $A C$ and $A B$ at $E$ and $F$ respectively. Note that $E$ and $F$ both lie on the circle with diameter $A I$ since $\\angle A E I=\\angle A F I=90^{\\circ}$. The key observation is that $D, E, P$ are collinear. To prove this, suppose that $P$ lies outside the triangle (the other case is analogous), then $\\angle P E A=\\angle P I A=\\angle I B A+\\angle I A B=\\frac{1}{2}(\\angle B+\\angle A)=90^{\\circ}-\\frac{1}{2} \\angle C=\\angle D E C$, which implies that $D, E, P$ are collinear. Similarly $D, F, Q$ are collinear. Then, by Power of a Point, $D E \\cdot D P=D F \\cdot D Q$. So $D P / D Q=D F / D E$.\nNow we compute $D F / D E$. Note that $D F=2 D B \\sin \\angle D B I=2 \\sqrt{6^{2}-3^{2}}\\left(\\frac{3}{6}\\right)=3 \\sqrt{3}$, and $D E=$ $2 D C \\sin \\angle D C I=2 \\sqrt{5^{2}-3^{2}}\\left(\\frac{3}{5}\\right)=\\frac{24}{5}$. Therefore, $D F / D E=\\frac{5 \\sqrt{3}}{8}$.", "metadata": {"resource_path": "HarvardMIT/segmented/en-112-2008-feb-geo-solutions.jsonl", "problem_match": "\n9. [7]", "solution_match": "\n## Answer: "}}
{"year": "2008", "tier": "T4", "problem_label": "10", "problem_type": "Geometry", "exam": "HMMT", "problem": "Let $A B C$ be a triangle with $B C=2007, C A=2008, A B=2009$. Let $\\omega$ be an excircle of $A B C$ that touches the line segment $B C$ at $D$, and touches extensions of lines $A C$ and $A B$ at $E$ and $F$, respectively (so that $C$ lies on segment $A E$ and $B$ lies on segment $A F$ ). Let $O$ be the center of $\\omega$. Let $\\ell$ be the line through $O$ perpendicular to $A D$. Let $\\ell$ meet line $E F$ at $G$. Compute the length $D G$.", "solution": "2014024 Let line $A D$ meet $\\omega$ again at $H$. Since $A F$ and $A E$ are tangents to $\\omega$ and $A D H$ is a secant, we see that $D E H F$ is a harmonic quadrilateral. This implies that the pole of $A D$ with respect to $\\omega$ lies on $E F$. Since $\\ell \\perp A D$, the pole of $A D$ lies on $\\ell$. It follows that the pole of $A D$ is $G$.\n![](https://cdn.mathpix.com/cropped/2025_01_24_684c7884f8823da5eea3g-5.jpg?height=649&width=928&top_left_y=239&top_left_x=645)\n\nThus, $G$ must lie on the tangent to $\\omega$ at $D$, so $C, D, B, G$ are collinear. Furthermore, since the pencil of lines $(A E, A F ; A D, A G)$ is harmonic, by intersecting it with the line $B C$, we see that $(C, B ; D, G)$ is harmonic as well. This means that\n\n$$\n\\frac{B D}{D C} \\cdot \\frac{C G}{G B}=-1\n$$\n\n(where the lengths are directed.) The semiperimeter of $A B C$ is $s=\\frac{1}{2}(2007+2008+2009)=3012$. So $B D=s-2009=1003$ and $C D=s-2008=1004$. Let $x=D G$, then the above equations gives\n\n$$\n\\frac{1003}{1004} \\cdot \\frac{x+1004}{x-1003}=1\n$$\n\nSolving gives $x=2014024$.\nRemark: If you are interested to learn about projective geometry, check out the last chapter of Geometry Revisited by Coxeter and Greitzer or Geometric Transformations III by Yaglom.", "metadata": {"resource_path": "HarvardMIT/segmented/en-112-2008-feb-geo-solutions.jsonl", "problem_match": "\n10. [7]", "solution_match": "\nAnswer: "}}